GSEB Class 12 Physics Solutions Chapter 4 Moving Charges and Magnetism

Question 1. A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?
Answer: To find the magnetic field at the center of a circular coil, we use the formula \( B = \frac{\mu_{0} \mathrm{nI}}{2 \mathrm{r}} \).
Given values are: number of turns \( \mathrm{n} = 100 \), current \( \mathrm{I} = 0.40 \, \mathrm{A} \), and radius \( \mathrm{r} = 8.0 \, \mathrm{cm} = 8.0 \times 10^{-2} \, \mathrm{m} \).
Plugging these into the formula:
\( B = \frac{4 \pi \times 10^{-7} \times 100 \times 0.4}{2 \times 8 \times 10^{-2}} \)
\( B = \pi \times 10^{-4} \, \mathrm{T} \)
In simple words: We calculated the magnetic field at the center of a circular coil by using a specific formula that considers the number of turns, current, and radius of the coil.

🎯 Exam Tip: Remember the formula for the magnetic field at the center of a circular coil. Ensure correct unit conversion (cm to m) and careful calculation of exponents for full marks.

Question 2. A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?
Answer: To determine the magnetic field produced by a long straight wire, the formula \( B = \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \) is used.
The given current is \( \mathrm{I} = 35 \, \mathrm{A} \), and the distance from the wire is \( \mathrm{r} = 20 \, \mathrm{cm} = 20 \times 10^{-2} \, \mathrm{m} \).
Substituting these values into the formula:
\( B = \frac{4 \pi \times 10^{-7} \times 35}{2 \pi \times 20 \times 10^{-2}} \)
\( B = 3.5 \times 10^{-5} \, \mathrm{T} \)
In simple words: We found the magnetic field strength near a long wire by using the wire's current and the distance from it in a specific physics formula.

🎯 Exam Tip: Know the formula for the magnetic field of a long straight wire. Always convert distances to meters before calculation and be precise with scientific notation.

Question 3. A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.
Answer: To calculate the magnetic field magnitude, we use the formula for a long straight wire: \( B = \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \).
Here, the current is \( \mathrm{I} = 50 \, \mathrm{A} \), and the distance is \( \mathrm{r} = 2.5 \, \mathrm{m} \).
Substituting the values:
\( B = \frac{4 \pi \times 10^{-7} \times 50}{2 \pi \times 2.5} \)
\( B = 4 \times 10^{-6} \, \mathrm{T} \)
According to the right-hand thumb rule, if the current flows from north to south, the magnetic field at a point 2.5 m east of the wire will be directed upwards.
In simple words: We calculated the strength of the magnetic field using the current in the wire and the distance from it, and then used a rule to find that the field points upwards.

🎯 Exam Tip: For problems involving current direction and magnetic field, always use the right-hand thumb rule to correctly determine the field's direction. Ensure calculations are accurate and units are consistent.

Question 4. A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?
Answer: The magnetic field strength for a long straight current-carrying wire is given by \( B = \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \).
Given current \( \mathrm{I} = 90 \, \mathrm{A} \) and distance \( \mathrm{r} = 1.5 \, \mathrm{m} \).
Substituting these into the formula:
\( B = \frac{4 \pi \times 10^{-7} \times 90}{2 \pi \times 1.5} \)
\( B = 1.2 \times 10^{-5} \, \mathrm{T} \)
Using the right-hand thumb rule, if the current flows from east to west, the magnetic field below the line will be directed towards the South.
In simple words: We found the magnetic field's strength below a power line using the current and distance, and determined its direction is South by applying a hand rule.

🎯 Exam Tip: Accurately apply the right-hand thumb rule to determine the magnetic field direction. Remember to state both magnitude and direction for such questions.

Question 5. What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15T?
Answer: The magnetic force on a current-carrying wire in a magnetic field is given by \( \mathrm{F} = \mathrm{BIL} \sin \theta \). To find the force per unit length, we divide by L, so \( \frac{\mathrm{F}}{\mathrm{L}} = \mathrm{BI} \sin \theta \).
We are given: current \( \mathrm{I} = 8 \, \mathrm{A} \), magnetic field \( \mathrm{B} = 0.15 \, \mathrm{T} \), and angle \( \theta = 30^\circ \).
Substituting these values:
Force per unit length \( \frac{\mathrm{F}}{\mathrm{L}} = 0.15 \times 8 \times \sin 30^\circ \)
Since \( \sin 30^\circ = 0.5 \):
\( \frac{\mathrm{F}}{\mathrm{L}} = 0.15 \times 8 \times 0.5 \)
\( \frac{\mathrm{F}}{\mathrm{L}} = 0.6 \, \mathrm{Nm}^{-1} \)
In simple words: We calculated the magnetic force on a wire for each unit of its length by multiplying the magnetic field strength, the current, and the sine of the angle between the wire and the field.

🎯 Exam Tip: Ensure you use the correct formula \( F = BIL \sin \theta \) and remember that force *per unit length* means \( F/L \). Pay attention to the angle in the sine function.

Question 6. A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?
Answer: The magnetic force acting on a current-carrying wire within a magnetic field is calculated using the formula \( \mathrm{F} = \mathrm{BIL} \sin \theta \). Since the wire is placed perpendicular to the solenoid's axis, the angle \( \theta \) is \( 90^\circ \), and \( \sin 90^\circ = 1 \), so \( \mathrm{F} = \mathrm{BIL} \).
Given values are: length of the wire \( \mathrm{L} = 3.0 \, \mathrm{cm} = 3.0 \times 10^{-2} \, \mathrm{m} \), current \( \mathrm{I} = 10 \, \mathrm{A} \), and magnetic field \( \mathrm{B} = 0.27 \, \mathrm{T} \).
Substituting these values:
\( \mathrm{F} = 0.27 \times 10 \times 3.0 \times 10^{-2} \)
\( \mathrm{F} = 8.1 \times 10^{-2} \, \mathrm{N} \)
In simple words: We found the magnetic force on a wire inside a solenoid by multiplying the magnetic field strength, the current, and the length of the wire, since the wire was placed at a right angle to the field.

🎯 Exam Tip: When a wire is perpendicular to the magnetic field, \( \sin \theta = 1 \). Always convert length to meters before performing calculations. This simplifies the force formula.

Question 7. Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.
Answer: The force between two parallel current-carrying wires is given by \( \mathrm{F} = \frac{\mu_{0} \mathrm{I}_{1} \mathrm{I}_{2} \mathrm{L}}{2 \pi \mathrm{r}} \).
Given: current in wire A \( \mathrm{I}_{1} = 8.0 \, \mathrm{A} \), current in wire B \( \mathrm{I}_{2} = 5.0 \, \mathrm{A} \).
The separation distance is \( \mathrm{r} = 4.0 \, \mathrm{cm} = 4.0 \times 10^{-2} \, \mathrm{m} \).
The length of wire A section is \( \mathrm{L} = 10 \, \mathrm{cm} = 0.1 \, \mathrm{m} \).
Substituting these values:
\( \mathrm{F} = \frac{2 \times 10^{-7} \times 8 \times 5 \times 0.1}{4 \times 10^{-2}} \)
\( \mathrm{F} = 2 \times 10^{-5} \, \mathrm{N} \)
Since the currents are in the same direction, the force will be attractive, meaning wire A experiences a force towards wire B.
In simple words: We calculated the magnetic force on a part of wire A due to wire B, considering their currents, distance, and length, finding it to be an attractive force because currents flow in the same direction.

🎯 Exam Tip: Remember that parallel currents in the same direction attract, while opposite directions repel. Always convert all lengths to meters before calculation.

Question 8. A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.
Answer: The magnetic field inside a solenoid is given by \( \mathrm{B} = \mu_{0} \mathrm{nI} \), where \( \mathrm{n} \) is the number of turns per unit length.
First, calculate the total number of turns: The solenoid has 5 layers, each with 400 turns, so total turns \( \mathrm{N} = 5 \times 400 = 2000 \).
The length of the solenoid is \( \mathrm{L} = 80 \, \mathrm{cm} = 80 \times 10^{-2} \, \mathrm{m} \).
The number of turns per unit length is \( \mathrm{n} = \frac{\mathrm{N}}{\mathrm{L}} = \frac{2000}{80 \times 10^{-2}} = \frac{2000}{0.8} = 2500 \, \mathrm{turns/m} \).
The current carried is \( \mathrm{I} = 8.0 \, \mathrm{A} \).
Now, substitute into the magnetic field formula:
\( \mathrm{B} = 4 \pi \times 10^{-7} \times 2500 \times 8 \)
\( \mathrm{B} = 8 \pi \times 10^{-3} \, \mathrm{T} \)
In simple words: We calculated the magnetic field inside the solenoid by first finding the total number of wire turns and then using a formula that includes the current and the turns per meter of the solenoid.

🎯 Exam Tip: For solenoids with multiple layers, always calculate the total number of turns first. Ensure the length is in meters when calculating turns per unit length.

Question 9. A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
Answer: The magnitude of torque experienced by a coil in a magnetic field is given by \( \tau = \mathrm{nBIA} \sin \theta \).
Given: current \( \mathrm{I} = 12 \, \mathrm{A} \), number of turns \( \mathrm{n} = 20 \), magnetic field \( \mathrm{B} = 0.80 \, \mathrm{T} \).
The side of the square coil is \( 10 \, \mathrm{cm} \), so its area \( \mathrm{A} = (10 \times 10^{-2})^2 \, \mathrm{m}^2 = 100 \times 10^{-4} \, \mathrm{m}^2 \).
The angle \( \theta \) between the normal to the coil and the magnetic field is \( 30^\circ \).
Substituting these values:
\( \tau = 20 \times 0.80 \times 12 \times (100 \times 10^{-4}) \times \sin 30^\circ \)
\( \tau = 20 \times 0.80 \times 12 \times 0.01 \times 0.5 \)
\( \tau = 0.96 \, \mathrm{Nm} \)
In simple words: We calculated the turning force (torque) on the coil by multiplying its number of turns, the magnetic field strength, the current, the coil's area, and the sine of the angle it makes with the field.

🎯 Exam Tip: Carefully identify the angle \( \theta \) in the torque formula; it's the angle between the normal to the coil and the magnetic field. Ensure area units are converted to square meters.

Question 10. Two moving coil meters, M\(_{1}\) and M\(_{2}\) have the following particulars:
R\(_{1}\) = 10 \( \Omega \), N\(_{1}\) = 30, A\(_{1}\) = 3.6 x 10\(^{-3}\) m\(^{2}\), B\(_{1}\) = 0.25 T R\(_{2}\) = 14 \( \Omega \), N\(_{2}\) = 42, A\(_{2}\) = 1.8 x 10\(^{-3}\) m\(^{2}\), B\(_{2}\) = 0.50 T
(The spring constants are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M\(_{2}\) and M\(_{1}\).
Answer:
(a) Current sensitivity \( \left( \frac{\theta}{\mathrm{I}} \right) \) is given by \( \frac{\mathrm{BAN}}{\mathrm{C}} \). Since the spring constants (C) are identical, the ratio of current sensitivities for M\(_{2}\) and M\(_{1}\) is:
\( \frac{(\mathrm{Current \, sensitivity})_{2}}{(\mathrm{Current \, sensitivity})_{1}} = \frac{\frac{\mathrm{B}_{2}\mathrm{A}_{2}\mathrm{N}_{2}}{\mathrm{C}}}{\frac{\mathrm{B}_{1}\mathrm{A}_{1}\mathrm{N}_{1}}{\mathrm{C}}} = \frac{\mathrm{B}_{2}\mathrm{A}_{2}\mathrm{N}_{2}}{\mathrm{B}_{1}\mathrm{A}_{1}\mathrm{N}_{1}} \)
Substituting the given values:
\( \frac{\mathrm{B}_{2}\mathrm{A}_{2}\mathrm{N}_{2}}{\mathrm{B}_{1}\mathrm{A}_{1}\mathrm{N}_{1}} = \frac{0.5 \times 1.8 \times 10^{-3} \times 42}{0.25 \times 3.6 \times 10^{-3} \times 30} = \frac{7}{5} = 1.4 \)

(b) Voltage sensitivity \( \left( \frac{\theta}{\mathrm{V}} \right) \) is given by \( \frac{\mathrm{BAN}}{\mathrm{CR}} = \frac{\mathrm{Current \, sensitivity}}{\mathrm{R}} \).
The ratio of voltage sensitivities for M\(_{2}\) and M\(_{1}\) is:
\( \frac{(\mathrm{Voltage \, sensitivity})_{2}}{(\mathrm{Voltage \, sensitivity})_{1}} = \frac{\frac{(\mathrm{Current \, sensitivity})_{2}}{\mathrm{R}_{2}}}{\frac{(\mathrm{Current \, sensitivity})_{1}}{\mathrm{R}_{1}}} = \frac{(\mathrm{Current \, sensitivity})_{2}}{(\mathrm{Current \, sensitivity})_{1}} \times \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}} \)
Using the ratio from part (a) and the given resistances:
\( \frac{(\mathrm{Voltage \, sensitivity})_{2}}{(\mathrm{Voltage \, sensitivity})_{1}} = 1.4 \times \frac{10}{14} = 1 \)
In simple words: For current sensitivity, we compared the product of magnetic field, area, and number of turns for both meters. For voltage sensitivity, we multiplied the current sensitivity ratio by the inverse ratio of their resistances to find the overall ratio.

🎯 Exam Tip: Understand the definitions of current and voltage sensitivity. For ratios, often common factors (like spring constant C here) cancel out, simplifying calculations.

Question 11. In a chamber, a uniform magnetic field of 6.5G (1G = 10\(^{-4}\) T) is maintained. An electron is shot into the field with a speed of 4.8 x 10\(^{6}\) ms\(^{-1}\) normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.6 x 10\(^{-19}\) C, m\(_{e}\) = 9.1 x 10\(^{-31}\) kg)
Answer: When a charged particle enters a uniform magnetic field perpendicular to its velocity, the magnetic force acts as a centripetal force. This force is always perpendicular to both the velocity and the magnetic field, causing the particle to move in a circular path without changing its speed, only its direction. This centripetal force \( \mathrm{Bqv} \) balances the centrifugal force \( \frac{\mathrm{mv}^{2}}{\mathrm{r}} \).
So, \( \mathrm{Bqv} = \frac{\mathrm{mv}^{2}}{\mathrm{r}} \).
From this, the radius of the circular orbit \( \mathrm{r} = \frac{\mathrm{mv}}{\mathrm{qB}} \).
Given: magnetic field \( \mathrm{B} = 6.5 \, \mathrm{G} = 6.5 \times 10^{-4} \, \mathrm{T} \), speed \( \mathrm{v} = 4.8 \times 10^{6} \, \mathrm{m/s} \).
Charge of electron \( \mathrm{e} = 1.6 \times 10^{-19} \, \mathrm{C} \), mass of electron \( \mathrm{m}_{e} = 9.1 \times 10^{-31} \, \mathrm{kg} \).
Substituting these values into the radius formula:
\( \mathrm{r} = \frac{9.1 \times 10^{-31} \times 4.8 \times 10^{6}}{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}} \)
\( \mathrm{r} = 4.2 \times 10^{-2} \, \mathrm{m} \)
\( \mathrm{r} = 4.2 \, \mathrm{cm} \)
In simple words: An electron moves in a circle in a magnetic field because the magnetic force constantly pushes it towards the center, changing its direction but not its speed. We found the circle's size using the electron's mass, speed, charge, and the magnetic field strength.

🎯 Exam Tip: Understand the fundamental reason for circular motion in a magnetic field (magnetic force provides centripetal force). Always convert Gauss to Tesla for calculations and carefully handle exponents.

Question 12. In Exercise. 11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.
Answer: The frequency of revolution for a charged particle in a magnetic field is given by \( \mathrm{\nu} = \frac{\mathrm{qB}}{2 \pi \mathrm{m}} \).
Using the values from Question 11: \( \mathrm{q} = 1.6 \times 10^{-19} \, \mathrm{C} \), \( \mathrm{B} = 6.5 \times 10^{-4} \, \mathrm{T} \), \( \mathrm{m} = 9.1 \times 10^{-31} \, \mathrm{kg} \).
Substituting these values:
\( \mathrm{\nu} = \frac{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}{2 \times \pi \times 9.1 \times 10^{-31}} \)
\( \mathrm{\nu} \approx 18 \times 10^{6} \, \mathrm{Hz} \)
\( \mathrm{\nu} = 18 \, \mathrm{MHz} \)
The frequency of revolution, also known as cyclotron frequency, \( \mathrm{\nu} = \frac{\mathrm{qB}}{2 \pi \mathrm{m}} \), does not depend on the speed of the electron. This is because both the magnetic force (which provides the centripetal force) and the centripetal acceleration are proportional to speed, so speed cancels out in the equation for frequency.
In simple words: We calculated how many times the electron goes around in a second using its charge, the magnetic field, and its mass. The electron's speed doesn't change how often it orbits because the stronger the push, the faster it goes in a larger circle, keeping the time per circle the same.

🎯 Exam Tip: Remember that cyclotron frequency is independent of particle speed and orbital radius. This is a key concept for devices like cyclotrons. Show your calculations clearly.

Question 13. (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter-torque that must be applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)
Answer:
(a) The torque experienced by a current-carrying coil in a magnetic field is \( \tau = \mathrm{N I A B \sin \theta} \). The counter-torque required to prevent turning will have the same magnitude.
Given: number of turns \( \mathrm{N} = 30 \), radius \( \mathrm{r} = 8.0 \, \mathrm{cm} = 8.0 \times 10^{-2} \, \mathrm{m} \), current \( \mathrm{I} = 6.0 \, \mathrm{A} \), magnetic field \( \mathrm{B} = 1.0 \, \mathrm{T} \), angle \( \theta = 60^\circ \).
The area of the circular coil \( \mathrm{A} = \pi \mathrm{r}^2 = \pi \times (8.0 \times 10^{-2})^2 \, \mathrm{m}^2 \).
Substituting the values:
\( \tau = 30 \times 6.0 \times \pi \times (8.0 \times 10^{-2})^2 \times 1.0 \times \sin 60^\circ \)
\( \tau = 30 \times 6.0 \times \pi \times (64 \times 10^{-4}) \times 1.0 \times \frac{\sqrt{3}}{2} \)
\( \tau \approx 3.1 \, \mathrm{Nm} \)

(b) No, the answer would not change. The torque on a planar current loop in a uniform magnetic field depends only on the number of turns, current, magnetic field strength, the area enclosed by the coil, and the angle between the area vector and the magnetic field. It does not depend on the specific shape of the coil, as long as the enclosed area remains the same. The formula \( \vec{\tau} = \mathrm{N I} (\vec{\mathrm{A}} \times \vec{\mathrm{B}}) \) confirms this, where \( \vec{\mathrm{A}} \) is the area vector.
In simple words: (a) We calculated the turning force needed to stop the coil from rotating by using a formula that includes the number of turns, current, magnetic field, coil area, and the angle of the coil. (b) The shape of the coil doesn't matter, only the area it covers affects the turning force in a uniform magnetic field.

🎯 Exam Tip: The torque formula \( \tau = \mathrm{N I A B \sin \theta} \) applies to any planar coil. The key is the *area enclosed*, not the specific shape. Pay close attention to unit conversions for radius and angle for part (a).

Question 14. Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो संकेंद्रित वृत्ताकार कुंडलियों X और Y को दर्शाता है। कुंडली X का व्यास 20 इकाई और 16 A का धारा है, जबकि कुंडली Y का व्यास 25 इकाई और 18 A का धारा है। दोनों कुंडलियाँ एक ही ऊर्ध्वाधर तल में हैं और उनके केंद्र पर चुंबकीय क्षेत्र की गणना की जानी है। बाहरी कुंडली X उत्तर-दक्षिण दिशा में 16 cm त्रिज्या पर है और आंतरिक कुंडली Y उत्तर-दक्षिण दिशा में 10 cm त्रिज्या पर है।
Answer: The magnetic field at the center of a circular coil is given by \( \mathrm{B} = \frac{\mu_{0} \mathrm{nI}}{2 \mathrm{r}} \).
For coil X:
Radius \( \mathrm{r}_{X} = 16 \, \mathrm{cm} = 16 \times 10^{-2} \, \mathrm{m} \)
Turns \( \mathrm{N}_{X} = 20 \)
Current \( \mathrm{I}_{X} = 16 \, \mathrm{A} \) (anticlockwise, when viewed from west). By right-hand rule, this creates a magnetic field towards the East.
\( \mathrm{B}_{X} = \frac{4 \pi \times 10^{-7} \times 20 \times 16}{2 \times 16 \times 10^{-2}} = 4 \pi \times 10^{-4} \, \mathrm{T} \) (towards East)

For coil Y:
Radius \( \mathrm{r}_{Y} = 10 \, \mathrm{cm} = 10 \times 10^{-2} \, \mathrm{m} \)
Turns \( \mathrm{N}_{Y} = 25 \)
Current \( \mathrm{I}_{Y} = 18 \, \mathrm{A} \) (clockwise, when viewed from west). By right-hand rule, this creates a magnetic field towards the West.
\( \mathrm{B}_{Y} = \frac{4 \pi \times 10^{-7} \times 25 \times 18}{2 \times 10 \times 10^{-2}} = 9 \pi \times 10^{-4} \, \mathrm{T} \) (towards West)

Since the magnetic fields are in opposite directions (East and West), the net magnetic field will be the difference between their magnitudes.
Net field \( \mathrm{B}_{\text{net}} = \mathrm{B}_{Y} - \mathrm{B}_{X} \) (because \( \mathrm{B}_{Y} \) is larger than \( \mathrm{B}_{X} \))
\( \mathrm{B}_{\text{net}} = (9 \pi \times 10^{-4}) - (4 \pi \times 10^{-4}) = 5 \pi \times 10^{-4} \, \mathrm{T} \)
The direction of the net field will be towards the West, as \( \mathrm{B}_{Y} \) is greater.
In simple words: We calculated the magnetic field for each coil at the center using their turns, currents, and radii. Since the currents flowed in opposite directions when viewed from west, their magnetic fields at the center opposed each other. We subtracted the smaller field from the larger one to find the final magnetic field and its direction.

🎯 Exam Tip: Use the right-hand rule consistently to determine the direction of the magnetic field for each coil. Pay attention to whether currents are in the same or opposite directions to correctly add or subtract fields.

Question 15. For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by, \( \mathrm{B} = \frac{\mu_{0} \mathrm{I R}^{2} \mathrm{N}}{2\left(\mathrm{x}^{2}+\mathrm{R}^{2}\right)^{3 / 2}} \)
(a) Show that this reduces to the familiar result for the field at the centre of the coil.
(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by, \( \mathrm{B} = 0.72 \frac{\mu_{0} \mathrm{NI}}{\mathrm{R}} \), approximately.
[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]
Answer:
(a) The general expression for the magnetic field on the axis of a circular coil at a distance x from its center is: \( \mathrm{B} = \frac{\mu_{0} \mathrm{I R}^{2} \mathrm{N}}{2\left(\mathrm{x}^{2}+\mathrm{R}^{2}\right)^{3 / 2}} \).
At the center of the coil, the distance \( \mathrm{x} = 0 \).
Substitute \( \mathrm{x} = 0 \) into the equation:
\( \mathrm{B}_{\text{center}} = \frac{\mu_{0} \mathrm{I R}^{2} \mathrm{N}}{2\left(0^{2}+\mathrm{R}^{2}\right)^{3 / 2}} = \frac{\mu_{0} \mathrm{I R}^{2} \mathrm{N}}{2(\mathrm{R}^{2})^{3 / 2}} = \frac{\mu_{0} \mathrm{I R}^{2} \mathrm{N}}{2 \mathrm{R}^{3}} = \frac{\mu_{0} \mathrm{IN}}{2 \mathrm{R}} \)
This is the standard formula for the magnetic field at the center of a circular coil.

(b) For two parallel co-axial coils (Helmholtz coils) separated by distance R, and both carrying current I in the same direction, consider a point on the axis at a distance \( \mathrm{x} \) from the midpoint between the coils.
The distance of this point from the center of the first coil is \( \mathrm{R}/2 - \mathrm{x} \).
The distance of this point from the center of the second coil is \( \mathrm{R}/2 + \mathrm{x} \).
The total magnetic field at this point is the sum of fields from both coils:
\( \mathrm{B}_{\text{total}} = \frac{\mu_{0} \mathrm{I R}^{2} \mathrm{N}}{2\left( (\mathrm{R}/2 - \mathrm{x})^{2}+\mathrm{R}^{2}\right)^{3 / 2}} + \frac{\mu_{0} \mathrm{I R}^{2} \mathrm{N}}{2\left( (\mathrm{R}/2 + \mathrm{x})^{2}+\mathrm{R}^{2}\right)^{3 / 2}} \)
When \( \mathrm{x} \) is very small compared to \( \mathrm{R} \), and by taking the derivative of B with respect to x and setting it to zero to find uniformity, it can be shown that the field is nearly uniform at the midpoint and its value is approximately:
\( \mathrm{B} = \frac{8}{5 \sqrt{5}} \frac{\mu_{0} \mathrm{NI}}{\mathrm{R}} \)
\( \mathrm{B} = \frac{8}{5 \times 2.236} \frac{\mu_{0} \mathrm{NI}}{\mathrm{R}} \approx \frac{8}{11.18} \frac{\mu_{0} \mathrm{NI}}{\mathrm{R}} \approx 0.715 \frac{\mu_{0} \mathrm{NI}}{\mathrm{R}} \)
Rounding this gives: \( \mathrm{B} \approx 0.72 \frac{\mu_{0} \mathrm{NI}}{\mathrm{R}} \)
In simple words: (a) We showed that the general formula for a magnetic field along a coil's axis simplifies to the magnetic field at its center when the distance from the center is zero. (b) For two specific coils (Helmholtz coils), we found that the magnetic field in the middle is almost steady and has a certain value, which helps create a uniform magnetic field over a small area.

🎯 Exam Tip: Remember the condition for the field at the center (\( x=0 \)). For Helmholtz coils, the key is the uniform field near the midpoint, which is achieved when the coil separation equals the radius. No need to derive the complex Helmholtz field; just state the result and its approximation.

Question 16. A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field
(i) outside the toroid
(ii) inside the core of the toroid
(iii) in the empty space surrounded by the toroid.
Answer: A toroid is a donut-shaped coil. Ampere's circuital law is used to find the magnetic field in different regions.
(i) **Outside the toroid:** For any Amperian loop outside the toroid, the net current enclosed is zero, because the current flowing into the toroid winding is equal and opposite to the current flowing out. Therefore, the magnetic field outside the toroid is zero.

(ii) **Inside the core of the toroid:** The magnetic field inside the core of a toroid is given by \( \mathrm{B} = \frac{\mu_{0} \mathrm{NI}}{2 \pi \mathrm{r}_{\text{avg}}} \), where \( \mathrm{N} \) is the total number of turns, \( \mathrm{I} \) is the current, and \( \mathrm{r}_{\text{avg}} \) is the average radius of the toroid.
Given: inner radius \( \mathrm{r}_{1} = 25 \, \mathrm{cm} \), outer radius \( \mathrm{r}_{2} = 26 \, \mathrm{cm} \).
Average radius \( \mathrm{r}_{\text{avg}} = \frac{25 + 26}{2} \, \mathrm{cm} = 25.5 \, \mathrm{cm} = 25.5 \times 10^{-2} \, \mathrm{m} \).
Number of turns \( \mathrm{N} = 3500 \).
Current \( \mathrm{I} = 11 \, \mathrm{A} \).
Substituting these values:
\( \mathrm{B} = \frac{4 \pi \times 10^{-7} \times 3500 \times 11}{2 \pi \times 25.5 \times 10^{-2}} \)
\( \mathrm{B} \approx 3.02 \times 10^{-2} \, \mathrm{T} \)

(iii) **In the empty space surrounded by the toroid (the hollow region inside the windings):** For any Amperian loop taken within the empty space (where the core is not), the net current enclosed is zero. Therefore, the magnetic field in this empty space is also zero.
In simple words: (i) The magnetic field is zero outside the toroid because the total current going through any path outside it is zero. (ii) Inside the core, we found the magnetic field using the number of turns, current, and the average radius. (iii) The magnetic field is also zero in the empty space inside the toroid because no net current passes through that region.

🎯 Exam Tip: Remember Ampere's circuital law for toroids: the magnetic field is zero outside and in the empty space inside, but non-zero within the core. Always use the average radius for calculations within the core.

Question 17. Answer the following questions.
(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?
(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?
(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in the north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight-line path.
Answer:
(a) The magnetic force on a charged particle is given by \( \vec{\mathrm{F}} = \mathrm{q}(\vec{\mathrm{v}} \times \vec{\mathrm{B}}) \). For the particle to travel undeflected along a straight path, the magnetic force must be zero. This happens if the angle \( \theta \) between the velocity vector \( \vec{\mathrm{v}} \) and the magnetic field vector \( \vec{\mathrm{B}} \) is \( 0^\circ \) or \( 180^\circ \) (i.e., \( \sin \theta = 0 \)).
Since the magnetic field has a constant direction from east to west, the initial velocity of the particle must be either parallel (west to east) or anti-parallel (east to west) to the magnetic field. This means the particle's initial velocity is along the east-west line.

(b) Yes, its final speed would equal its initial speed, assuming no collisions. A magnetic field does not perform any work on a charged particle because the magnetic force is always perpendicular to the particle's velocity (\( \mathrm{W} = \vec{\mathrm{F}} \cdot \vec{\mathrm{s}} = \vec{\mathrm{F}} \cdot (\vec{\mathrm{v}} \Delta \mathrm{t}) = 0 \)). Since no work is done by the magnetic field, the kinetic energy of the particle remains constant, and therefore, its speed remains constant.

(c) The electron travels from west to east. The uniform electrostatic field is directed from north to south. Since the electron is negatively charged, the electric force on the electron (\( \vec{\mathrm{F}}_{\mathrm{E}} = \mathrm{q}\vec{\mathrm{E}} \)) will be opposite to the electric field direction, i.e., from south to north.
To prevent deflection, a magnetic force \( \vec{\mathrm{F}}_{\mathrm{B}} = \mathrm{e}(\vec{\mathrm{v}} \times \vec{\mathrm{B}}) \) must be created that precisely cancels out the electric force. This means the magnetic force must be directed from north to south.
The electron's velocity \( \vec{\mathrm{v}} \) is from west to east. To get a magnetic force \( \vec{\mathrm{F}}_{\mathrm{B}} \) from north to south, using the right-hand rule (for positive charge, then reverse for electron), if \( \vec{\mathrm{v}} \) is East and \( \vec{\mathrm{F}}_{\mathrm{B}} \) is South, then \( \vec{\mathrm{B}} \) must be in the downward direction.
In simple words: (a) The particle's initial speed must be in the same direction as the magnetic field (east to west or west to east) for it to travel straight without being pushed sideways. (b) The particle's speed won't change even in a complex magnetic field because the magnetic force only changes its direction, not how fast it moves. (c) To stop the electron from bending, a magnetic field should point downwards, so its force cancels out the upward push from the electric field.

🎯 Exam Tip: For undeflected motion in a magnetic field, the velocity must be parallel or anti-parallel to the field. Magnetic fields do no work, so kinetic energy and speed remain constant. Use the right-hand rule carefully, remembering to reverse the direction for negative charges like electrons.

Question 18. An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with the uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field
(a) is transverse to its initial velocity
(b) makes an angle of 30° with the initial velocity.
Answer: First, calculate the speed of the electron after acceleration. The kinetic energy gained by the electron is equal to the work done by the potential difference: \( \mathrm{KE} = \mathrm{eV} = \frac{1}{2} \mathrm{mv}^2 \).
So, \( \mathrm{v} = \sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}} \).
Given: potential difference \( \mathrm{V} = 2.0 \, \mathrm{kV} = 2.0 \times 10^3 \, \mathrm{V} \), magnetic field \( \mathrm{B} = 0.15 \, \mathrm{T} \).
Charge of electron \( \mathrm{e} = 1.6 \times 10^{-19} \, \mathrm{C} \), mass of electron \( \mathrm{m}_{e} = 9.1 \times 10^{-31} \, \mathrm{kg} \).
\( \mathrm{v} = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 2.0 \times 10^3}{9.1 \times 10^{-31}}} = \sqrt{\frac{6.4 \times 10^{-16}}{9.1 \times 10^{-31}}} \approx 8.3 \times 10^7 \, \mathrm{m/s} \).

(a) If the field is transverse (perpendicular) to the initial velocity (\( \theta = 90^\circ \)), the electron will move in a circular trajectory. The magnetic force \( \mathrm{Bqv} \) provides the centripetal force \( \frac{\mathrm{mv}^2}{\mathrm{r}} \).
So, \( \mathrm{r} = \frac{\mathrm{mv}}{\mathrm{Bq}} \).
\( \mathrm{r} = \frac{9.1 \times 10^{-31} \times 8.3 \times 10^7}{0.15 \times 1.6 \times 10^{-19}} \approx 3.1 \times 10^{-3} \, \mathrm{m} \approx 3.1 \, \mathrm{mm} \)
The path is a circle with a radius of approximately 3.1 mm.

(b) If the field makes an angle of \( 30^\circ \) with the initial velocity, the electron's velocity can be resolved into two components: one parallel to the magnetic field (\( \mathrm{v}_{\parallel} = \mathrm{v} \cos 30^\circ \)) and one perpendicular to it (\( \mathrm{v}_{\perp} = \mathrm{v} \sin 30^\circ \)).
The parallel component \( \mathrm{v}_{\parallel} \) causes linear motion along the field direction, while the perpendicular component \( \mathrm{v}_{\perp} \) causes circular motion. The combination of these two motions results in a helical (spiral) trajectory.
The radius of the helical path is determined by the perpendicular velocity component: \( \mathrm{r} = \frac{\mathrm{mv}_{\perp}}{\mathrm{Bq}} = \frac{\mathrm{mv} \sin 30^\circ}{\mathrm{Bq}} \).
\( \mathrm{r} = \frac{9.1 \times 10^{-31} \times 8.3 \times 10^7 \times 0.5}{0.15 \times 1.6 \times 10^{-19}} \approx 1.55 \times 10^{-3} \, \mathrm{m} \approx 1.55 \, \mathrm{mm} \).
The pitch of the helix (distance traveled along the field during one revolution) would be \( \mathrm{p} = \mathrm{v}_{\parallel} \mathrm{T} = \mathrm{v} \cos 30^\circ \times \frac{2 \pi \mathrm{m}}{\mathrm{qB}} \).
In simple words: First, we found the electron's speed from how much it was accelerated. (a) If the magnetic field is straight across its path, the electron will move in a perfect circle. (b) If the field is at an angle, the electron will follow a spiral (helical) path because part of its motion is straight and part is circular.

🎯 Exam Tip: Remember that perpendicular velocity components lead to circular motion, and parallel components lead to linear motion. Their combination results in a helical path. Always calculate the velocity from the potential difference first.

Question 19. A magnetic field set up using Helmholtz coils (described in Exercise 15) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 x 10\(^{5}\) Vm\(^{-1}\), make a simple guess as to what the beam contains. Why is the answer not unique?
Answer: For a charged particle to pass through a region with both electric and magnetic fields undeflected, the electric force (\( \mathrm{qE} \)) must be balanced by the magnetic force (\( \mathrm{qvB} \)). This means \( \mathrm{qE} = \mathrm{qvB} \), which simplifies to \( \mathrm{v} = \frac{\mathrm{E}}{\mathrm{B}} \).
Given: magnetic field \( \mathrm{B} = 0.75 \, \mathrm{T} \), electrostatic field \( \mathrm{E} = 9.0 \times 10^{5} \, \mathrm{Vm}^{-1} \).
The velocity of the undeflected beam is:
\( \mathrm{v} = \frac{9.0 \times 10^{5}}{0.75} = 1.2 \times 10^{6} \, \mathrm{m/s} \)

The particles are accelerated through a potential difference of \( \mathrm{V}_{\text{accel}} = 15 \, \mathrm{kV} = 15 \times 10^3 \, \mathrm{V} \). The kinetic energy gained is \( \mathrm{KE} = \mathrm{qV}_{\text{accel}} = \frac{1}{2} \mathrm{mv}^2 \).
From this, we can find the charge-to-mass ratio \( \frac{\mathrm{q}}{\mathrm{m}} \):
\( \frac{\mathrm{q}}{\mathrm{m}} = \frac{\mathrm{v}^2}{2 \mathrm{V}_{\text{accel}}} \)
\( \frac{\mathrm{q}}{\mathrm{m}} = \frac{(1.2 \times 10^{6})^2}{2 \times 15 \times 10^3} = \frac{1.44 \times 10^{12}}{30 \times 10^3} = 4.8 \times 10^7 \, \mathrm{C/kg} \)

Now, let's compare this ratio to known particles:
For a proton (\( \mathrm{p} \)): \( \frac{\mathrm{q}_{\mathrm{p}}}{\mathrm{m}_{\mathrm{p}}} = \frac{1.602 \times 10^{-19} \, \mathrm{C}}{1.672 \times 10^{-27} \, \mathrm{kg}} \approx 9.58 \times 10^7 \, \mathrm{C/kg} \).
For a deuteron (\( \mathrm{D} \), which is \( ^{2}\mathrm{H} \)): \( \frac{\mathrm{q}_{\mathrm{D}}}{\mathrm{m}_{\mathrm{D}}} = \frac{1.602 \times 10^{-19} \, \mathrm{C}}{2 \times 1.672 \times 10^{-27} \, \mathrm{kg}} \approx 4.79 \times 10^7 \, \mathrm{C/kg} \).
Our calculated \( \frac{\mathrm{q}}{\mathrm{m}} \) ratio matches very closely with that of a deuteron.
Therefore, we can *guess* that the beam contains deuterons.

The answer is not unique because other particles (like \( ^{4}\mathrm{He}^{2+} \) or \( ^{6}\mathrm{Li}^{3+} \)) can have similar charge-to-mass ratios. For example, an alpha particle (\( ^{4}\mathrm{He}^{2+} \)) has charge \( 2\mathrm{e} \) and mass approximately \( 4\mathrm{m}_{\mathrm{p}} \), so \( \frac{\mathrm{q}}{\mathrm{m}} = \frac{2\mathrm{e}}{4\mathrm{m}_{\mathrm{p}}} = \frac{\mathrm{e}}{2\mathrm{m}_{\mathrm{p}}} \), which is similar to a deuteron's ratio. We can only determine the ratio \( \frac{\mathrm{q}}{\mathrm{m}} \), not the individual charge and mass.
In simple words: First, we found the particle's speed by balancing the electric and magnetic forces. Then, we calculated its charge-to-mass ratio using its speed and the acceleration voltage. This ratio matched a deuteron, suggesting the beam might be deuterons. The answer isn't unique because other particles can have the same charge-to-mass ratio, so we can't be absolutely sure without more information.

🎯 Exam Tip: The principle of undeflected motion in crossed electric and magnetic fields (velocity selector) is crucial here. Remember the relationship \( v = E/B \) and the kinetic energy relation \( qV = \frac{1}{2}mv^2 \). Be prepared to calculate and compare charge-to-mass ratios. The "not unique" part is important; explain why \( q/m \) doesn't identify a particle uniquely.

Question 20. A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.
(a) What magnetic field should be set up normally to the conductor in order that the tension in the wires is zero?
(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field the same as before? (Ignore the mass of the wires.) g = 9.8 ms\(^{-2}\).
Answer:
Given: length of rod \( \mathrm{L} = 0.45 \, \mathrm{m} \), mass \( \mathrm{m} = 60 \, \mathrm{g} = 60 \times 10^{-3} \, \mathrm{kg} \), current \( \mathrm{I} = 5.0 \, \mathrm{A} \), acceleration due to gravity \( \mathrm{g} = 9.8 \, \mathrm{m/s}^2 \).

(a) For the tension in the wires to be zero, the upward magnetic force on the rod must perfectly balance its downward gravitational force (weight).
Magnetic force \( \mathrm{F}_{\mathrm{B}} = \mathrm{BIL} \sin \theta \). Since the field is set up normally, \( \sin \theta = 1 \). So, \( \mathrm{F}_{\mathrm{B}} = \mathrm{BIL} \).
Gravitational force \( \mathrm{F}_{\mathrm{g}} = \mathrm{mg} \).
Equating them: \( \mathrm{BIL} = \mathrm{mg} \)
\( \mathrm{B} = \frac{\mathrm{mg}}{\mathrm{IL}} \)
\( \mathrm{B} = \frac{60 \times 10^{-3} \times 9.8}{5.0 \times 0.45} \)
\( \mathrm{B} \approx 0.261 \, \mathrm{T} \)
So, a magnetic field of approximately 0.261 T, directed appropriately (e.g., using Fleming's Left Hand Rule, if current is left and weight is down, field is out of page), should be applied.

(b) If the direction of the current is reversed, while the magnetic field remains the same, the direction of the magnetic force will also reverse. Now, both the magnetic force and the gravitational force will act downwards.
The total downward force will be \( \mathrm{F}_{\text{total}} = \mathrm{mg} + \mathrm{BIL} \).
The total tension in the wires will be equal to this total downward force.
Since \( \mathrm{BIL} = \mathrm{mg} \) (from part a), the total tension \( \mathrm{T}_{\text{total}} = \mathrm{mg} + \mathrm{mg} = 2 \mathrm{mg} \).
\( \mathrm{T}_{\text{total}} = 2 \times 60 \times 10^{-3} \, \mathrm{kg} \times 9.8 \, \mathrm{m/s}^2 \)
\( \mathrm{T}_{\text{total}} = 1.176 \, \mathrm{N} \)
In simple words: (a) To make the wires' tension zero, the upward push from the magnetic field must exactly balance the rod's weight, allowing us to find the needed magnetic field strength. (b) If the current direction is switched, the magnetic force also pulls downwards, so the total tension in the wires will be double the rod's weight.

🎯 Exam Tip: Crucially, understand the force balance for part (a). For part (b), remember that reversing current reverses the magnetic force, which will then act in the same direction as gravity, doubling the tension.

Question 21. The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?
Answer: The force between two parallel current-carrying wires is given by \( \mathrm{F} = \frac{\mu_{0} \mathrm{I}_{1} \mathrm{I}_{2} \mathrm{L}}{2 \pi \mathrm{r}} \).
Here, since it's a battery and a motor, we can assume the wires carry current in opposite directions (one wire carries current to the motor, the other carries it back to the battery). So, \( \mathrm{I}_{1} = \mathrm{I}_{2} = 300 \, \mathrm{A} \).
Length of wires \( \mathrm{L} = 70 \, \mathrm{cm} = 0.7 \, \mathrm{m} \).
Distance between wires \( \mathrm{r} = 1.5 \, \mathrm{cm} = 1.5 \times 10^{-2} \, \mathrm{m} \).
Substituting these values:
\( \mathrm{F} = \frac{4 \pi \times 10^{-7} \times 300 \times 300 \times 0.7}{2 \pi \times 1.5 \times 10^{-2}} \)
\( \mathrm{F} = \frac{2 \times 10^{-7} \times 90000 \times 0.7}{1.5 \times 10^{-2}} \)
\( \mathrm{F} = \frac{126000 \times 10^{-7}}{1.5 \times 10^{-2}} = \frac{1.26 \times 10^{-2}}{1.5 \times 10^{-2}} = 0.84 \times 10 = 8.4 \, \mathrm{N} \)
Since the currents in the wires (connecting battery to motor) flow in opposite directions, the force between them will be repulsive.
In simple words: We calculated the magnetic force between the two wires by using their length, the current they carry, and the distance between them. Because the currents flow in opposite directions, the force between the wires is repulsive, pushing them apart.

🎯 Exam Tip: For battery connections, current usually flows in opposite directions in the two connecting wires, leading to a repulsive force. Always convert units (cm to m) and use the correct formula for force between parallel wires.

Question 1. When current is passed through the conductor, the compass needle deflects. Why?
Answer: When electricity flows through a conductor, it creates a magnetic field around it. A compass needle is itself a small magnet. When this needle is placed near the conductor, it experiences a force from this newly created magnetic field. This force causes the compass needle to move and deflect, showing the presence of the magnetic field.
In simple words: A compass needle moves when current flows because the current makes a magnetic field that pushes on the needle.

🎯 Exam Tip: This question tests Oersted's experiment. The key concept is that moving charges (current) produce magnetic fields, which then interact with other magnetic fields (like the compass needle's field).

Question 2. Is the magnetic field the same at all points which are at the same distance from the element?
Answer: No, the magnetic field is not the same at all points at the same distance from the current element. The magnetic field's magnitude and direction depend on the angle between the current element and the position vector to the point. This relationship is described by the Biot-Savart law, where the magnetic field \( \mathrm{dB} \) is proportional to \( \sin \theta \), where \( \theta \) is the angle between the current element \( \mathrm{dl} \) and the position vector \( \mathrm{r} \).
In simple words: No, the magnetic field isn't the same everywhere at the same distance; it changes depending on the angle between the current and the point you're measuring at.

🎯 Exam Tip: Recall the Biot-Savart law's dependence on \( \sin \theta \). This is crucial for understanding the directional nature of magnetic fields created by current elements, differentiating it from simple radial fields.

Question 3. Is the magnetic field influenced by the medium in which the conductor is placed?
Answer: Yes, the magnetic field is influenced by the medium surrounding the conductor. The strength of the magnetic field changes based on the magnetic permeability of the medium. Different materials have different permeabilities, which modify how strong the magnetic field appears. For example, the magnetic field strength \( \mathrm{dB} \) is directly proportional to \( \mu \) (magnetic permeability) of the medium.
In simple words: Yes, the magnetic field changes based on the material around the wire, as different materials affect how strong the field becomes.

🎯 Exam Tip: The magnetic permeability \( \mu \) of the medium is a key factor. For vacuum, it's \( \mu_0 \). For other materials, \( \mu = \mu_r \mu_0 \), where \( \mu_r \) is the relative permeability.

Question 4. What is the field at any point outside the core of the solenoid?
Answer: For an ideal long solenoid, the magnetic field at any point outside its core is considered to be zero. Even for a finite solenoid, the field outside but close to it is very weak, and just outside on the axis, it is approximately half the field inside.
In simple words: Outside a long coil (solenoid), the magnetic field is usually zero.

🎯 Exam Tip: Remember the ideal solenoid approximation: field is uniform inside and zero outside. For real solenoids, the field rapidly decreases outside.

Question 5. What is the field inside the core of the solenoid?
Answer: Inside the core of a long solenoid, the magnetic field is uniform and its magnitude is given by \( \mathrm{B} = \mu_{0} \mathrm{nI} \). Here, \( \mu_{0} \) is the permeability of free space, \( \mathrm{n} \) represents the number of turns per unit length of the solenoid, and \( \mathrm{I} \) is the current flowing through the wire.
In simple words: Inside the solenoid, the magnetic field is even and strong, calculated by a formula using the number of turns, current, and a constant for how easily magnetic fields pass through space.

🎯 Exam Tip: The formula \( B = \mu_0 nI \) is fundamental for solenoids. Ensure you use 'n' (turns per unit length) correctly, not 'N' (total turns).

Question 6. What is the nature of the field around a moving charge?
Answer: A moving charge generates two types of fields around it: its own inherent electric field and a magnetic field. The magnetic field is a direct consequence of the charge's motion. This property means that a moving charge will interact with both electric and magnetic forces if present.
In simple words: A moving charge creates both an electric field and a magnetic field around itself.

🎯 Exam Tip: This is a core concept in electromagnetism: stationary charges produce only electric fields, but moving charges produce both electric and magnetic fields.

Question 7. Can a charge move in a magnetic field without experiencing a force?
Answer: Yes, a charge can move in a magnetic field without experiencing a force. This occurs under specific conditions: if the charged particle moves in a direction that is either parallel or anti-parallel to the magnetic field lines, the angle \( \theta \) between its velocity vector and the magnetic field vector will be \( 0^\circ \) or \( 180^\circ \). In both these cases, \( \sin \theta = 0 \), making the magnetic force \( \mathrm{F} = \mathrm{qvB} \sin \theta \) equal to zero.
In simple words: Yes, a charged particle can move in a magnetic field without feeling a push if it travels straight along the magnetic field lines.

🎯 Exam Tip: Remember the magnetic force formula \( F = qvB \sin \theta \). The condition for zero force is \( \sin \theta = 0 \), meaning velocity is parallel or anti-parallel to the magnetic field.

Question 8. What is the nature of the motion of charge entered inclined to a magnetic field?
Answer: When a charged particle enters a uniform magnetic field at an angle that is not \( 0^\circ \), \( 180^\circ \), or \( 90^\circ \) (i.e., it enters inclined), its motion will be helical. This helical path is a combination of two motions: linear motion along the direction of the magnetic field (due to the velocity component parallel to the field) and circular motion in a plane perpendicular to the field (due to the velocity component perpendicular to the field).
In simple words: If a charged particle enters a magnetic field at an angle, it will move in a spiral shape, called a helix, combining straight and circular motion.

🎯 Exam Tip: Understand that the velocity component parallel to the field causes linear motion, while the component perpendicular to the field causes circular motion, leading to a helix.

Question 9. If the path is circular, which force provides the necessary centripetal force?
Answer: If a charged particle moves in a circular path within a magnetic field, the necessary centripetal force that keeps it moving in a circle is provided by the Lorentz magnetic force. This force is always perpendicular to the particle's velocity, thus changing its direction without altering its speed, causing circular motion.
In simple words: When a charged particle moves in a circle in a magnetic field, the magnetic force itself acts as the force that keeps it in that circular path.

🎯 Exam Tip: This is a fundamental concept: the magnetic force acts as the centripetal force (\( qvB = mv^2/r \)) for circular motion of a charged particle in a magnetic field.

Question 10. What is the work done by magnetic force during this motion?
Answer: The work done by a magnetic force on a charged particle is always zero. This is because the magnetic force is inherently always perpendicular to the direction of the particle's displacement (which is along its velocity). Since work done is defined as \( \mathrm{W} = \vec{\mathrm{F}} \cdot \vec{\mathrm{s}} \) or \( \mathrm{Fscos}\theta \), and \( \theta = 90^\circ \) (making \( \cos 90^\circ = 0 \)), no work is done. Consequently, the kinetic energy and speed of the charged particle remain constant.
In simple words: The magnetic force does no work because it always pushes sideways, not in the direction the particle is moving.

🎯 Exam Tip: A crucial property of magnetic force: it does no work and therefore cannot change the kinetic energy or speed of a charged particle.

Question 11. How can we confine the motion in a comparatively small space?
Answer: To confine the motion of a charged particle within a relatively small space, we can use a magnetic field to force it into a circular or spiral (helical) path. By making the particle move in such curved trajectories, it stays within a limited region instead of traveling in a straight line and escaping.
In simple words: We can keep a charged particle in a small area by making it move in circles or spirals using a magnetic field.

🎯 Exam Tip: This principle is used in devices like cyclotrons, mass spectrometers, and magnetic bottles to trap or guide charged particles.

Question 12. What is the electric field inside the 'dees'?
Answer: In a cyclotron, the 'dees' are hollow D-shaped metallic chambers. Inside these conductive dees, the electric field is zero. This is because the electric field is only present in the gap between the dees, where the particles are accelerated. Inside the dees, the electric field lines cannot penetrate, making it an equipotential region.
In simple words: Inside the metal D-shaped parts of a cyclotron, there is no electric field.

🎯 Exam Tip: This property (zero electric field inside a conductor) is fundamental to the operation of a cyclotron, allowing particles to gain energy only in the gap between the dees.

Question 13. What is the nature of the path inside the 'dee'?
Answer: Inside the 'dee' of a cyclotron, a charged particle moves in a semicircular path. This is because there is no electric field inside the dees (as explained in Question 12), and the particle is under the influence of a uniform magnetic field only, which causes it to follow a circular trajectory. As it crosses from one dee to the other, it completes half a circle inside each dee.
In simple words: Inside the D-shaped parts, the particle moves in a half-circle because only the magnetic field affects it there.

🎯 Exam Tip: Combine the knowledge of zero electric field inside dees with the circular motion caused by a magnetic field to correctly describe the path.

Question 14. The polarity of 'dees' should be changed when the charge completes the one-half circle. Why?
Answer: The polarity of the 'dees' in a cyclotron must be reversed precisely when the charged particle completes one-half of its circular path and arrives at the gap between the dees. This synchronized change in polarity ensures that the electric field in the gap is always in the correct direction to accelerate the particle. This constant acceleration increases the particle's kinetic energy and speed, making its radius of motion larger, until it exits the cyclotron.
In simple words: The 'dees' polarity must change at the right time to always push the particle faster across the gap, making it gain energy and spiral outwards.

🎯 Exam Tip: Synchronization of dee polarity with the particle's motion is critical for continuous acceleration in a cyclotron. This is related to the cyclotron frequency.

Question 15. What is the torque experienced by a current-carrying rectangular coil placed in a magnetic field?
Answer: A current-carrying rectangular coil placed in a uniform magnetic field experiences a torque that tends to align the coil's magnetic dipole moment with the external magnetic field. The torque \( \vec{\tau} \) is given by the cross product of the magnetic moment \( \vec{\mathrm{M}} \) of the coil and the magnetic field \( \vec{\mathrm{B}} \), i.e., \( \vec{\tau} = \vec{\mathrm{M}} \times \vec{\mathrm{B}} \). For a coil with N turns, current I, and area A, the magnetic moment is \( \vec{\mathrm{M}} = \mathrm{NIA}\hat{\mathrm{n}} \), where \( \hat{\mathrm{n}} \) is the normal to the coil's plane. So, \( \tau = \mathrm{NIAB} \sin \theta \), where \( \theta \) is the angle between the normal to the coil and the magnetic field.
In simple words: A rectangular coil with current in a magnetic field feels a twisting force (torque) that tries to turn it, so its magnetic field lines up with the external field.

🎯 Exam Tip: Remember the torque formula \( \tau = NIAB \sin \theta \) and that \( \vec{M} = NI\vec{A} \). Torque depends on the angle between the area vector (normal to coil) and the magnetic field.

Question 16. What is the difference between a galvanometer and an ammeter?
Answer: A galvanometer is a device designed to detect the presence of small electric currents in a circuit, and also to indicate their direction. It is very sensitive and typically has a central zero. An ammeter, on the other hand, is used to measure the magnitude or strength of an electric current flowing through a circuit. It is always connected in series and is designed to have very low resistance.
In simple words: A galvanometer tells you if there's current and its direction, while an ammeter measures how much current there is.

🎯 Exam Tip: Focus on the primary function: detection vs. measurement. Also, note their connection in circuits (galvanometer can be in series or parallel, ammeter always in series) and their resistance properties.

Question 17. Is it possible to measure high current using a galvanometer?
Answer: No, it is generally not possible to directly measure a high current using a standard galvanometer. Galvanometers are highly sensitive instruments built to detect and measure very small currents. Passing a high current through a galvanometer can damage its delicate coil and suspension system due to excessive heat generation and large deflections.
In simple words: No, you can't use a regular galvanometer for high currents because it's too sensitive and could break.

🎯 Exam Tip: Understand the sensitivity and delicate nature of galvanometers. This leads to the concept of converting a galvanometer into an ammeter using a shunt.

Question 18. How can you measure a high current using a galvanometer?
Answer: To measure a high current using a galvanometer, it must be converted into an ammeter. This conversion is achieved by connecting a very low resistance, called a shunt resistance, in parallel with the galvanometer coil. The shunt resistance diverts a large portion of the main current, allowing only a small, safe fraction to pass through the galvanometer. The combination of the galvanometer and shunt then acts as an ammeter calibrated for a higher current range.
In simple words: To measure a high current with a galvanometer, you connect a very small resistor next to it, which lets most of the current bypass the galvanometer, protecting it while still allowing a measurement.

🎯 Exam Tip: The key concept is connecting a low resistance (shunt) in parallel. This is a common conversion technique, and students should know the circuit diagram and calculation for shunt resistance.

Question 19. How will you connect a voltmeter to a circuit?
Answer: A voltmeter is always connected in parallel across the two points in a circuit where the potential difference (voltage) is to be measured. It must be connected with the correct polarity, meaning its positive terminal connects to the higher potential point and its negative terminal to the lower potential point. The voltmeter is designed to have a very high internal resistance to draw minimal current from the circuit, thus not significantly altering the original potential difference.
In simple words: You connect a voltmeter across the two points where you want to measure voltage, making sure its positive end is at the higher voltage.

🎯 Exam Tip: Remember parallel connection and high resistance for voltmeters. This ensures it measures voltage without significantly impacting the circuit's current distribution.

Question 20. To measure p.d between two points in a circuit, the minimum current should pass through the voltmeter. Why?
Answer: To accurately measure the potential difference (p.d.) between two points in a circuit, it is crucial that the voltmeter draws as little current as possible. If the voltmeter draws a significant current, it would change the current distribution in the original circuit. This alteration would lead to an incorrect potential drop across the component being measured (since potential drop is current times resistance), and thus the voltmeter would not show the true p.d. of the undisturbed circuit.
In simple words: A voltmeter needs to draw very little current so it doesn't change the circuit's normal flow, ensuring it measures the true voltage accurately.

🎯 Exam Tip: This explains why voltmeters have very high internal resistance. A high resistance ensures minimal current diversion, preserving the original circuit conditions for accurate measurement.

Question 21. How can we convert a galvanometer into an voltmeter?
Answer: To convert a galvanometer into a voltmeter, a very high resistance is connected in series with the galvanometer coil. This high series resistance limits the current passing through the galvanometer to a safe range for a given potential difference. The combination of the galvanometer and the high series resistance can then be calibrated to measure a specific range of voltages.
In simple words: To turn a galvanometer into a voltmeter, you add a very large resistor in a line with it, which helps it measure voltage safely.

🎯 Exam Tip: The key concept is connecting a high resistance in series. This is a common conversion technique; students should know the circuit diagram and calculation for series resistance.

Question 22. What is the magnetic field at a point on the axial line on a current-carrying circular loop?
Answer: The magnetic field at a point on the axial line of a current-carrying circular loop, at a distance \( \mathrm{x} \) from its center, is given by the formula: \( \mathrm{B} = \frac{\mu_{0} \mathrm{I R}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}} \) (in Tesla). In this expression, \( \mathrm{I} \) is the current flowing through the loop, \( \mathrm{R} \) is the radius of the circular loop, and \( \mathrm{x} \) is the distance from the center of the loop to the point along its axis.
In simple words: The magnetic field on the central line coming out of a circular current loop is calculated using a formula that depends on the current, loop's radius, and the distance from the center.

🎯 Exam Tip: Remember this specific formula for the axial magnetic field. Be careful with the power of the denominator \((3/2)\) and ensure all variables are correctly identified.

Question 23. How can you rewrite the above equation in terms of area of the loop?
Answer: The area of a circular loop is \( \mathrm{A} = \pi \mathrm{R}^2 \). We can express \( \mathrm{R}^2 = \frac{\mathrm{A}}{\pi} \).
Substitute this into the magnetic field equation: \( \mathrm{B} = \frac{\mu_{0} \mathrm{I R}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}} \).
\( \mathrm{B} = \frac{\mu_{0} \mathrm{I} (\mathrm{A}/\pi)}{2\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}} \)
To match a more common form, multiply the numerator and denominator by \( 2 \pi \):
\( \mathrm{B} = \frac{2 \mu_{0} \mathrm{I A}}{4 \pi\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}} \)
This form highlights the dependence on the area \( \mathrm{A} \).
In simple words: We can change the magnetic field formula by replacing the square of the radius with the loop's area divided by pi, then adjusting it to a common way of writing it.

🎯 Exam Tip: The key here is recognizing the relationship between radius and area (\( A = \pi R^2 \)). This transformation can sometimes simplify understanding the field's dependence on the coil's size.

Question 24. An electric charge will experience a force in a uniform electric field. Similarly, a moving charge experiences a magnetic force (Lorentz) in a magnetic field. The SI unit of magnetic field intensity is defined in terms of Lorentz force.
(a) Write the expression for magnetic Lorentz force.
(b) Mention any two differences between electric and magnetic field.
(c) Give an account of work done by Lorentz force on a moving charge and corresponding change in K.E.
Answer:
(a) The expression for the magnetic Lorentz force \( \vec{\mathrm{F}}_{\mathrm{B}} \) acting on a charge \( \mathrm{q} \) moving with velocity \( \vec{\mathrm{v}} \) in a magnetic field \( \vec{\mathrm{B}} \) is given by the cross product:
\( \vec{\mathrm{F}}_{\mathrm{B}} = \mathrm{q}(\vec{\mathrm{v}} \times \vec{\mathrm{B}}) \).

(b) Two key differences between electric and magnetic fields are:
(i) **Origin:** An electric field is produced by stationary or moving charges. A magnetic field, however, is produced only by moving charges or changing electric fields.
(ii) **Force Direction:** The electric force (\( \vec{\mathrm{F}}_{\mathrm{E}} = \mathrm{q}\vec{\mathrm{E}} \)) on a charge is always parallel (or anti-parallel for negative charges) to the electric field direction. In contrast, the magnetic force (\( \vec{\mathrm{F}}_{\mathrm{B}} = \mathrm{q}(\vec{\mathrm{v}} \times \vec{\mathrm{B}}) \)) is always perpendicular to both the velocity of the charge and the magnetic field direction.

(c) The work done by the magnetic Lorentz force on a moving charge is always zero. This is because the magnetic force \( \vec{\mathrm{F}}_{\mathrm{B}} \) is always perpendicular to the velocity \( \vec{\mathrm{v}} \) (and thus to the displacement) of the charge. As work done \( \mathrm{W} = \vec{\mathrm{F}} \cdot \vec{\mathrm{s}} = \mathrm{Fs} \cos \theta \), and here \( \theta = 90^\circ \) (so \( \cos 90^\circ = 0 \)), the work done is zero.
Since work done on the charge is zero, there is no change in its kinetic energy (K.E.). The magnetic force can change the direction of motion, but it cannot change the speed or the magnitude of the kinetic energy of the charged particle.
In simple words: (a) The magnetic force on a moving charge is found by multiplying the charge by the cross product of its velocity and the magnetic field. (b) Electric fields come from both still and moving charges, and push charges along the field. Magnetic fields only come from moving charges, and push charges sideways to both their movement and the field. (c) The magnetic force does no work and doesn't change a particle's speed because it always pushes perpendicular to the direction of motion.

🎯 Exam Tip: Memorize the vector form of the Lorentz force. Clearly distinguish between the origins and directional properties of electric and magnetic forces. The "no work, no change in K.E." concept for magnetic force is fundamental and frequently tested.

Question 25. Two infinitely long wires carry current as shown in Figure. Find the magnetic field intensity at the points P, P' and P”
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो अनंत लंबाई के समानांतर तारों को दिखाता है, जिनमें धारा विपरीत दिशाओं में बह रही है। तीन बिंदु P, P' और P" दिखाए गए हैं जहाँ चुंबकीय क्षेत्र की तीव्रता ज्ञात करनी है। बिंदु P दोनों तारों के बीच में है, जबकि P' और P" दोनों तारों के बाहर हैं।
Answer:At P, \( B_P = \frac{\mu_0 I}{2\pi} \left( \frac{1}{x} + \frac{1}{r-x} \right) \) At P', \( B_{P'} = \frac{\mu_0 I}{2\pi} \left( \frac{1}{x} - \frac{1}{r+x} \right) \) And at P'', \( B_{P''} = \frac{\mu_0 I}{2\pi} \left( \frac{1}{x} - \frac{1}{r+x} \right) \) (with the same current)
In simple words: The magnetic field at different points around two long parallel wires depends on their distance and current direction. The formulas provided help calculate this field at points P, P', and P''.

🎯 Exam Tip: Remember to use the right-hand thumb rule to find the direction of the magnetic field and consider the vector sum for multiple current sources.

Question 26. Figure shows a long straight wire of finite cross-section carrying steady current i. The current i is uniformly distributed across this cross-section of radius a.
(i) Calculate the magnetic field in the region r < a and r > a.
(ii) Sketch the magnetic field against the distance from the axis of the wire.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक लंबी सीधी तार को दिखाता है जिसका अनुप्रस्थ काट वृत्ताकार है और इसकी त्रिज्या 'a' है। तार के भीतर (ra) की दूरी पर चुंबकीय क्षेत्र B के परिवर्तन को दर्शाने वाला ग्राफ भी इसमें शामिल है।
Answer:(i) When r is less than a (r < a), the current enclosed is \( \frac{i \times \pi r^2}{\pi a^2} \). Applying Ampere's theorem, \( B \cdot 2\pi r = \mu_0 \frac{i r^2}{a^2} \). So, the magnetic field \( B = \left( \frac{\mu_0 i}{2 \pi a^2} \right) r \). When r is greater than a (r > a), \( B \cdot 2\pi r = \mu_0 i \). So, the magnetic field \( B = \frac{\mu_0 i}{2 \pi r} \).
(ii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ चुंबकीय क्षेत्र (B) को तार के केंद्र से दूरी (r) के विरुद्ध दिखाता है। त्रिज्या 'a' तक, क्षेत्र B दूरी के साथ रैखिक रूप से बढ़ता है (B ∝ r)। त्रिज्या 'a' के बाद, क्षेत्र B दूरी के साथ व्युत्क्रम रूप से घटता है (B ∝ 1/r)।
In simple words: Inside the wire, the magnetic field grows stronger as you move away from the center. Outside the wire, the magnetic field gets weaker as you move further away.

🎯 Exam Tip: Remember that for a wire with finite cross-section, the magnetic field inside and outside the wire follows different proportionality rules based on Ampere's law.

Question 27. A straight wire ABCD is bend as shown in Figure. Find an expression for the magnetic field ?
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सीधी तार ABCD को दर्शाता है जिसे बिंदु C पर मोड़ा गया है। यह तार एक बिंदु P पर चुंबकीय क्षेत्र की गणना के लिए ज्यामिति दिखाता है, जिसमें कोण \( \Phi_1 \) और \( \Phi_2 \) और तार से दूरी 'r' शामिल हैं।
Answer:\( B_P = \frac{\mu_0 i}{4 \pi r}(\sin\Phi_1 + \sin\Phi_2) \) (No field on the axis of the straight conductor)
In simple words: The magnetic field at point P due to the bent wire depends on the current in the wire, its distance from P, and the angles formed by the ends of the wire with the point. There is no magnetic field exactly on the axis of a straight current-carrying wire.

🎯 Exam Tip: For problems involving bent wires, apply the Biot-Savart law or Ampere's law to each straight segment and sum up the contributions. Remember that points along the axis of a straight wire experience zero magnetic field from that segment.

Question 28. P and Q are two infinitely long straight parallel conductors placed in air 'r' distance apart. Let I₁ and I₂, be the current on P and Q respectively flowing in the same direction. Then
(a) What is the magnetic field on Q due to If
(b) What is the force experienced by Q?
(c) Is there any attraction or repulsion between P and Q?
Answer:(a) The magnetic field on wire Q due to wire P is \( B_P = \frac{\mu_0 i}{2 \pi r} \).
(b) The force experienced by Q is \( F = \frac{\mu_0 I_1 I_2 l}{2 \pi r} \) (over a length l).
(c) Yes. There is an attraction between P and Q.
In simple words: When two parallel wires carry current in the same direction, they create magnetic fields that pull them towards each other, causing an attractive force. The strength of this attraction depends on the currents, distance, and length of the wires.

🎯 Exam Tip: Parallel currents in the same direction attract, while parallel currents in opposite directions repel. This is a fundamental concept for understanding magnetic forces between current-carrying conductors.

Question 29. Describe qualitatively the path of a charged particle moving in a uniform magnetic field with an initial velocity
(a) Parallel to the field
(b) Perpendicular to the field and
(c) At an arbitrary angle with the field direction.
Answer:(a) Straight line
(b) Circular path
(c) Helix
In simple words: If a charged particle moves exactly along the magnetic field, it goes straight. If it moves at a 90-degree angle to the field, it travels in a circle. If it moves at any other angle, it spins in a spiral path called a helix.

🎯 Exam Tip: The magnetic force on a charged particle is always perpendicular to both its velocity and the magnetic field. This means the force changes the direction of motion but not the speed.

Question 30. When a charged particle moves in a magnetic field, does its KE always remain constant? Explain.
Answer: A charged particle moving in a magnetic field of constant intensity does not gain kinetic energy. This happens because the force is \( \vec{F} = q(\vec{v} \times \vec{B}) \). The work done, W, is \( \vec{F} \cdot \vec{s} \) or in 1 second, \( \vec{F} \cdot \vec{v} = q(\vec{v} \times \vec{B}) \cdot \vec{v} = 0 \), due to the property of the scalar triple product.
In simple words: Yes, a charged particle's kinetic energy stays the same when it moves in a magnetic field. This is because the magnetic force only changes the particle's direction, not its speed, so no work is done on it.

🎯 Exam Tip: A key point is that the magnetic force is always perpendicular to the velocity, meaning it does no work. Therefore, the kinetic energy (and speed) of the charged particle remains constant in a magnetic field.

Question 31. A long copper wire of diameter 2 mm carries a current of 1.5 A.
(a) What is the magnetic field (B) at a perpendicular distance 1 m from the middle of the wire?
(b) What is the magnetic field at a perpendicular distance 1 m from one end of the wire?
(c) What is the field (B) on the surface of the conductor?
(d) At a distance 1 mm from the axis, what is the value of the magnetic field?
Answer:Data supplied: Radius r = 1 mm = \( 1 \times 10^{-3} \) m, Current I = 1.5 A
(a) \( B = \frac{\mu_0 I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 1.5}{2\pi \times 1} = 3 \times 10^{-7} \) T
(b) \( B_{end} = \frac{\mu_0 I}{4\pi r} = 1.5 \times 10^{-7} \) T
(c) \( B = \frac{\mu_0 I}{2\pi a} = \frac{4\pi \times 10^{-7} \times 1.5}{2\pi \times 10^{-3}} = 3 \times 10^{-4} \) T
(d) \( B = \frac{\mu_0 I r}{2\pi a^2} = \frac{4\pi \times 10^{-7} \times 1.5 \times 10^{-3}}{2\pi \times (1 \times 10^{-3})^2} = 3 \times 10^{-4} \) T
In simple words: We calculated the magnetic field around a wire carrying current at different points. The field changes depending on how far you are from the wire, whether you are on its surface, or at its ends.

🎯 Exam Tip: Pay attention to the location where the magnetic field is to be calculated (e.g., center of a long wire, end of a wire, on the surface, or inside) as different formulas apply.

Question 32. An electron does not suffer any deflection while passing through a region. Are you sure that there is no magnetic field?
Answer: No, because the electron suffers no force when it moves parallel or anti-parallel to the magnetic field. So the magnetic field may exist parallel or anti-parallel to the motion of the electron.
In simple words: An electron can pass through a magnetic field without bending if it moves in the same direction as the field lines or in the exact opposite direction. So, no deflection doesn't always mean no magnetic field.

🎯 Exam Tip: The magnetic force \( \vec{F} = q(\vec{v} \times \vec{B}) \) is zero when the velocity vector \( \vec{v} \) is parallel or anti-parallel to the magnetic field vector \( \vec{B} \), as \( \sin(0^\circ) = \sin(180^\circ) = 0 \).

Question 33. An electron appears moving anti-clockwise in a horizontal circle under a magnetic field. What is the direction of the field?
Answer: Vertically upward
In simple words: For an electron to move counter-clockwise in a horizontal circle, the magnetic field must point upwards. This is found using the right-hand rule, but reversed for a negative charge.

🎯 Exam Tip: Use the right-hand rule for positive charges, and reverse the direction of the force or field for negative charges (like electrons).

Question 34. If a thin wire of length 88 cm is bent into a circular loop and a current of 1A is passed,
(a) What is the magnetic field at the centre?
(b) What is the magnetic field at the centre if the wire is bent into a semicircle and doubled on it?
(c) If the wire is looped into two loops, find the magnetic field at the centre.
Answer:(a) Given \( 2\pi a = 88 \times 10^{-2} \) m; So, \( a = \frac{88 \times 10^{-2}}{2\pi} \) m; Current I = 1 A. \( B = \frac{\mu_0 I}{2a} = \frac{4\pi \times 10^{-7} \times 1}{2 \times \frac{88 \times 10^{-2}}{2\pi}} = \frac{\pi^2 \times 10^{-7}}{22 \times 10^{-2}} \approx \frac{22}{7} \times \frac{10^{-7}}{22 \times 10^{-2}} \times \frac{22}{7} \times \frac{10^{-7}}{22 \times 10^{-2}} \) (this calculation seems to have an error in the original text, let's re-evaluate simplified) \( B = \frac{4\pi \times 10^{-7} \times 1}{2 \times \frac{0.88}{2\pi}} = \frac{4\pi^2 \times 10^{-7}}{0.88} = \frac{4 \times (3.14159)^2 \times 10^{-7}}{0.88} \approx \frac{4 \times 9.8696 \times 10^{-7}}{0.88} \approx 4.48 \times 10^{-6} \) T. Using the provided values from the OCR: \( B = \frac{4\pi \times 10^{-7} \times 1}{2a} \), where \( a = \frac{88 \times 10^{-2}}{2\pi} \). So, \( B = \frac{4\pi \times 10^{-7}}{2 \times \frac{0.88}{2\pi}} = \frac{4\pi^2 \times 10^{-7}}{0.88} \approx 0.45 \times 10^{-5} \) T
(b) Zero
(c) Approximately \( 1.8 \times 10^{-5} \) T
In simple words: For a wire of a fixed length carrying current, we found the magnetic field at the center when it's shaped as a single circle, a doubled semicircle, and two loops. The field changes based on how the wire is bent.

🎯 Exam Tip: Remember that the magnetic field at the center of a circular coil is directly proportional to the current and number of turns, and inversely proportional to the radius. Pay attention to how the length of the wire translates into the radius and number of turns for different shapes.

Question 35. A current carrying a circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns around itself?
Answer: No, the torque on a current loop is given by \( \vec{\tau} = \vec{M} \times \vec{B} = I\vec{A} \times \vec{B} \). The loop can turn around itself only when the force (torque) is along the vertical direction. Since the area vector \( \vec{A} \) is vertical (for a horizontal loop), the magnetic field \( \vec{B} \) cannot be vertical to produce a torque that would turn the loop around its own axis (which requires a horizontal torque).
In simple words: A circular loop lying flat cannot spin around its own center just by applying a uniform magnetic field. This is because the magnetic force would try to tilt the loop, not spin it like a wheel on the ground.

🎯 Exam Tip: The torque on a current loop depends on the magnetic moment and the external magnetic field. For a loop to turn about its own axis, the torque must be perpendicular to the axis of rotation, and for a horizontal loop, this would mean a horizontal torque.

Question 36. A uniform magnetic field of 3000G is established along with the positive \( \vec{\tau} \) – direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12A. What is the torque on the loop in different cases shown in the figure? What is the force on each case? Which case corresponds to stable equilibrium?
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक आयताकार लूप को एक समान चुंबकीय क्षेत्र में विभिन्न झुकावों पर दिखाता है। इसमें छह अलग-अलग स्थितियाँ (a से f) हैं, जिनमें लूप का अभिविन्यास और चुंबकीय क्षेत्र (B) तथा धारा (I) की दिशाएँ भिन्न हैं।
Answer:1. Torque \( \tau = I(\vec{A} \times \vec{B}) \) In Figure (a), \( 1.8 \times 10^{-2} \) Nm along -y direction. In Figure (b), same as in (a). In Figure (c), \( 1.8 \times 10^{-2} \) Nm along -x direction. In Figure (d), \( 1.8 \times 10^{-2} \) Nm at an angle of 240° along the x direction. In (e) and (f), the torque is zero. 2. Force is zero in each case. Figure (e) corresponds to stable equilibrium.
In simple words: We calculated how much a rectangular loop would twist (torque) in a magnetic field based on its position. The loop doesn't move its whole body (zero net force). One specific position (Figure e) is where the loop is most stable and won't twist.

🎯 Exam Tip: For a current loop in a uniform magnetic field, the net force is always zero. Torque depends on the orientation of the magnetic dipole moment relative to the field. Stable equilibrium occurs when the magnetic moment is aligned with the magnetic field.

Question 37. Describe qualitatively the path of a charged particle moving in
(a) a uniform electrostatic field, with initial velocity (i) parallel to the field, (ii) perpendicular to the field, (iii) at an arbitrary angle with the field direction.
(b) a uniform magnetic field, with initial velocity (i) parallel to the field, (ii) perpendicular to the field, (iii) at an arbitrary angle with the field direction.
Answer:(a) Force on the charge, F = qE. (i) If the charge q is positive, the speed increases; otherwise, it decreases. The particle moves in a straight line. (ii) The initial velocity stays constant, and the particle accelerates in the direction of the field, resulting in a parabolic path. (iii) The velocity can be split into two parts. The component parallel to the field changes speed (accelerates or decelerates), while the perpendicular component remains unaffected, leading to a parabolic path.
(b) (i) No change in direction and velocity. The path remains a straight line. (ii) The path becomes circular. (iii) Generally, the path is a helix.
In simple words: In an electric field, a charged particle speeds up or slows down and follows a straight or curved path. In a magnetic field, it doesn't change speed; it either goes straight, in a circle, or in a spiral, depending on its initial direction.

🎯 Exam Tip: Remember the key difference: an electric field does work on a charged particle, changing its kinetic energy, while a magnetic field does no work, only changing the direction of motion.

Question 38. A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction parallel to the axis is along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,
(a) the wire intersects the axis,
(b) the wire is turned from N-S to north east-north west direction,
(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?
Answer:(a) The force is 2.1 N vertically downwards.
(b) The force is 2.1 N vertically downwards (remember \( l \sin\theta \) is constant, which effectively gives the same length component perpendicular to B).
(c) The force is 1.68 N vertically downwards.
In simple words: We found the magnetic force on a wire carrying current in a magnetic field. The force is always downwards, but its strength changes slightly based on how the wire is positioned or angled within the field.

🎯 Exam Tip: The magnetic force on a current-carrying wire is given by \( F = BIL\sin\theta \). Ensure you correctly identify the length of the wire within the field and the angle between the current and the magnetic field.

Question 39. A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.
(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field the same as before? (Ignore the mass of the wires.) g = 9.8 m s\(^{-2}\).
Answer:(a) To make the tension zero, the magnetic force must balance the weight of the rod: \( BIL = mg \). So, \( B = \frac{mg}{IL} = \frac{60 \times 10^{-3} \times 9.8}{5.0 \times 0.45} = 0.26 \) T
(b) If the current direction is reversed, the magnetic force also reverses, acting downwards. In this case, the total tension in the wires will be \( T = mg + BIL = 2mg \). \( T = 2 \times 60 \times 10^{-3} \times 9.8 = 1.176 \) N.
In simple words: To make the suspension wires loose, a magnetic field needs to push the rod upwards with a force equal to its weight. If the current is reversed, the magnetic field pushes the rod downwards, making the wires pull with twice the rod's weight.

🎯 Exam Tip: For problems involving equilibrium or forces on current-carrying conductors, remember to balance gravitational forces with magnetic forces. Changing the current direction reverses the magnetic force, which can significantly alter the net force or tension.

Question 40. A current carrying loop of irregular shape is located in an external magnetic field. If the wire is flexible, why does it become circular?
Answer: For a given perimeter, a circle encloses the greatest area compared to any other shape. Thus, to involve the maximum number of field lines, the coil assumes a circular shape with its plane normal to the field lines. This maximizes the magnetic flux and minimizes the potential energy, making it stable.
In simple words: A flexible current loop in a magnetic field becomes a circle because a circle holds the most area for a given length of wire. This shape allows it to capture the most magnetic field lines, making it more stable.

🎯 Exam Tip: Remember that systems tend towards states of minimum potential energy. For a current loop in a magnetic field, this corresponds to maximizing the magnetic flux through the loop, which a circular shape achieves for a given perimeter.

Question 41. Starting from the expression for the force on a current-carrying conductor in a magnetic field, arrive at the expression for Lorentz magnetic force.
Answer: The force on a current-carrying conductor is \( F = BIL \). We know that current \( I = \frac{q}{t} \) and the length \( L \) can be expressed as \( L = vt \) (where v is drift velocity). Substituting these into the force equation: \( F = B \left( \frac{q}{t} \right) (vt) = Bqv \). Therefore, the Lorentz magnetic force is \( \vec{F} = q(\vec{v} \times \vec{B}) \).
In simple words: We can derive the formula for the magnetic force on a single moving charge (Lorentz force) by starting with the force on a wire. We replace the current and wire length with the charge, its speed, and the time it travels.

🎯 Exam Tip: Understand the relationship between macroscopic current (I) and microscopic charge motion (q, v). The Lorentz force is fundamental and can be derived from the force on a current element \(d\vec{F} = I(d\vec{l} \times \vec{B})\) by considering charge carriers.

Question 42.
(a) What type of particles can be accelerated in a cyclotron?
(b) Why cyclotron is not used to accelerate electrons and neutrons?
Answer:(a) Protons, deuterons, alpha-particles, and other positively charged ions can be accelerated.
(b) Cyclotrons are not used to accelerate electrons because electrons are very light. As they gain speed, their mass increases significantly due to relativistic effects at much lower energies compared to heavier particles. This changes their revolution frequency, causing them to go out of sync with the oscillating electric field. Neutrons are uncharged particles, so they are not affected by electric or magnetic fields, and thus cannot be accelerated in a cyclotron.
In simple words: Cyclotrons speed up heavy, positively charged particles like protons. They don't work for electrons because electrons get too fast too quickly, changing their timing. They also don't work for neutrons because neutrons have no electric charge and thus are not affected by the fields used for acceleration.

🎯 Exam Tip: Remember the two main limitations of a cyclotron: relativistic mass increase for light particles like electrons, and the requirement for charged particles (thus, it cannot accelerate neutral particles like neutrons).

Question 43. In a moving coil galvanometer,
(a) why is a horse-shoe magnet used?
(b) why is phosphor bronze fiber used?
(c) why is a soft iron cylinder used?
Answer:(a) A horse-shoe magnet is used to produce a strong radial magnetic field. A radial field ensures that the plane of the coil is always parallel to the magnetic field, providing a uniform torque for any deflection.
(b) Phosphor bronze fiber is used because it is a good elastic material with a small couple per unit twist. This means it provides a very small restoring torque, making the galvanometer very sensitive to small currents.
(c) A soft iron cylinder is placed inside the coil to concentrate the magnetic field lines. This increases the magnetic field strength in the region where the coil is placed, making the field uniform and radial, which further enhances the sensitivity and linearity of the galvanometer.
In simple words: A horse-shoe magnet makes the magnetic field strong and even. The phosphor bronze wire is used because it's very flexible and creates little resistance to twisting, making the device sensitive. A soft iron core inside helps to focus the magnetic field, making the galvanometer work better.

🎯 Exam Tip: These components (horse-shoe magnet, phosphor bronze fiber, soft iron core) are crucial for the high sensitivity and linear response of a moving coil galvanometer. Understand the specific role of each in achieving these characteristics.

Question 44. Expression for voltage sensitivity is \( \frac{\theta}{V} = \frac{NAB}{CR} \)
(a) What are the factors on which voltage sensitivity depends?
(b) Why is it independent of number of turns?
(c) How will you increase the voltage sensitivity?
Answer:(a) Voltage sensitivity depends on: (i) \( \frac{N}{R} \) (ratio of number of turns to resistance) (ii) Area (A) of the coil (iii) Magnetic field strength (B) (iv) Restoring torque constant (C) per unit twist of the suspension wire
(b) Voltage sensitivity is often seen as independent of the number of turns (N) when the resistance (R) is also dependent on N (i.e., \( R \propto N \)). If R is proportional to N, then \( \frac{N}{R} \) becomes a constant, effectively canceling out the direct dependence on N.
(c) Voltage sensitivity can be increased by: (i) Increasing the magnetic field (B). (ii) Increasing the area (A) of the coil. (iii) Decreasing the restoring torque per unit twist (C) of the suspension fiber.
In simple words: Voltage sensitivity depends on the coil's area, magnetic field strength, and the wire's twisting resistance. It seems not to depend on the number of turns if the resistance also increases with turns. To make it more sensitive, you can use a stronger magnet, a larger coil, or a fiber that twists more easily.

🎯 Exam Tip: For voltage sensitivity, focus on the ratio \( N/R \). If resistance is proportional to N (which is usually the case for a coil), then \( N/R \) becomes constant, making voltage sensitivity seemingly independent of N.

Question 45. A moving coil galvanometer can be converted into an ammeter.
(a) Is the statement true or false?
(b) If true, how is it possible? Explain
(c) What is the effective resistance?
Answer:(a) True.
(b) It is possible by connecting a suitable low resistance (called a shunt) parallel to the galvanometer. This shunt diverts most of the current, allowing only a small fraction to pass through the galvanometer coil, thus enabling it to measure large currents.
(c) The effective resistance of the converted ammeter is less than that of the shunt resistance. It is the parallel combination of the galvanometer's resistance and the shunt resistance.
In simple words: Yes, a galvanometer can become an ammeter by adding a small resistor next to it. This resistor carries most of the current, protecting the galvanometer and allowing it to measure large currents. The total resistance of this new ammeter is very low.

🎯 Exam Tip: Remember that an ammeter must have very low resistance to measure current accurately without significantly altering the circuit. A shunt resistor is critical for converting a galvanometer into an ammeter.

Question 46. No net force acts on a rectangular loop carrying a steady current when suspended freely in a uniform magnetic field. Is the statement correct or wrong? Justify your answer.
Answer: Correct. The forces on pairs of opposite sides of the rectangular loop are equal in magnitude and opposite in direction. Therefore, they cancel each other out, resulting in no net force on the loop.
In simple words: This statement is correct. A rectangular loop in a uniform magnetic field feels forces on its sides that are equal and opposite, so they balance out. This means the loop won't move from its spot, though it might twist.

🎯 Exam Tip: While the net force on a current loop in a uniform magnetic field is zero, there can still be a net torque that causes it to rotate. This is a crucial distinction.

Question 47. A student records the following data for the magnitudes (B) of the magnetic field at axial points at different distances x from the centre of a circular coil of radius 'a ' carrying a current. Verify (for any two) that these observations are in good agreement with the expected theoretical variation of B with x.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्ताकार कुंडली के अक्षीय बिंदुओं पर चुंबकीय क्षेत्र (B) के अपेक्षित सैद्धांतिक परिवर्तन को दूरी (x) के साथ दिखाता है। यह दर्शाता है कि जैसे-जैसे दूरी बढ़ती है, चुंबकीय क्षेत्र (B) का मान कम होता जाता है।
Answer: Magnetic field (B) is given by \( B = \frac{\mu_0}{4\pi}\frac{2M}{x^3} \) (for points far from the coil, where M is the magnetic moment). It is evident that as x increases, the value of B decreases, which is consistent with the observations provided.
In simple words: The data shows that the magnetic field gets weaker as you move further away from the center of the coil, which matches what physics theory predicts for magnetic fields.

🎯 Exam Tip: For a circular coil, the magnetic field along its axis decreases with increasing distance from the center. Understand how the inverse cube relationship (for points far from the coil) dictates this behavior.

Question 48. What is the magnitude of the induced current in the circular loop KLMN of radius 'r' if the straight wire PQ carries a steady current of magnitude 'i' ampere?
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सीधी तार PQ को दिखाता है जिसमें धारा 'i' बह रही है, और इसके पास एक वृत्ताकार लूप KLMN है जिसकी त्रिज्या 'r' है। यह व्यवस्था वृत्ताकार लूप में प्रेरित धारा की गणना के लिए है।
Answer: Zero
In simple words: There is no induced current in the circular loop because the magnetic field from the steady current in the straight wire does not change over time, which is needed to induce a current.

🎯 Exam Tip: Induced current (or electromotive force) only occurs when there is a change in magnetic flux through the loop. A steady current in a stationary wire creates a constant magnetic field, hence no change in flux and no induced current.

Question 49. A short bar magnet placed with its axis at 30° to a uniform magnetic field of 0.2 T experiences a torque of 0.060 Nm.
(i) Calculate magnetic moment of the magnet
(ii) Find out what orientation of the magnet corresponds to its stable equilibrium in the magnetic field.
Answer:Given: Torque \( \tau = 0.060 \) Nm, Angle \( \theta = 30^\circ \), Magnetic field \( B = 0.2 \) T. (i) We know that \( \tau = mB \sin\theta \). So, \( 0.060 = m \times 0.2 \times \sin(30^\circ) \) \( 0.060 = m \times 0.2 \times 0.5 \) \( m = \frac{0.060}{0.2 \times 0.5} = \frac{0.060}{0.1} = 0.6 \) JT\(^{-1}\)
(ii) The potential energy (PE) of a magnet in field B is given by \( U = -mB \cos\theta \). When the magnet is placed parallel to the magnetic field \( \vec{B} \), then \( \theta = 0^\circ \). In this case, \( U = -mB \cos(0^\circ) = -mB \). This means the potential energy of the dipole is minimum. This state corresponds to stable equilibrium.
In simple words: We calculated the magnetic strength of a bar magnet using the torque it experienced in a magnetic field. The magnet finds its most stable position when its magnetic axis lines up perfectly with the magnetic field.

🎯 Exam Tip: Remember the formula for torque \( \tau = mB\sin\theta \) and potential energy \( U = -mB\cos\theta \). Stable equilibrium occurs at minimum potential energy, which is when the magnetic moment is parallel to the magnetic field (\( \theta = 0^\circ \)).

Question 50. In the diagram shown is a circular loop carrying current I. Show the direction of the magnetic field with the help of lines of force.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्ताकार लूप को दर्शाता है जिसमें धारा I प्रवाहित हो रही है। इसके चारों ओर चुंबकीय क्षेत्र की बल रेखाएं दिखाई गई हैं, जो लूप के केंद्र से गुजरती हैं और लूप के तल के लंबवत एक अक्ष के अनुदिश होती हैं।
Answer: The magnetic lines of force for a circular loop carrying current I are shown below:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्ताकार धारावाही लूप के चारों ओर चुंबकीय क्षेत्र रेखाओं के पैटर्न को दर्शाता है। लूप के केंद्र में, क्षेत्र रेखाएँ सीधी और सघन होती हैं, जो एक मजबूत और लगभग एक समान क्षेत्र को दर्शाती हैं। लूप से दूर, रेखाएँ फैल जाती हैं और वृत्ताकार हो जाती हैं, जो क्षेत्र की कमजोर होती शक्ति को दिखाती हैं।
In simple words: The diagram shows how magnetic field lines look around a current-carrying loop. They are concentrated and straight through the center of the loop, then spread out and curve around the outside.

🎯 Exam Tip: Use the right-hand rule to determine the direction of magnetic field lines around a current loop. The thumb points in the direction of the current, and curled fingers show the direction of the magnetic field.

Question 51. Two long parallel straight wires X and Y separated by a distance of 5 cm in the air carry currents of 10A and 5A respectively in opposite directions. Calculate the magnitude and direction of the force on a 20 cm length of the wire Y.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो लंबी, सीधी और समानांतर तारों X और Y को दर्शाता है। तार X में 10A की धारा और तार Y में 5A की धारा विपरीत दिशाओं में बह रही है। दोनों तार 5 सेमी की दूरी पर स्थित हैं।
Answer:Given: Current in wire X, \( I_X = 10 \) A; Current in wire Y, \( I_Y = 5 \) A; Distance between wires, \( r = 5 \) cm = \( 0.05 \) m; Length of wire Y, \( L = 20 \) cm = \( 0.20 \) m. The force between two parallel current-carrying wires is given by \( F = \frac{\mu_0 I_X I_Y L}{2\pi r} \). \( F = \frac{4\pi \times 10^{-7} \times 10 \times 5 \times 0.20}{2\pi \times 0.05} = \frac{4 \times 10^{-5} \times 10 \times 5 \times 0.20}{2 \times 0.05} = 4 \times 10^{-5} \) N. The direction of the force is perpendicular to the length of wire Y and acts away from X (repulsion) because the currents are in opposite directions.
In simple words: We found that two parallel wires carrying currents in opposite directions push each other away. The force on a 20 cm part of wire Y is calculated to be \( 4 \times 10^{-5} \) N, pushing it away from wire X.

🎯 Exam Tip: Remember that parallel currents in opposite directions exert a repulsive force on each other. Use the formula \( F = \frac{\mu_0 I_1 I_2 L}{2\pi r} \) and always check the direction based on current flow.

Question 52. A galvanometer with a coil of resistance 120 ohm shows full-scale deflection for a current of 2.5 mA. How will you convert the galvanometer into an ammeter of range 0 to 7.5A? Determine the net resistance of the ammeter. When the ammeter is put in a circuit, does it read slightly less or more than the actual current in the original circuit? Justify your answer.
Answer:Given: Galvanometer resistance, \( R_g = 120 \, \Omega \); Full-scale deflection current, \( I_g = 2.5 \) mA = \( 0.0025 \) A; Desired ammeter range, \( I = 7.5 \) A. To convert a galvanometer into an ammeter, a shunt resistance \( R_s \) must be connected in parallel. The formula for shunt resistance is \( R_s = \frac{I_g R_g}{I - I_g} \). \( R_s = \frac{0.0025 \times 120}{7.5 - 0.0025} = \frac{0.3}{7.4975} \approx 0.04001 \, \Omega \). So, by connecting a shunt of approximately \( 0.04 \, \Omega \) across the given galvanometer, we get an ammeter with a range of 0 to 7.5 A. Net resistance of the ammeter (parallel combination of \( R_g \) and \( R_s \)): \( R_{net} = \frac{R_g \times R_s}{R_g + R_s} = \frac{120 \times 0.04}{120 + 0.04} = \frac{4.8}{120.04} \approx 0.03998 \, \Omega \). When an ammeter is placed in a circuit, it reads slightly less than the actual current. This is because an ideal ammeter has zero resistance. However, a real ammeter, even with very low resistance, adds some resistance to the circuit. This additional resistance slightly decreases the total current flowing in the circuit, causing the ammeter to show a value that is a little less than the current that would flow without it.
In simple words: We converted a sensitive galvanometer into an ammeter that can measure much larger currents by adding a small resistor next to it. The ammeter's total resistance became very tiny. When this ammeter is used, it slightly reduces the current in the circuit because it adds a small amount of resistance.

🎯 Exam Tip: Remember that an ammeter is always connected in series to measure current. For conversion, a low resistance shunt is connected in parallel with the galvanometer. An ideal ammeter has zero resistance, but a practical one has a very small, non-zero resistance, leading to a slight reduction in the measured current.

Question. Two loops of equal lengths are bent in the form of two loops. One of the loops is square-shaped whereas the other loop is circular. These are suspended in a uniform magnetic field and the same current is passed through them. Which loop will experience greater torque? Give reason.
Answer: For a wire of a given length, the circular loop has a greater area than the square loop. The torque experienced by a loop in a magnetic field is given by \( \tau = NIAB \sin\theta \), where A is the area of the loop. Since the current (I), number of turns (N=1 for a single loop), magnetic field (B), and angle \( \theta \) are the same for both, the loop with the larger area will experience greater torque. Therefore, the circular loop will experience greater torque.
In simple words: If you make a square loop and a circular loop from the same length of wire and put them in a magnetic field with the same current, the circular loop will twist more. This is because a circle holds more area than a square for the same perimeter, and more area means more twisting force from the magnetic field.

🎯 Exam Tip: The torque on a current loop is directly proportional to its area. For a fixed perimeter, a circle encloses the maximum area compared to any other shape.