GSEB Class 12 Physics Solutions Chapter 6 Electromagnetic Induction

GSEB Class 12 Physics Electromagnetic Induction Text Book Questions and Answers

 

Question 1. Predict the direction of induced current in the situations described by the following figures (a) to (f).
ℹ️ चित्र व्याख्या (Diagram Explanation):(a) A bar magnet with its S-pole facing a coil is moving towards the coil. (b) A bar magnet with its N-pole facing a coil is moving away from the coil. (c) A coil connected to a battery with a tapping key is shown next to another coil. The tapping key is just closed. (d) A coil connected to a battery with a rheostat is shown next to another coil. The rheostat setting is being changed. (e) A coil connected to a battery with a tapping key is shown next to another coil. The tapping key is just released. (f) A straight current-carrying wire is placed above a coil, and the current in the wire is decreasing at a steady rate.
Answer:(a) Based on Lenz's law, the coil end facing the magnet's S-pole becomes an S-pole itself to push away the coming magnet. To make this happen, the current in the coil flows in the QRPQ direction.
(b) For the coil on the left side: For the same reason as in (a), the current in this coil will flow in the PRQP direction.
For the coil on the right side: According to Lenz's law, the end of the coil facing the magnet's N-pole will become an S-pole to pull the magnet moving away. For this to happen, the current in this coil flows in the YZXY direction.
(c) When the tapping key is just closed, current flows in the coil, and its magnetic field grows. Lenz's law states that the induced current in the nearby coil should create a magnetic field that opposes this growth. This happens if the induced magnetic field is from right to left, meaning the induced current flows in the YZX direction.
(d) As the rheostat setting changes (resistance decreases), the current in the coil and its magnetic field increase. Lenz's law dictates that the induced current in the nearby coil will flow in a direction that opposes this increase. This occurs if the induced magnetic field is from right to left, meaning the induced current flows in the ZYX direction.
(e) Before releasing the tapping key, the right side of the core is an N-pole, and the magnetic field runs from left to right. When the key is released, the magnetic field linked with the coil starts to decrease. Lenz's law says that the induced current in the nearby coil will flow in a way that helps the magnetic field from decreasing. This happens if the induced magnetic field is from left to right, meaning the induced current flows in the XRY direction.
(f) The current in the straight conductor produces a magnetic field that is parallel to the plane of the coil. Because no magnetic field lines pass through the coil, no induced current will be created in it.
In simple words: Lenz's law helps us find the direction of current made by a changing magnetic field. The induced current always makes a magnetic field that works against the change that caused it. For diagrams (a) to (e), we see how a moving magnet or changing current creates an opposing force, leading to current flow in a specific direction. In diagram (f), since the magnetic field doesn't go through the coil, no current is induced.

🎯 Exam Tip: Understanding Lenz's law is crucial for these types of questions. Always think about how the induced current will create a magnetic field that opposes the initial change in magnetic flux. Clearly state the direction of current based on this principle.

 

Question 2. Use Lenz's law to determine the direction of induced current in the situations described by the following figures (a) to (b).
ℹ️ चित्र व्याख्या (Diagram Explanation):(a) A wire of irregular shape is shown in a region with a magnetic field pointing inwards (marked by 'x'). The wire is deforming to become a circular shape. (b) A circular loop is shown in a region with a magnetic field pointing inwards (marked by 'x'). The loop is deforming to become a narrow straight wire.
Answer:(a) Using the right-hand thumb rule, the current direction will be along adcba. (b) In this case, the current direction is opposite to what is shown in figure (a); it will be along a'd'c'b'a'.
In simple words: When a wire changes its shape in a magnetic field, current is made. For the irregular wire becoming a circle in (a), the current flows in one direction (adcba). For the circle becoming a straight wire in (b), the current flows in the opposite direction (a'd'c'b'a').

🎯 Exam Tip: For problems involving changing loop shapes in a magnetic field, apply Lenz's law by determining whether the magnetic flux through the loop is increasing or decreasing. The induced current's magnetic field will then act to oppose this change. Right-hand thumb rule is key for direction.

 

Question 3. A long solenoid with 15 turns per cm has a small loop area 2.0 cm\(^2\) placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?
Answer:Here, the number of turns per unit length (n') is 15 turns/cm. Convert n' to turns/meter: \(n' = 15 \text{ cm}^{-1} = \frac{15}{10^{-2}} \text{ m}^{-1} = 1500 \text{ m}^{-1}\) The small loop area (A) is 2.0 cm\(^2\). Convert A to m\(^2\): \(A = 2.0 \times 10^{-4} \text{ m}^2\) The current changes from \(I_1 = 2.0 \text{ A}\) to \(I_2 = 4.0 \text{ A}\). The change in current is \(\Delta I = I_2 - I_1 = 4.0 \text{ A} - 2.0 \text{ A} = 2.0 \text{ A}\). The time taken for the change (\(\Delta t\)) is 0.1 s. The magnetic field (B) inside a solenoid is given by \(B = \mu_0 n' I\), where \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \text{ T} \cdot \text{m/A}\)). The change in magnetic field (\(\Delta B\)) is \(\mu_0 n' \Delta I\). The induced electromotive force (\(\mathscr{E}\)) in the loop is given by Faraday's law of induction: \(\mathscr{E} = -N \frac{d\Phi}{dt}\). Since there is only one loop, N=1. The magnetic flux (\(\Phi\)) through the loop is \(B \cdot A\). So, \(\mathscr{E} = A \frac{dB}{dt}\). Given a steady change, we can write \(\mathscr{E} = A \frac{\Delta B}{\Delta t}\). Substituting \(\Delta B = \mu_0 n' \Delta I\): \(\mathscr{E} = A \frac{\mu_0 n' \Delta I}{\Delta t}\) \(\mathscr{E} = (2.0 \times 10^{-4} \text{ m}^2) \times \frac{(4\pi \times 10^{-7} \text{ T} \cdot \text{m/A}) \times (1500 \text{ m}^{-1}) \times (2.0 \text{ A})}{0.1 \text{ s}}\) \(\mathscr{E} = (2.0 \times 10^{-4}) \times \frac{(4\pi \times 10^{-7}) \times 1500 \times 2}{0.1}\) \(\mathscr{E} = (2.0 \times 10^{-4}) \times (4\pi \times 10^{-7}) \times 1500 \times 20\) \(\mathscr{E} = 7.536 \times 10^{-6} \text{ V}\) (approximately \(7.51 \times 10^{-6} \text{ V}\) using \(\pi \approx 3.14\)) Final Answer: The induced emf is \(7.51 \times 10^{-6} \text{ V}\).
In simple words: A wire coil (solenoid) has current flowing through it, which creates a magnetic field. When this current changes, the magnetic field also changes. A small loop inside the solenoid feels this changing magnetic field, and this change creates an electric voltage (emf) in the small loop. We calculate this voltage using the coil's size, how tightly wound the solenoid is, and how fast the current changes.

🎯 Exam Tip: Remember the formula for the magnetic field inside a solenoid (\(B = \mu_0 n' I\)) and Faraday's law of induction (\(\mathscr{E} = A \frac{dB}{dt}\)). Pay close attention to unit conversions (cm to m, cm\(^2\) to m\(^2\)) and ensure all values are in SI units before calculation. Accuracy in calculation is key.

 

Question 4. A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of the uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1cm s\(^{-1}\) in a direction normal to the (i) longer side, (ii) shorter side of the loop? For how long does the induced voltage last in each case?
Answer:The magnetic field (B) is 0.3 T. The dimensions of the loop are length (L) = 8 cm and width (W) = 2 cm. Convert to meters: \(L = 8 \times 10^{-2} \text{ m}\) and \(W = 2 \times 10^{-2} \text{ m}\). The velocity (v) is 1 cm s\(^{-1}\). Convert to m/s: \(v = 1 \times 10^{-2} \text{ m/s}\). (i) **Direction of motion normal to the longer side:** When the loop moves out with its velocity perpendicular to the longer side, the cutting length is the longer side. Length (l) = longer side = \(8 \text{ cm} = 8 \times 10^{-2} \text{ m}\). The induced emf (e) is given by \(e = Blv\). \(e = (0.3 \text{ T}) \times (8 \times 10^{-2} \text{ m}) \times (1 \times 10^{-2} \text{ m/s})\) \(e = 0.24 \times 10^{-3} \text{ V} = 0.24 \text{ mV}\) The emf will last until the entire loop moves out of the magnetic field. This happens when the loop travels a distance equal to its shorter side. Distance to travel = shorter side = \(2 \text{ cm} = 2 \times 10^{-2} \text{ m}\). Time (t) = \(\frac{\text{distance}}{\text{velocity}} = \frac{2 \times 10^{-2} \text{ m}}{1 \times 10^{-2} \text{ m/s}} = 2 \text{ s}\). (ii) **Direction of motion normal to the shorter side:** When the loop moves out with its velocity perpendicular to the shorter side, the cutting length is the shorter side. Length (l) = shorter side = \(2 \text{ cm} = 2 \times 10^{-2} \text{ m}\). The induced emf (e) is given by \(e = Blv\). \(e = (0.3 \text{ T}) \times (2 \times 10^{-2} \text{ m}) \times (1 \times 10^{-2} \text{ m/s})\) \(e = 0.06 \times 10^{-3} \text{ V} = 0.06 \text{ mV}\) The emf will last until the entire loop moves out of the magnetic field. This happens when the loop travels a distance equal to its longer side. Distance to travel = longer side = \(8 \text{ cm} = 8 \times 10^{-2} \text{ m}\). Time (t) = \(\frac{\text{distance}}{\text{velocity}} = \frac{8 \times 10^{-2} \text{ m}}{1 \times 10^{-2} \text{ m/s}} = 8 \text{ s}\).
In simple words: When a wire loop moves out of a magnetic field, a voltage (emf) is created across its cut. The size of this voltage depends on the strength of the magnetic field, the speed of the loop, and the length of the wire segment cutting through the field lines. How long this voltage lasts depends on the loop's other side length and its speed, specifically how long it takes for the entire loop to leave the field.

🎯 Exam Tip: Remember the motional EMF formula \(e = Blv\). Ensure all units are consistent (SI units are preferred). For the duration of the induced voltage, identify which side's length determines the distance the loop needs to travel to completely exit the magnetic field. Clearly differentiate between the "cutting length" (l) and the "distance to travel" for each case.

 

Question 5. A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s\(^{-1}\) about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.
Answer:The length of the metallic rod (L) is 1.0 m. The angular frequency (\(\omega\)) is 400 rad s\(^{-1}\). The magnetic field (B) is 0.5 T. The induced emf (\(\mathscr{E}\)) between the center and the ring for a rotating rod in a uniform magnetic field parallel to the axis of rotation is given by: \(\mathscr{E} = \frac{1}{2} B L^2 \omega\) \(\mathscr{E} = \frac{1}{2} \times (0.5 \text{ T}) \times (1.0 \text{ m})^2 \times (400 \text{ rad s}^{-1})\) \(\mathscr{E} = \frac{1}{2} \times 0.5 \times 1 \times 400\) \(\mathscr{E} = 0.25 \times 400\) \(\mathscr{E} = 100 \text{ V}\) The emf developed between the center and the ring is 100 V.
In simple words: Imagine a metal rod spinning like a clock hand. If there's a magnetic field all around it, this spinning motion creates a voltage between the center where it spins and its outer edge. The faster it spins, the longer the rod, and the stronger the magnetic field, the bigger this voltage will be.

🎯 Exam Tip: For problems involving a rotating rod in a magnetic field, use the formula for motional EMF: \(\mathscr{E} = \frac{1}{2} B L^2 \omega\). Ensure all given quantities (length, magnetic field, angular frequency) are converted to SI units before performing calculations.

 

Question 6. A circular coil of radius 8.0 cm and 20 turns is rotated about an vertical diameter with an angular speed of 50 rad s\(^{-1}\) in a uniform horizontal magnetic field of magnitude 3.0 x 10\(^{-2}\) T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?
Answer:Radius of the coil (r) = 8.0 cm = \(8.0 \times 10^{-2} \text{ m}\). Number of turns (N) = 20. Angular speed (\(\omega\)) = 50 rad s\(^{-1}\). Magnetic field (B) = \(3.0 \times 10^{-2} \text{ T}\). Resistance of the coil (R) = 10 Ω. First, calculate the area (A) of the coil: \(A = \pi r^2 = \pi \times (8.0 \times 10^{-2} \text{ m})^2 = \pi \times 64 \times 10^{-4} \text{ m}^2\) The maximum induced emf (\(\mathscr{E}_{\text{max}}\)) in a rotating coil is given by: \(\mathscr{E}_{\text{max}} = NBA\omega\) \(\mathscr{E}_{\text{max}} = (20) \times (3.0 \times 10^{-2} \text{ T}) \times (\pi \times 64 \times 10^{-4} \text{ m}^2) \times (50 \text{ rad s}^{-1})\) \(\mathscr{E}_{\text{max}} = 20 \times 50 \times \pi \times 64 \times 10^{-4} \times 3 \times 10^{-2}\) \(\mathscr{E}_{\text{max}} = 1000 \times \pi \times 192 \times 10^{-6}\) \(\mathscr{E}_{\text{max}} = 192 \pi \times 10^{-3} \text{ V}\) \(\mathscr{E}_{\text{max}} \approx 192 \times 3.14159 \times 10^{-3} \text{ V}\) \(\mathscr{E}_{\text{max}} \approx 0.603 \text{ V}\) The average emf over one complete cycle for a sinusoidal emf is zero. So, the average emf (\(\mathscr{E}_{\text{average}}\)) = 0 V. Next, calculate the maximum current (\(I_{\text{max}}\)) in the coil: \(I_{\text{max}} = \frac{\mathscr{E}_{\text{max}}}{R}\) \(I_{\text{max}} = \frac{0.603 \text{ V}}{10 \text{ Ω}} = 0.0603 \text{ A}\) Now, calculate the average power loss due to Joule heating. For an AC circuit where current and voltage are sinusoidal, the average power is given by: \(P_{\text{average}} = \frac{1}{2} \mathscr{E}_{\text{max}} I_{\text{max}}\) \(P_{\text{average}} = \frac{1}{2} \times (0.603 \text{ V}) \times (0.0603 \text{ A})\) \(P_{\text{average}} = \frac{1}{2} \times 0.0363609\) \(P_{\text{average}} \approx 0.01818 \text{ W}\) (approximately \(0.018 \text{ W}\)) The source of this power loss is the external rotor, which provides the necessary torque to keep the coil rotating. This torque overcomes the opposing torque from the induced current.
In simple words: A coil spinning in a magnetic field creates a maximum voltage (emf). The average voltage over a full spin is zero. This voltage also causes a maximum current to flow if the coil is a closed loop. The energy lost as heat (Joule heating) in the coil comes from the external force that keeps the coil spinning.

🎯 Exam Tip: Remember that for a rotating coil, the maximum emf is given by \(NBA\omega\), while the average emf over a full cycle is zero. For average power loss due to heating in AC circuits, use \(P_{\text{average}} = \frac{1}{2} \mathscr{E}_{\text{max}} I_{\text{max}}\). Be careful with units and conversions, especially for radius to area calculation. The source of power must be identified as the external agent causing rotation.

 

Question 7. A straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s\(^{-1}\), at right angles to the horizontal component of the earth's magnetic field, 0.30 x 10\(^{-4}\) Wb m\(^{-2}\). (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of the emf? (c) Which end of the wire is at the higher electrical potential?
Answer:Length of the wire (l) = 10 m. Velocity of the wire (V) = 5.0 m s\(^{-1}\). Horizontal component of Earth's magnetic field (B\(_{\text{H}}\)) = \(0.30 \times 10^{-4} \text{ Wb m}^{-2}\) (which is \(0.30 \times 10^{-4} \text{ T}\)). (a) The instantaneous induced emf (e) is given by the motional emf formula: \(e = B_{\text{H}} l V\) \(e = (0.30 \times 10^{-4} \text{ T}) \times (10 \text{ m}) \times (5.0 \text{ m s}^{-1})\) \(e = 1.5 \times 10^{-3} \text{ V}\) (b) The wire is falling vertically downwards, and the horizontal component of the Earth's magnetic field is from South to North. The wire itself extends from East to West. Using Fleming's right-hand rule (or the Lorentz force on charge carriers), the direction of the induced emf will be from the West end to the East end. (c) Since the induced emf is directed from the West end to the East end, it means that positive charges accumulate at the East end and negative charges at the West end. Therefore, the eastern end of the wire will be at a higher electrical potential.
In simple words: When a long wire falls through the Earth's magnetic field, a small voltage (emf) is created across it. This voltage's direction depends on how the wire moves and where the magnetic field points. The end where positive charges gather will have a higher electrical potential.

🎯 Exam Tip: For motional EMF problems, use \(e = Blv\), ensuring B is the component of the magnetic field perpendicular to both the wire's length (l) and its velocity (v). Use Fleming's right-hand rule (for induced current/emf direction) or the Lorentz force (\(\vec{F} = q(\vec{v} \times \vec{B})\)) to determine the polarity and direction of the induced emf.

 

Question 8. Current in a circuit falls from 5.0 A to 0.0 A in 0.1s. If an average emf of 200 V is induced, give an estimate of the self-inductance of the circuit.
Answer:The initial current (\(I_1\)) is 5.0 A. The final current (\(I_2\)) is 0.0 A. The change in current (\(dI\)) = \(I_2 - I_1 = 0.0 \text{ A} - 5.0 \text{ A} = -5.0 \text{ A}\). The time taken for this change (\(dt\)) is 0.1 s. The average induced emf (\(\mathscr{E}\)) is 200 V. The induced emf in a circuit due to self-inductance (L) is given by: \(\mathscr{E} = -L \frac{dI}{dt}\) We are given the magnitude of the average emf, so we use: \(|\mathscr{E}| = L \left| \frac{dI}{dt} \right|\) Rearranging for L: \(L = \frac{|\mathscr{E}|}{\left| \frac{dI}{dt} \right|} = \frac{|\mathscr{E}| \cdot dt}{|dI|}\) \(L = \frac{200 \text{ V} \times 0.1 \text{ s}}{|-5.0 \text{ A}|}\) \(L = \frac{200 \times 0.1}{5.0}\) \(L = \frac{20}{5}\) \(L = 4 \text{ H}\) The self-inductance of the circuit is 4 H.
In simple words: When the current in an electrical circuit changes very quickly, the circuit itself creates a voltage (emf) to oppose this change. The size of this opposing voltage is related to a property called self-inductance. If we know how much the current changes, how fast it changes, and the voltage produced, we can figure out the circuit's self-inductance.

🎯 Exam Tip: Remember the formula for induced emf due to self-inductance: \(\mathscr{E} = -L \frac{dI}{dt}\). The negative sign indicates that the induced emf opposes the change in current (Lenz's law). When calculating self-inductance, use the magnitude of the change in current and emf unless specific directionality is required.

 

Question 9. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
Answer:Mutual inductance (M) = 1.5 H. Initial current (\(I_1\)) = 0 A. Final current (\(I_2\)) = 20 A. Change in current (\(dI\)) = \(I_2 - I_1 = 20 \text{ A} - 0 \text{ A} = 20 \text{ A}\). Time taken for current change (dt) = 0.5 s. The change in magnetic flux linkage (\(d\Phi\)) with the second coil due to a change in current in the first coil is given by the formula: \(d\Phi = M \cdot dI\) \(d\Phi = (1.5 \text{ H}) \times (20 \text{ A})\) \(d\Phi = 30 \text{ Wb}\) The change of flux linkage with the other coil is 30 Wb.
In simple words: When two coils are near each other, a change in current in one coil creates a magnetic field that affects the other coil, causing a change in its magnetic flux. This relationship is measured by mutual inductance. If we know the mutual inductance and how much the current changes, we can calculate the total change in magnetic flux.

🎯 Exam Tip: For mutual inductance problems, the key relationship is \(\Phi = MI\), where \(\Phi\) is the magnetic flux linkage and I is the current. Thus, the change in flux linkage is \(\Delta\Phi = M \Delta I\). Ensure all quantities are in SI units for accurate results.

 

Question 10. A jet plane is traveling towards the west at a speed of 1800 km/h. What is the voltage developed between the ends of the wing having a span of 25 m, if the earth's magnetic field at the location has a magnitude of 5 x 10\(^{-4}\) T and the dip angle is 30°?
Answer:Speed of the jet plane (v) = 1800 km/h. Convert v to m/s: \(v = 1800 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 500 \text{ m/s}\). Length of the wing (l) = 25 m. Magnitude of Earth's magnetic field (B) = \(5 \times 10^{-4} \text{ T}\). Dip angle (\(\delta\)) = 30°. To find the induced emf, we need the vertical component of the Earth's magnetic field (B\(_{\text{v}}\)), which is perpendicular to both the wing's length and the plane's velocity. The vertical component of the Earth's magnetic field is given by: \(B_{\text{v}} = B \sin \delta\) \(B_{\text{v}} = (5 \times 10^{-4} \text{ T}) \times \sin(30^\circ)\) \(B_{\text{v}} = (5 \times 10^{-4} \text{ T}) \times 0.5\) \(B_{\text{v}} = 2.5 \times 10^{-4} \text{ T}\) The induced emf (e) developed across the wing is given by the motional emf formula: \(e = B_{\text{v}} l v\) \(e = (2.5 \times 10^{-4} \text{ T}) \times (25 \text{ m}) \times (500 \text{ m/s})\) \(e = 2.5 \times 25 \times 500 \times 10^{-4}\) \(e = 31250 \times 10^{-4}\) \(e = 3.125 \text{ V}\) The voltage developed between the ends of the wing is 3.125 V.
In simple words: When a plane flies through the Earth's magnetic field, a voltage is created across its wings. This voltage depends on the plane's speed, the wing's length, and the part of Earth's magnetic field that points up or down (vertical component). We use the dip angle to find this vertical part of the magnetic field.

🎯 Exam Tip: For problems involving induced emf in moving conductors in the Earth's magnetic field, remember to use the component of the magnetic field that is perpendicular to both the velocity and the length of the conductor. The vertical component (\(B_{\text{v}} = B \sin \delta\)) is often relevant for horizontal motion. Always convert speed to m/s for consistency.

 

Question 11. Suppose the loop in Exercise.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s\(^{-1}\). If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat? What is the source of this power?
Answer:From Exercise 4, the dimensions of the rectangular loop are 8 cm by 2 cm. Area of the loop (A) = \(8 \text{ cm} \times 2 \text{ cm} = 16 \text{ cm}^2 = 16 \times 10^{-4} \text{ m}^2\). Initial magnetic field = 0.3 T. Rate of decrease of magnetic field (\(\frac{dB}{dt}\)) = \(0.02 \text{ T s}^{-1}\). Resistance of the loop (R) = 1.6 Ω. The induced emf (\(\mathscr{E}\)) in the loop is given by Faraday's law of induction: \(\mathscr{E} = A \left| \frac{dB}{dt} \right|\) (since the loop is stationary and N=1) \(\mathscr{E} = (16 \times 10^{-4} \text{ m}^2) \times (0.02 \text{ T s}^{-1})\) \(\mathscr{E} = 32 \times 10^{-6} \text{ V} = 3.2 \times 10^{-5} \text{ V}\) The induced current (I) in the loop is given by Ohm's law: \(I = \frac{\mathscr{E}}{R}\) \(I = \frac{3.2 \times 10^{-5} \text{ V}}{1.6 \text{ Ω}}\) \(I = 2 \times 10^{-5} \text{ A}\) The power dissipated by the loop as heat (P) is given by Joule's law: \(P = I^2 R\) or \(P = \mathscr{E} I\) Using \(P = \mathscr{E} I\): \(P = (3.2 \times 10^{-5} \text{ V}) \times (2 \times 10^{-5} \text{ A})\) \(P = 6.4 \times 10^{-10} \text{ W}\) The source of this power is the external agency that causes the change in the magnetic field (i.e., the electromagnet whose current is being reduced). This external agency performs work to reduce the magnetic field, and this energy is converted into electrical energy (induced emf) and then dissipated as heat in the loop.
In simple words: When a magnetic field around a stationary wire loop gets weaker, a small voltage (emf) is created in the loop. This voltage makes a current flow, and because the loop has resistance, some energy is lost as heat. The energy for this heat comes from the outside system that is making the magnetic field weaker.

🎯 Exam Tip: When dealing with induced emf due to a changing magnetic field, use \(\mathscr{E} = A \frac{dB}{dt}\). Then apply Ohm's law (\(I = \frac{\mathscr{E}}{R}\)) and the power dissipation formula (\(P = I^2 R\) or \(P = \mathscr{E} I\)). Remember that the power dissipated as heat ultimately comes from the external agent causing the change in flux.

 

Question 12. A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s\(^{-1}\) in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10\(^{-3}\) T cm\(^{-1}\) along the negative x-direction (that is it increases by 10\(^{-3}\) T cm\(^{-1}\) as one move in the negative x-direction), and it is decreasing in time at the rate of 10\(^{-3}\) T s\(^{-1}\). Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 m Ω.
Answer:Side of the square loop (l) = 12 cm = \(0.12 \text{ m}\). Area of the loop (A) = \(l^2 = (0.12 \text{ m})^2 = 0.0144 \text{ m}^2 = 144 \times 10^{-4} \text{ m}^2\). Velocity of the loop (v) = 8 cm s\(^{-1}\) in positive x-direction = \(0.08 \text{ m s}^{-1}\). Magnetic field (B) is in the positive z-direction. Gradient of B along negative x-direction = \(10^{-3} \text{ T cm}^{-1} = 10^{-3} \times 10^2 \text{ T m}^{-1} = 0.1 \text{ T m}^{-1}\). Rate of change of B with time (\(\frac{dB}{dt}\)) = \(10^{-3} \text{ T s}^{-1}\) (decreasing, so \(-10^{-3} \text{ T s}^{-1}\)). Resistance of the loop (R) = 4.50 mΩ = \(4.50 \times 10^{-3} \text{ Ω}\). The total induced emf has two components: one due to the time variation of the magnetic field and another due to the spatial variation of the magnetic field (motional emf). 1. **Emf due to time variation of B (\(\mathscr{E}_{\text{time}}\)):** \(\mathscr{E}_{\text{time}} = -A \frac{dB}{dt}\) \(\mathscr{E}_{\text{time}} = -(144 \times 10^{-4} \text{ m}^2) \times (-10^{-3} \text{ T s}^{-1})\) (since B is decreasing, \(\frac{dB}{dt}\) is negative) \(\mathscr{E}_{\text{time}} = 144 \times 10^{-7} \text{ V} = 1.44 \times 10^{-5} \text{ V}\) By Lenz's law, as B (into z-direction) decreases, the induced current will try to produce B in the positive z-direction. This means current will be clockwise when viewed from positive z-axis (or viewed from top). Let's call this the positive direction for emf. 2. **Emf due to spatial variation of B (motional emf, \(\mathscr{E}_{\text{motional}}\)):** The loop moves in the positive x-direction. The magnetic field increases along the negative x-direction. This means that the side of the loop entering the stronger field experiences a larger force. Let B be the magnetic field at position x. Then \(B(x) = B_0 - kx\) where k is the gradient. Or, more simply, \(dB/dx = -10^{-3} \text{ T cm}^{-1}\). Since the gradient is \(10^{-3} \text{ T cm}^{-1}\) along the negative x-direction, it means B increases as x decreases. When the loop moves in the positive x-direction, it enters regions of *weaker* magnetic field. The rate of change of flux due to motion is \(\frac{d\Phi_{\text{motional}}}{dt} = \frac{d}{dt} (B A) = A \frac{dB}{dx} \frac{dx}{dt} = A (\frac{dB}{dx}) v\). Here, the change in flux due to motion in a non-uniform field is more accurately found by considering the difference in EMF across the leading and trailing edges. However, the provided solution uses a more direct flux rate approach. Let's consider the magnetic field \(B_x = B_0 - \alpha x\), where \(\alpha\) is the gradient (magnitude \(10^{-3} \text{ T cm}^{-1}\)). The change in flux due to motion is \(\Delta\Phi_{\text{motional}} = B_{\text{change}} \times \text{Area}\). The change in B across the width of the loop due to motion: \(\Delta B = (\text{gradient}) \times (\text{distance moved}) = (10^{-3} \text{ T cm}^{-1}) \times (8 \text{ cm s}^{-1} \times 1 \text{ s}) = 8 \times 10^{-3} \text{ T}\) for a 1-second interval. The rate of change of flux due to motion is \(A \times (\text{rate of change of B due to motion})\). Rate of change of B due to motion: \((\text{gradient}) \times v = (10^{-3} \text{ T cm}^{-1}) \times (8 \text{ cm s}^{-1}) = 8 \times 10^{-3} \text{ T s}^{-1}\). \(\mathscr{E}_{\text{motional}} = A \times (\text{gradient}) \times v\) \(\mathscr{E}_{\text{motional}} = (144 \times 10^{-4} \text{ m}^2) \times (0.1 \text{ T m}^{-1}) \times (0.08 \text{ m s}^{-1})\) \(\mathscr{E}_{\text{motional}} = 144 \times 0.1 \times 0.08 \times 10^{-4} \text{ V}\) \(\mathscr{E}_{\text{motional}} = 1.152 \times 10^{-5} \text{ V}\) Since the loop moves in the positive x-direction and the field increases in the negative x-direction, the field is *decreasing* in the direction of motion. So, flux is decreasing. The induced current will try to produce B in the positive z-direction (clockwise). This means the motional emf also causes a clockwise current (positive direction). 3. **Net induced emf (\(\mathscr{E}_{\text{net}}\)):** Both effects contribute to current in the same direction (clockwise, maintaining the flux in +z direction). \(\mathscr{E}_{\text{net}} = \mathscr{E}_{\text{time}} + \mathscr{E}_{\text{motional}}\) \(\mathscr{E}_{\text{net}} = (1.44 \times 10^{-5} \text{ V}) + (1.152 \times 10^{-5} \text{ V})\) \(\mathscr{E}_{\text{net}} = 12.96 \times 10^{-5} \text{ V}\) 4. **Magnitude of induced current (I):** \(I = \frac{\mathscr{E}_{\text{net}}}{R}\) \(I = \frac{12.96 \times 10^{-5} \text{ V}}{4.50 \times 10^{-3} \text{ Ω}}\) \(I = 2.88 \times 10^{-2} \text{ A}\) 5. **Direction of induced current:** As explained above, both effects (decreasing field with time and decreasing field with motion) lead to an induced current that tries to maintain the magnetic flux in the positive z-direction. This means the induced current will flow in a clockwise direction when viewed from the positive z-axis. The problem states "increases the magnetic flux linked with the loop in the positive direction." This confirms the clockwise current (to counteract the decrease in flux).
In simple words: A square wire loop moves in a magnetic field that is changing both in space and over time. Both changes work together to create an electric voltage (emf) in the loop. This voltage then drives a current. The direction of this current is such that it tries to keep the magnetic field strong in the direction it was already pointing, to fight against the changes. We calculate the total voltage from both effects and then use the loop's resistance to find the current.

🎯 Exam Tip: When the magnetic field changes both in time and space, consider two components of induced emf: \(\mathscr{E}_{\text{time}} = -A \frac{dB}{dt}\) and \(\mathscr{E}_{\text{motional}} = A \frac{dB}{dx} v\). Determine the direction of each induced emf using Lenz's law. If they are in the same direction, add them; if opposite, subtract. Finally, use Ohm's law to find the current. Pay attention to signs and unit conversions (cm to m).

 

Question 13. It is desired to measure the magnitude of the field between the poles of a powerful loudspeaker magnet. A small flat search coil of area 2 cm\(^2\) with 25 closely wound turns, is positioned normal to the field direction and then quickly snatched out of the field region. Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction. The total charge flew in the coil (measured by a ballistic galvanometer connected to the coil) is 7.5mC. The combined resistance of the coil and the galvanometer is 0.50 Ω. Estimate the field strength of the magnet.
Answer:Area of the search coil (A) = 2 cm\(^2\) = \(2 \times 10^{-4} \text{ m}^2\). Number of turns (N) = 25. Total charge flown (Q) = 7.5 mC = \(7.5 \times 10^{-3} \text{ C}\). Combined resistance (R) = 0.50 Ω. When the search coil is positioned normal to the field, the initial magnetic flux (\(\Phi_i\)) through it is maximum: \(\Phi_i = NBA\) When the coil is quickly snatched out or turned 90° to be parallel to the field, the final magnetic flux (\(\Phi_f\)) is zero. \(\Phi_f = 0\) The change in magnetic flux (\(\Delta\Phi\)) = \(\Phi_i - \Phi_f = NBA - 0 = NBA\). The relationship between induced charge (Q), change in magnetic flux (\(\Delta\Phi\)), number of turns (N), and resistance (R) is: \(Q = \frac{N \Delta\Phi}{R}\) Substituting \(\Delta\Phi = NBA\): \(Q = \frac{N (NBA)}{R} = \frac{N^2 BA}{R}\) (Correction: The standard formula for induced charge is \(Q = \frac{N \Delta\Phi}{R}\), and \(\Delta\Phi\) is the change in flux through *one* turn. So, if the initial flux *through the coil* (N turns) is \(NBA\), and final is 0, then \(\Delta\Phi_{\text{coil}} = NBA\). The induced emf is \(\mathscr{E} = - \frac{d\Phi_{\text{coil}}}{dt} = -N A \frac{dB}{dt}\). The charge Q is \(\int I dt = \int \frac{\mathscr{E}}{R} dt = \frac{1}{R} \int \mathscr{E} dt = \frac{1}{R} \int (-N A \frac{dB}{dt}) dt = -\frac{NA}{R} \Delta B\). This formula is for a single turn in terms of change in B. For N turns, \(\Delta\Phi = N \times (B \times A)\), so the total flux linkage change is \(N(BA)\). Therefore, \(Q = \frac{N (BA)}{R}\)) From \(Q = \frac{N \Delta\Phi}{R}\), we have \(\Delta\Phi = \frac{QR}{N}\). Since \(\Delta\Phi = BA\), we get \(BA = \frac{QR}{N}\). Therefore, the magnetic field strength (B) is: \(B = \frac{QR}{NA}\) \(B = \frac{(7.5 \times 10^{-3} \text{ C}) \times (0.50 \text{ Ω})}{(25) \times (2 \times 10^{-4} \text{ m}^2)}\) \(B = \frac{3.75 \times 10^{-3}}{50 \times 10^{-4}}\) \(B = \frac{3.75 \times 10^{-3}}{5 \times 10^{-3}}\) \(B = \frac{3.75}{5} = 0.75 \text{ T}\) The field strength of the magnet is 0.75 T.
In simple words: We can find the strength of a magnet's field by using a small coil. If we quickly move the coil out of the magnetic field, a tiny burst of electricity (charge) flows through it. By measuring this charge, along with the coil's size, number of turns, and its electrical resistance, we can calculate how strong the magnetic field was.

🎯 Exam Tip: The key formula for ballistic galvanometers is \(Q = \frac{N \Delta\Phi}{R}\), where \(\Delta\Phi\) is the change in magnetic flux through *one* turn of the coil. If the coil starts normal to the field and is removed, \(\Delta\Phi = BA\). Thus, \(B = \frac{QR}{NA}\). Ensure all units are in SI before calculation.

 

Question 14. An air-cored solenoid with a length of 30 cm, area of cross-section 25 cm\(^2\) and a number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10\(^{-3}\) s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in the magnetic field near the ends of the solenoid.
Answer:Length of the solenoid (l) = 30 cm = \(0.30 \text{ m}\). Area of cross-section (A) = 25 cm\(^2\) = \(25 \times 10^{-4} \text{ m}^2\). Number of turns (N) = 500. Initial current (\(I_1\)) = 2.5 A. Final current (\(I_2\)) = 0 A (current switched off). Change in current (\(\Delta I\)) = \(I_2 - I_1 = 0 - 2.5 \text{ A} = -2.5 \text{ A}\). Time taken (\(\Delta t\)) = \(10^{-3} \text{ s}\). Permeability of free space (\(\mu_0\)) = \(4\pi \times 10^{-7} \text{ T} \cdot \text{m/A}\). First, calculate the initial magnetic field (B) inside the solenoid: \(B = \mu_0 n I_1\), where \(n = \frac{N}{l}\) is the number of turns per unit length. \(n = \frac{500}{0.30 \text{ m}}\) \(B = (4\pi \times 10^{-7}) \times \left(\frac{500}{0.30}\right) \times 2.5 \text{ T}\) \(B = \frac{4\pi \times 10^{-7} \times 500 \times 2.5}{0.30}\) \(B = 5235.98 \times 10^{-7} \text{ T} \approx 5.236 \times 10^{-4} \text{ T}\) Now, calculate the initial magnetic flux linkage (\(\Phi_1\)) through the solenoid: \(\Phi_1 = NBA\) \(\Phi_1 = (500) \times (5.236 \times 10^{-4} \text{ T}) \times (25 \times 10^{-4} \text{ m}^2)\) \(\Phi_1 = 500 \times 5.236 \times 25 \times 10^{-8}\) \(\Phi_1 = 65450 \times 10^{-8} \text{ Wb} = 6.545 \times 10^{-4} \text{ Wb}\) Since the current is switched off, the final magnetic flux linkage (\(\Phi_2\)) is 0. The change in magnetic flux linkage (\(\Delta\Phi\)) = \(\Phi_2 - \Phi_1 = 0 - 6.545 \times 10^{-4} \text{ Wb} = -6.545 \times 10^{-4} \text{ Wb}\). The average induced emf (\(\mathscr{E}\)) is given by: \(\mathscr{E} = -\frac{\Delta\Phi}{\Delta t}\) \(\mathscr{E} = -\frac{(-6.545 \times 10^{-4} \text{ Wb})}{10^{-3} \text{ s}}\) \(\mathscr{E} = \frac{6.545 \times 10^{-4}}{10^{-3}}\) \(\mathscr{E} = 6.545 \times 10^{-1} \text{ V} = 0.6545 \text{ V}\) Alternatively, using the self-inductance (L) of the solenoid: \(L = \frac{\mu_0 N^2 A}{l}\) \(L = \frac{(4\pi \times 10^{-7}) \times (500)^2 \times (25 \times 10^{-4})}{0.30}\) \(L = \frac{4\pi \times 10^{-7} \times 250000 \times 25 \times 10^{-4}}{0.30}\) \(L = \frac{4\pi \times 25 \times 25}{0.30} \times 10^{-7} \times 10^{-4} \times 10^4\) \(L = \frac{7853.98}{0.30} \times 10^{-7} \approx 2.618 \times 10^{-2} \text{ H}\) Average induced emf (\(\mathscr{E}\)) = \(-L \frac{\Delta I}{\Delta t}\) \(\mathscr{E} = -(2.618 \times 10^{-2} \text{ H}) \times \frac{(-2.5 \text{ A})}{10^{-3} \text{ s}}\) \(\mathscr{E} = (2.618 \times 10^{-2}) \times 2500\) \(\mathscr{E} = 65.45 \text{ V}\) (There seems to be a significant difference in results. Let's re-check the original solution's method.) The provided solution calculates \(\Phi_B = BAN\) first. \(\Phi_B = (\mu_0 \frac{N}{l} I) A N = \mu_0 \frac{N^2}{l} A I\) Initial flux linkage \(\Phi_1 = \frac{4\pi \times 10^{-7} \times (500)^2 \times (25 \times 10^{-4}) \times 2.5}{0.30} = 6.545 \times 10^{-4} \text{ Wb}\). Final flux linkage \(\Phi_2 = 0\). Change in flux linkage \(\Delta\Phi = \Phi_2 - \Phi_1 = -6.545 \times 10^{-4} \text{ Wb}\). Induced emf \(\mathscr{E} = -\frac{\Delta\Phi}{\Delta t} = - \frac{-6.545 \times 10^{-4}}{10^{-3}} = 6.545 \text{ V}\). My calculation based on flux change matches the second part of the solution when using \(\mu_0 \approx 4\pi \times 10^{-7}\). The first method where I used B first then \(\Phi_1 = NBA\) also yields the same. The calculation using L seems to be off. Let's recheck the L formula. The formula for self-inductance L of a solenoid is \(L = \frac{\mu_0 N^2 A}{l}\). \(L = \frac{(4 \times 3.14159 \times 10^{-7}) \times (500)^2 \times (25 \times 10^{-4})}{0.30}\) \(L = \frac{(12.566 \times 10^{-7}) \times (250000) \times (0.0025)}{0.30}\) \(L = \frac{12.566 \times 250000 \times 0.0025 \times 10^{-7}}{0.30} = \frac{7853.75 \times 10^{-7}}{0.30}\) \(L = 26179.16 \times 10^{-7} = 2.6179 \times 10^{-3} \text{ H}\) Now, using \(L = 2.6179 \times 10^{-3} \text{ H}\) \(\mathscr{E} = -L \frac{\Delta I}{\Delta t} = -(2.6179 \times 10^{-3} \text{ H}) \times \frac{(-2.5 \text{ A})}{10^{-3} \text{ s}}\) \(\mathscr{E} = (2.6179 \times 10^{-3}) \times 2500\) \(\mathscr{E} = 6.54475 \text{ V}\) Both methods give approximately 6.545 V. The error was in my L calculation initially. Final Answer: The average back emf induced is approximately 6.545 V.
In simple words: When the current in a solenoid (a coil of wire) is suddenly turned off, the magnetic field inside it quickly disappears. This rapid change in the magnetic field creates a voltage (back emf) across the ends of the solenoid. We calculate this voltage by knowing how many turns the coil has, its size, and how fast the current (and thus the magnetic field) changes.

🎯 Exam Tip: You can calculate induced emf using two primary methods for a solenoid: 1) direct flux change (\(\mathscr{E} = -\frac{\Delta\Phi}{\Delta t}\) where \(\Phi = NBA\)) or 2) self-inductance (\(\mathscr{E} = -L \frac{\Delta I}{\Delta t}\) where \(L = \frac{\mu_0 N^2 A}{l}\)). Both methods should yield consistent results if calculations are precise. Pay careful attention to the negative sign, which indicates the opposing nature of the induced emf (back emf).

 

GSEB Class 12 Physics Electromagnetic Induction Additional Important Questions and Answers

 

Question 1. A solenoid with a large number of turns is in a closed circuit (See Fig.E.3) and a short bar magnet is dropped through each with its length along the axis. State the acceleration of the falling magnet when it is: (a) Well above A (b) At the end A (c) At the middle (d) At the end B (e) Far away, down, from B
ℹ️ चित्र व्याख्या (Diagram Explanation):The diagram shows a solenoid (a coil of wire) with many turns, connected in a closed circuit (implied by the galvanometer G). A bar magnet is positioned above the solenoid, with its North pole (N) facing downwards. The solenoid has two ends labeled A and B. A galvanometer (G) is connected to the solenoid, indicating any induced current. The magnet is shown falling through the solenoid along its central axis.
Answer:(a) **Well above A:** When the magnet is far above the solenoid (well above point A), it is not yet interacting with the coil's magnetic field. Thus, there is no induced current. The magnet's acceleration is simply due to gravity. Therefore, its acceleration is equal to 'g'.
(b) **At the end A:** As the magnet approaches and enters the solenoid at end A, the magnetic flux through the coil changes. According to Lenz's law, the induced current will create a magnetic field that opposes the magnet's motion. This means the solenoid acts as a North pole to repel the approaching North pole of the magnet. This repulsive force reduces the net downward force on the magnet. So, the magnet's acceleration will be less than 'g'.
(c) **At the middle:** When the magnet is exactly in the middle of the solenoid, the rate of change of magnetic flux is maximum but momentarily zero if its velocity is constant (or the direction of flux change reverses). However, considering the continuous change as it passes through, the induced current is still opposing the motion, but the effect on acceleration depends on whether the flux is increasing or decreasing. Generally, the acceleration would still be less than 'g' due to the ongoing induced current opposing motion. For a very symmetrical passage, the acceleration might momentarily approach 'g' if the rate of change of flux is minimal at the exact center, but this is a complex dynamic. Often, the opposing force is present throughout the passage.
(d) **At the end B:** As the magnet leaves the solenoid at end B, the magnetic flux through the coil changes again. Now, the North pole of the magnet is moving away. According to Lenz's law, the induced current will create a magnetic field that attracts the magnet, trying to resist its departure. This means the solenoid acts as a South pole to attract the receding North pole. This attractive force again opposes the magnet's motion. So, the magnet's acceleration will be less than 'g'.
(e) **Far away, down, from B:** When the magnet has completely passed through the solenoid and is far below end B, it is no longer interacting with the coil. There is no induced current, and no magnetic force acts on it from the solenoid. Therefore, its acceleration is again equal to 'g'.
In simple words: When a magnet falls through a wire coil, the coil tries to slow it down. If the magnet is far from the coil, it falls at normal gravity (g). As it enters or leaves the coil, the coil creates an opposing push or pull, making the magnet fall slower than 'g'. Once it's completely past the coil and far away, it falls at normal gravity again.

🎯 Exam Tip: The core concept here is Lenz's law. Remember that the induced current always opposes the *change* in magnetic flux. When the magnet approaches or leaves the coil, there's a change in flux, leading to an opposing force and thus an acceleration less than 'g'. When there's no interaction (far away), acceleration is 'g'.

 

Question 2. A rectangular closed loop moves horizontally in a uniform magnetic field. (i) Will there be any induced current in the loop if the loop is completely in the magnetic field? (ii) Will there be any induced current in the loop if the loop is partially out of the magnetic field? (iii) Will the induced current in the loop remain if the loop is stationary and the magnetic field changes with time?
Answer:(i) **No.** If the loop is completely inside a uniform magnetic field and moving horizontally, the magnetic flux through the loop does not change. According to Faraday's law of electromagnetic induction, induced emf (and thus induced current) is produced only when there is a change in magnetic flux. Since there is no change in flux, no induced current will be generated. (ii) **Yes.** If the loop is partially moving out of the magnetic field, the area of the loop exposed to the magnetic field changes. This change in area causes a change in the magnetic flux passing through the loop. Consequently, an induced emf and an induced current will be produced in the loop, according to Faraday's law. (iii) **Yes.** If the loop is stationary but the magnetic field strength itself changes with time (e.g., increases or decreases), then the magnetic flux through the stationary loop will change. A time-varying magnetic flux will induce an emf and a corresponding current in the loop, as per Faraday's law of induction.
In simple words: Current is only made in a wire loop if the magnetic field passing through it changes. If the loop is fully inside a steady magnetic field and just moves around, no current is made because the field isn't changing inside the loop. But if the loop enters or leaves the field, or if the magnetic field itself gets stronger or weaker over time, then current will flow.

🎯 Exam Tip: The fundamental principle is Faraday's Law: induced emf (and current) exists *only* when there is a change in magnetic flux (\(\Delta\Phi\)) through the coil. A uniform magnetic field means B is constant. For \(\Phi = BA\), flux changes if A changes (loop entering/leaving field) or if B changes (time-varying field).

 

Question 3. (a) How is the flux linkage through a coil related to current? (b) What is the constant of proportionality that appears in the above relation? (c) Define the constant of proportionality. (d) Give its dimensions and S. I. unit.
Answer:(a) For a current-carrying coil, the magnetic flux linked with the coil is directly proportional to the current flowing through it. This relationship is often written as \(\Phi_B \propto I\) or \(\Phi_B = LI\), where \(\Phi_B\) is the magnetic flux linkage and I is the current. (b) The constant of proportionality that appears in this relation is called the **Coefficient of Self-Induction**, denoted by 'L'. (c) **Self-inductance (L)** of a coil is defined as the magnetic flux linked with the coil when a unit current (1 Ampere) flows through it. In simpler terms, it measures how much magnetic flux is produced per unit current by the coil itself. (d) The dimensions of Self-inductance (L) are \([ML^2T^{-2}A^{-2}]\). The S.I. unit of self-inductance is **Henry (H)**.
In simple words: The amount of magnetic field lines (flux) that pass through a coil depends on how much current is flowing in that coil. The connection between them is called self-inductance (L). This 'L' tells us how good a coil is at creating its own magnetic field when current passes through it. Its unit is the Henry.

🎯 Exam Tip: Understand the direct proportionality: \(\Phi_B = LI\). The definition of self-inductance as flux linkage per unit current is essential. Remember its SI unit (Henry) and its dimensional formula, which is often asked in exams.

 

Question 4. The induced emf is sometimes called back emf Why?
Answer:The induced emf (electromotive force) is called "back emf" because, according to Lenz's law, its direction is always such that it opposes the cause that produced it. If the cause is a change in current in a circuit, the induced emf acts to oppose this change in current (or the applied voltage causing the current). This opposing nature, which tries to "push back" against the change, is why it's referred to as back emf.
In simple words: Induced voltage is called "back emf" because it always works against the change that created it. If current tries to increase, the back emf tries to stop it. If current tries to decrease, it tries to keep it flowing. It's like an electrical "push back."

🎯 Exam Tip: The key reason for the term "back emf" is Lenz's Law. Clearly state that the induced emf *opposes the cause* of its generation. This opposition to the change in magnetic flux (or current) is what "back" refers to.

 

Question 5. Why is spark produced in the switch of a fan when it is switched off?
Answer:When a fan (which contains an inductor, its motor winding) is switched off, the current flowing through its coil suddenly tries to drop to zero. Due to the property of self-inductance of the motor's coil, a large self-induced emf (back emf) is generated. This large induced emf tries to maintain the current flow. If the current path is suddenly broken by the switch opening, this high voltage can ionize the air in the small gap between the switch contacts, leading to an electrical spark.
In simple words: A fan motor acts like a coil (inductor). When you turn it off, the electric current stops suddenly. This sudden stop creates a very large opposing voltage (back emf) in the coil. This strong voltage is enough to make a spark jump across the opening switch gap in the air.

🎯 Exam Tip: Connect the fan's motor to the concept of an inductor. The sudden *break* in current flow (high \(\frac{dI}{dt}\)) across an inductor (L) generates a large induced emf (\(\mathscr{E} = -L \frac{dI}{dt}\)). This large emf is responsible for ionizing air and causing the spark.

 

Question 6. Fig. E. I shows magnet coil experiment of electromagnetic induction. What happens when (a) i. The number of turns of the coil is increased. ii. The strength of the magnet is increased. iii. The speed of motion of the magnet is increased. (b) An induced e.m.f. has no direction of its own. Why?
ℹ️ चित्र व्याख्या (Diagram Explanation):The diagram displays a simple experiment for electromagnetic induction. A bar magnet, labeled with North (N) and South (S) poles, is shown moving towards a coil of wire. The coil is connected to a galvanometer, which has a needle indicating induced current. The arrow labeled "Magnet moves up towards coil" indicates the relative motion. Magnetic field lines are shown emanating from the North pole of the magnet and passing through the coil.
Answer:(a) The induced electromotive force (emf) and current are directly proportional to the rate of change of magnetic flux. i. **If the number of turns of the coil is increased:** Increasing the number of turns (N) in the coil means that for the same change in magnetic field, the total flux linkage (\(\Phi = NBA\)) changes more significantly. Therefore, a larger induced emf and current will be generated. So, the induced emf increases. ii. **If the strength of the magnet is increased:** A stronger magnet produces a more intense magnetic field (higher B). When this stronger magnet moves, it causes a greater change in magnetic flux (\(\Delta\Phi\)) through the coil. Consequently, a larger induced emf and current will be generated. So, the induced emf increases. iii. **If the speed of motion of the magnet is increased:** Increasing the speed of the magnet means that the magnetic flux through the coil changes more rapidly (\(\frac{d\Phi}{dt}\)). A faster rate of change of magnetic flux leads to a larger induced emf and current. So, the induced emf increases. (b) The statement "An induced e.m.f. has no direction of its own" is misleading. In fact, an induced emf *always* has a definite direction, which is determined by Lenz's law. Lenz's law states that the direction of the induced emf (and current) is such that it opposes the change in magnetic flux that produced it. This means the direction is not arbitrary; it's always *against* the cause. For example, if a magnet moves towards a coil, the induced emf will try to repel it. If the magnet moves away, the induced emf will try to attract it. The direction is a direct consequence of the physical interaction opposing the change. If the direction of the magnet's motion reverses, the direction of the induced emf also reverses to continue opposing the new direction of change.
In simple words: (a) If you make a coil stronger (more turns), use a stronger magnet, or move the magnet faster, you'll get more induced voltage (emf). (b) Induced voltage always has a clear direction. It's not random; it always acts to stop whatever change caused it in the first place, like pushing against a magnet trying to move closer or pulling on one trying to move away.

🎯 Exam Tip: For part (a), remember that induced emf is proportional to N, B, and the rate of change of flux. For part (b), emphasize Lenz's Law: induced emf always has a direction that *opposes* the change in magnetic flux. Never state it has no direction of its own; it's precisely defined by the opposition.

 

Question 7. Take a strong cylindrical electromagnet connected to the AC source and place a light metallic disc at the top of it. When the current is switched 'on the disc is thrown up in the air. (a) Why does the disc go up? (b) How is the repulsive force produced?
Answer:(a) The metallic disc is thrown up because when the AC current is switched "on" in the electromagnet, it causes a rapidly changing magnetic field. This changing field induces eddy currents in the metallic disc. According to Lenz's law, these eddy currents will flow in a direction that creates a magnetic field opposing the change in flux produced by the electromagnet. This opposition manifests as a strong repulsive force between the electromagnet and the disc, causing the disc to be propelled upwards. (b) The repulsive force is produced due to the interaction of the magnetic fields. When the current is switched on in the electromagnet, the magnetic flux through the disc increases very quickly. By Lenz's law, the induced eddy currents in the disc generate their own magnetic field. This induced magnetic field is oriented in such a way that it opposes the electromagnet's increasing field. If the electromagnet's top surface becomes an N-pole as the current grows, the induced eddy currents in the disc will make its lower surface an N-pole too. Since like poles repel each other, a strong repulsive force acts on the disc, throwing it upwards.
In simple words: (a) A metal disc jumps up when an electromagnet is turned on because the changing magnetic field makes electric currents (eddy currents) in the disc. These currents create their own magnetic field that pushes against the electromagnet. (b) This push happens because the disc's magnetic field tries to stop the electromagnet's field from changing. If the electromagnet makes a "North" pole, the disc makes its own "North" pole to push away, causing repulsion.

🎯 Exam Tip: Focus on Lenz's Law and eddy currents. The key is that the induced magnetic field *opposes* the change in the electromagnet's field. This opposition leads to a repulsive force. Clearly explain the role of eddy currents and how their magnetic field direction generates the repulsion.

 

Question 8. A short bar magnet is dropped through a coil of wire of similar length (Fig.E. 2). Which one of the graphs below shows best how the current through the coil varies with time?
ℹ️ चित्र व्याख्या (Diagram Explanation):The diagram shows a coil of wire with a short bar magnet dropping through it. The magnet is falling from above the coil, through its center, and out below. Below this setup are four graphs (A), (B), (C), (D) showing current (P, y-axis) versus time (t, x-axis). Graph (A) shows a constant positive current pulse when entering and then zero. Graph (B) shows a sharp positive current spike when entering and then zero. Graph (C) shows a positive current peak when entering and a negative current peak when leaving, with both peaks having similar magnitudes but the second one being wider. Graph (D) shows a positive current peak when entering and a negative current peak when leaving, with the second peak having a larger magnitude and shorter duration, and the overall shape being somewhat distorted. (The OCR interpretation might be slightly different; focusing on common patterns for such problems). (Correction: based on typical physics problems, the key features are flux change rate (emf) and velocity change. As the magnet falls, its speed increases due to gravity, so the rate of change of flux (and thus induced emf/current) will be larger when it exits the coil than when it enters.)
Answer:**Answer: D.** **Explanation:** When the magnet enters the coil: The magnetic flux through the coil changes, inducing an emf and a current. According to Lenz's law, this current will oppose the entry of the magnet. When the magnet is inside the coil: As the magnet passes through the center, the rate of change of flux decreases, and then as it approaches the exit, the rate of change of flux increases again. When the magnet leaves the coil: The magnetic flux through the coil changes in the opposite direction, inducing an emf and current in the opposite direction to oppose its exit. Key points for the graph: 1. **Two peaks:** One positive and one negative, corresponding to entry and exit. 2. **Opposite directions:** The induced current will be in opposite directions for entry and exit. 3. **Increasing velocity:** As the magnet falls under gravity, its speed increases. Therefore, the rate of change of magnetic flux (and thus the induced emf and current) will be greater when it exits the coil than when it enters. This means the second peak (exit) should be taller than the first peak (entry). 4. **Symmetry/Shape:** The peaks are typically narrow and somewhat asymmetric, reflecting the varying speed and non-uniform rate of flux change. Graph (D) best represents these characteristics: two peaks in opposite directions, with the second peak (corresponding to the magnet leaving the coil) being taller than the first (corresponding to the magnet entering), reflecting the increased speed due to gravity.
In simple words: When a magnet falls through a coil, it creates two bursts of electricity (current). One burst happens when it enters, and another, in the opposite direction, happens when it leaves. Because the magnet speeds up as it falls, the second burst of current (when leaving) will be stronger and quicker than the first burst (when entering). Graph D shows this best.

🎯 Exam Tip: For magnet-dropping-through-coil problems, expect two current peaks (one positive, one negative) representing entry and exit. Crucially, the peak corresponding to *exit* will be larger in magnitude than the entry peak because the magnet accelerates due to gravity, leading to a faster rate of flux change on exit.

 

Question 9. When a metallic block moves in a magnetic field induced currents are developed in the body of the block. (a) What is this type of current called? (b) Which law gives the direction of such currents? (c) What is the strength of the induced current?
Answer:(a) This type of induced current developed in the bulk of a metallic block when it moves in a magnetic field is called **Eddy currents**. They are also known as Foucault currents. (b) The direction of these eddy currents is given by **Lenz's Law**. Lenz's law dictates that the eddy currents will flow in such a way that they create a magnetic field that opposes the motion or the change in magnetic flux that caused them. (c) The strength of the induced eddy current can be very large. Since metallic blocks typically have very low electrical resistance, even a small induced emf can lead to a significant current (I = V/R). This large current causes substantial energy dissipation as heat and can create a strong opposing force, which is why eddy currents are used in applications like magnetic braking.
In simple words: (a) When a metal block moves in a magnetic field, swirling electric currents called eddy currents form inside it. (b) Lenz's Law tells us that these currents flow in a direction that tries to stop the block's movement. (c) Because metals don't resist electricity much, these currents can be very strong, making a powerful opposing force and a lot of heat.

🎯 Exam Tip: Remember the specific name "Eddy currents" for currents induced within the bulk of a conductor. Associate their direction with Lenz's Law (opposing the cause) and their large strength with the low resistance of metals, leading to significant braking effects and heat generation.

 

Question 10. (a) Which law is demonstrated in the above figures? (b) State the law. (c) Explain the action that takes place in the above figures.
ℹ️ चित्र व्याख्या (Diagram Explanation):Figure E.4a shows a solenoid connected to a galvanometer. A bar magnet with its North pole (N) facing the coil is moving towards the coil. The galvanometer shows a deflection, indicating induced current. Figure E.4b shows the same solenoid and galvanometer setup. The bar magnet, with its North pole (N) facing the coil, is now moving *away* from the coil. The galvanometer shows a deflection in the opposite direction compared to E.4a, indicating induced current in the reverse direction.
Answer:(a) The law demonstrated in the above figures is **Lenz's Law**. (b) **Lenz's Law** states that the direction of the induced electromotive force (emf) or induced current is always such that it opposes the change in magnetic flux that produced it. This opposition refers to the cause of the induction. (c) **Explanation of the action in the figures:** In Figure E.4a: When the North pole of the magnet is moved *towards* the coil, the magnetic flux passing through the coil in a specific direction (let's say, to the right) increases. According to Lenz's law, an induced current flows in the coil to create a magnetic field that opposes this increase. This means the end of the coil facing the approaching North pole becomes a North pole itself, trying to repel the magnet. To create a North pole at that end, the induced current in the coil must flow in an anticlockwise direction (when viewed from the approaching magnet). In Figure E.4b: When the North pole of the magnet is moved *away* from the coil, the magnetic flux passing through the coil (to the right) decreases. According to Lenz's law, an induced current flows in the coil to create a magnetic field that opposes this decrease. This means the end of the coil facing the receding North pole becomes a South pole, trying to attract the magnet and prevent its departure. To create a South pole at that end, the induced current in the coil must flow in a clockwise direction (when viewed from the receding magnet), which is opposite to the direction in Figure E.4a.
In simple words: (a) These pictures show Lenz's Law in action. (b) This law says that any electricity (current or emf) made by a changing magnetic field will always try to stop that change. (c) So, when a magnet moves closer to a coil, the coil creates a magnetic push to stop it. When the magnet moves away, the coil creates a magnetic pull to stop it from leaving. The current changes direction depending on whether the magnet is coming or going.

🎯 Exam Tip: Clearly state Lenz's Law. For the explanation, use the "opposition" principle: identify the change in flux (increasing/decreasing, direction) and then determine how the induced current's magnetic field would act to oppose that change (repel approaching, attract receding). Always specify the direction of induced current based on this opposition.

 

Question 11. (a) Explain the cause of the spark. (b) Explain why a spark is also produced when the key is opened. (c) Why are no sparks produced when the key is left closed?
ℹ️ चित्र व्याख्या (Diagram Explanation):The diagram shows a primary coil wrapped around an iron core, connected to a battery ("Cell") and a "Key" (switch). A secondary coil is also wrapped around the same core, but its ends are connected to a "Spark gap" (two closely spaced metal rods). This setup demonstrates a basic induction coil or transformer. The "Spark gap" allows for observation of high induced voltage.
Answer:(a) **Cause of the spark:** The spark is produced in the secondary coil's spark gap due to a large induced electromotive force (emf) when the magnetic flux linked with the secondary coil changes rapidly. This happens because the current in the primary coil (and thus its magnetic field) is either switched on or off, causing a significant change in the magnetic flux through the core, which then links with the secondary coil. (b) **Spark when the key is opened:** When the key in the primary circuit is opened, the current flowing through the primary coil suddenly drops to zero. This rapid decrease in current causes a very quick collapse of the magnetic field produced by the primary coil. This sudden and large change in magnetic flux is linked with the secondary coil, inducing a very high emf in it. This high induced emf is enough to ionize the air in the spark gap, leading to an electrical spark. This is an example of mutual induction. (c) **No sparks when the key is left closed:** When the key in the primary circuit is left closed, a steady (constant) current flows through the primary coil. A constant current produces a constant magnetic field, which in turn means the magnetic flux linked with the secondary coil is also constant and not changing. According to Faraday's law of induction, an emf is induced only when there is a *change* in magnetic flux. Since there is no change in flux when the key is closed and current is steady, no emf is induced in the secondary coil, and therefore no spark is produced.
In simple words: (a) A spark happens when a very high voltage (induced emf) is created in the secondary coil because the magnetic field from the primary coil changes very quickly. (b) When you open the switch, the primary current suddenly stops, and its magnetic field collapses fast. This quick change creates a huge voltage in the secondary coil, causing a spark. (c) When the switch is closed, the current in the primary coil is steady, so the magnetic field doesn't change. No changing field means no induced voltage in the secondary coil, and thus no spark.

🎯 Exam Tip: The core idea is that an emf is induced *only* when magnetic flux *changes*. A sudden interruption (opening the key) causes a rapid flux collapse, leading to a large induced emf. A steady state (key closed) means constant flux, hence no induced emf. This illustrates both Faraday's law and the role of high \(\frac{d\Phi}{dt}\).

 

Question 12. You are given a solenoid of length 'l', number of turns per unit length 'n', and area of cross-section A. When a current 'I' passes through it, (a) What is the total magnetic flux through the solenoid? (b) What is the self-inductance of the solenoid? (c) How does the value of self-inductance get affected if some material of high relative permeability is filled inside the
Answer:Given: Length of solenoid = l, Number of turns per unit length = n, Area of cross-section = A, Current = I. (a) **Total magnetic flux through the solenoid (\(\Phi\)):** First, the magnetic field (B) inside a long solenoid is given by: \(B = \mu_0 n I\) The total number of turns in the solenoid is \(N = n \times l\). The magnetic flux through *one* turn is \(\Phi_{\text{turn}} = B A = (\mu_0 n I) A\). The total magnetic flux linked with the solenoid (which is the flux linkage) is the flux per turn multiplied by the total number of turns: \(\Phi = N \Phi_{\text{turn}} = (n l) \times (\mu_0 n I A)\) \(\Phi = \mu_0 n^2 l A I\) Alternatively, using \(\Phi = LI\), we can find L first and then \(\Phi\). The total magnetic flux is \(\mu_0 n^2 l A I\). (b) **Self-inductance of the solenoid (L):** From the relationship \(\Phi = LI\), we can find L: \(L = \frac{\Phi}{I} = \frac{\mu_0 n^2 l A I}{I}\) \(L = \mu_0 n^2 l A\) Alternatively, if N is the total turns, then \(n = N/l\). \(L = \mu_0 \left(\frac{N}{l}\right)^2 l A = \frac{\mu_0 N^2 A}{l}\). The self-inductance of the solenoid is \(\mu_0 n^2 l A\). (c) **Effect on self-inductance if a material of high relative permeability (\(\mu_r\)) is filled inside the solenoid:** If the air core of the solenoid is replaced by a material with relative permeability \(\mu_r\), the permeability of the medium becomes \(\mu = \mu_r \mu_0\). The self-inductance of the solenoid with the core material will be: \(L' = \mu n^2 l A = (\mu_r \mu_0) n^2 l A\) Since \(L = \mu_0 n^2 l A\) (for air core), we can write: \(L' = \mu_r L\) Therefore, if a material of high relative permeability is filled inside the solenoid, its self-inductance **increases** by a factor of \(\mu_r\).
In simple words: (a) The total magnetic field lines (flux) passing through a solenoid depend on how many turns it has per length, its size, the current flowing, and a constant (\(\mu_0\)). (b) The self-inductance (L) of the solenoid, which measures how much it opposes changes in its own current, is found using these same properties. (c) If you fill the solenoid with a material that helps magnetic fields (like iron), its self-inductance will become much bigger, multiplied by that material's relative permeability.

🎯 Exam Tip: Remember the formulas for magnetic field inside a solenoid (\(B = \mu_0 n I\)), total flux linkage (\(\Phi = NBA\)), and self-inductance (\(L = \mu_0 n^2 l A\)). For the effect of a core material, replace \(\mu_0\) with \(\mu_r \mu_0\) in the self-inductance formula, indicating that self-inductance increases by a factor of \(\mu_r\).

 

Question 13. The figure shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutually perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T resistance of the closed-loop containing the rod = 9.0 m Ω. Assume the field to be uniform.
ℹ️ चित्र व्याख्या (Diagram Explanation):The diagram illustrates a metal rod (PQ) placed on two parallel conducting rails (AB). A switch (K) and a galvanometer (G) are connected across the rails, forming a closed loop with the rod. The entire setup is positioned within a uniform magnetic field (B) created by the poles of a permanent magnet, such that the magnetic field is perpendicular to the plane of the rails and the rod. The North pole (N) is above and the South pole (S) is below, indicating the magnetic field points downwards. The rod PQ is shown in the middle of the rails.

 

Question 13. The figure shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutually perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T resistance of the closed-loop containing the rod = 9.0 m Ω. Assume the field to be uniform.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक धातु की छड़ PQ दिखाता है जो चिकनी रेल AB पर रखी है। यह छड़ एक स्थायी चुंबक के ध्रुवों के बीच स्थित है। रेलों को एक गैल्वेनोमीटर G और स्विच K से जोड़ा गया है। चुंबकीय क्षेत्र, छड़ और रेलें तीनों एक-दूसरे के लंबवत हैं।
(a) Suppose K is open and the rod is moved with a speed of 12 cm s^-1 in the direction shown. Give the polarity and magnitude of the induced emf.
Answer: When switch K is open and the rod moves at 12 cm/s in the given direction, an electromotive force (emf) is created. The voltage produced is \(9.0 \text{ mV}\). End P becomes positive, and end Q becomes negative because of the rod's motion.
(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed? With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience a magnetic force due to the motion of the rod. Explain.
Answer: Yes, when K is open, extra charge builds up at the rod's ends. If K is closed, current starts flowing, but the extra charge stays there as long as conditions change. When K is open and the rod moves at a steady speed, electrons in the rod PQ feel a magnetic force. However, this magnetic force is perfectly balanced by an opposing electric force. This electric force comes from the charges that build up with opposite signs at the rod's ends. So, the total force on the electrons is zero.
(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience a magnetic force due to the motion of the rod. Explain.
Answer: When switch K is open and the rod moves at a steady speed, the electrons inside the rod PQ feel a magnetic force. But there is no overall force on them. This is because the magnetic force is exactly balanced by an electric force. This electric force appears due to the positive and negative charges that gather at the rod's ends.
(d) What is the retarding force on the rod when K is closed?
Answer: When K is closed, a slowing-down force (retarding force) acts on the rod. Its value is \(75 \times 10^{-3} \text{ N}\).
(e) How much power is required (by an external agent) to keep the rod moving at the same speed (= 12 cm s^-1) when K is closed? How much power is required when K is open?
Answer: To keep the rod moving at the same speed (12 cm/s) when K is closed, an external force must provide power. This power is \(9 \times 10^{-3} \text{ W}\). When K is open, no current flows, so no power is needed to keep the rod moving.
(f) How much power is dissipated as heat in the closed-circuit? What is the source of this power?
Answer: When the circuit is closed, \(9.0 \times 10^{-3} \text{ W}\) of power is lost as heat. This heat comes from the energy supplied by the external agent that keeps the rod moving.
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?
Answer: If the magnetic field is placed parallel to the rails instead of straight up and down, no voltage (induced emf) will be created in the moving rod. This is because there will be no change in magnetic flux.
In simple words: This question explores how a metal rod moving in a magnetic field generates voltage and current, following Lenz's Law. It covers what happens when the circuit is open or closed, how much force and power are involved, and how the magnetic field's direction affects the outcome.

🎯 Exam Tip: Pay close attention to the conditions (switch open/closed, magnetic field direction) and apply Lenz's law and the motional EMF formula (\(\mathscr{E} = Blv\)) correctly to determine induced current direction, magnitude, and associated forces/power.

 

Question 14.
(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side an as shown in the figure.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक लंबा सीधा तार दिखाया गया है जिसमें धारा I बह रही है। इसके पास एक वर्गाकार लूप रखा है जिसकी प्रत्येक भुजा 'a' लंबाई की है। लूप की स्थिति को x और y अक्ष के संबंध में दिखाया गया है, जहाँ x तार से लूप की दूरी है। लूप दाहिनी ओर v गति से बढ़ रहा है।
Answer: To find the mutual inductance between a long straight wire and a square loop, we first consider a small part of the loop. The magnetic field at a distance 't' from the wire is \(B = \frac{\mu_0 I}{2 \pi t}\). If this small part has an area \(dA = a \cdot dt\), then the magnetic flux through it is \(d\phi_B = \frac{\mu_0 I a}{2 \pi t} dt\). Adding up the flux for the entire loop, we get the total flux \( \phi_B = \int_{t=x}^{t=a+x} \frac{\mu_0 I a}{2 \pi t} dt = \frac{\mu_0 I a}{2 \pi} \log_e \left(1 + \frac{a}{x}\right) \). Since flux is also \(\phi_B = MI\), the mutual inductance \(M\) is found to be \(M = \frac{\mu_0 a}{2 \pi} \log_e \left(1 + \frac{a}{x}\right)\).
(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s. Calculate the induced emf in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.
Answer: When the straight wire carries 50 A of current and the loop moves right at 10 m/s, we need to find the induced voltage (emf) when the loop is 0.2 m away from the wire. The side of the loop is 0.1 m. The induced emf is calculated using the formula \( \mathscr{E} = \frac{\mu_0 I_w a^2 v}{2 \pi x(x+a)} \). Substituting the given values: \(I_w = 50 \text{ A}\), \(v = 10 \text{ m/s}\), \(x = 0.2 \text{ m}\), \(a = 0.1 \text{ m}\), and \(\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}\). The calculated voltage is approximately \(1.66 \times 10^{-5} \text{ V}\). This voltage arises because the magnetic flux through the loop changes as it moves in the non-uniform magnetic field of the wire.
In simple words: This question helps us understand how the magnetic field from a straight wire affects a nearby moving loop. Part (a) shows how to find the mutual inductance, which is a measure of how much magnetic flux passes through the loop for a given current in the wire. Part (b) then uses this idea to calculate the voltage created in the loop as it moves away from the wire.

🎯 Exam Tip: When calculating mutual inductance or induced EMF for a moving loop near a current-carrying wire, remember that the magnetic field is not uniform. You must use integration to find the total flux and then differentiate with respect to time to find the induced EMF, paying attention to the relative velocity and changing distance `x`.