CBSE Class 12 Physics Magnetic Effects Of Current And Magnetism Frequently Asked Questions

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UNIT III

MAGNETIC EFFECTS OF CURRENT AND MAGNETISM

1 MARKS

 

1. A circular coil of radius 'R' carries a current 'I'. Write the expression for the magnetic field due to this coil at its centre. Find out the direction of the magnetic field. [1 Mark]

Answer:
1. The magnitude of magnetic field at the centre of the circular coil is given by \(B = \frac{\mu_0 I}{2R}\).
2. The direction of the magnetic field is given by the Right-Hand Thumb Rule.
3. If the current flows in the anti-clockwise direction, the magnetic field points normally outwards, and if clockwise, it points normally inwards.

Teacher's Note:
a) Biot-Savart law is applied to derive the magnetic field at the centre of a circular current-carrying loop.
b) Ensure proper sign convention for the direction of current and corresponding magnetic field vector.

 

2. Write the expression for the force on the charge moving in a magnetic field. Use this expression to define the SI unit of magnetic field. [1 Mark]

Answer:
1. The magnetic force on a charge q moving with velocity v in a magnetic field B is \(F = q(v \times B)\) or \(F = qvB \sin\theta\).
2. One tesla is defined as the magnetic field in which a charge of 1 coulomb moving with a speed of 1 m/s perpendicular to the magnetic field experiences a force of 1 newton.

Teacher's Note:
a) The Lorentz magnetic force formula forms the basis for defining the tesla.
b) Remember that magnetic force does no work on a moving charge as it is always perpendicular to velocity.

 

3. Define magnetic susceptibility of a material. Name two elements, one having positive susceptibility and the other having negative susceptibility. What does negative susceptibility signify? [1 Mark]

Answer:
1. Magnetic susceptibility is defined as the ratio of intensity of magnetization induced in the material to the applied magnetic intensity.
2. Aluminium has positive susceptibility, and Bismuth has negative susceptibility.
3. Negative susceptibility signifies that the material is diamagnetic and tends to expel magnetic field lines.

Teacher's Note:
a) Susceptibility is a dimensionless quantity representing the magnetic response of a medium.
b) Diamagnetic substances possess a small negative susceptibility independent of temperature.

 

4. Define the term magnetic dipole moment of a current loop. Write the expression for the magnetic moment when an electron revolves at a speed around an orbit of radius in hydrogen atom.. [1 Mark]

Answer:
1. The magnetic dipole moment of a current loop is the product of the current and the area vector of the loop, \(M = IA\).
2. The magnetic moment of a revolving electron is given by \(M = \frac{evr}{2}\), where e is electronic charge, v is speed, and r is radius.

Teacher's Note:
a) Current loop behaves like a magnetic dipole with a definite magnetic moment direction.
b) Gyromagnetic ratio relates the angular momentum and magnetic moment of an orbiting electron.

 

5. Explain with the help of a diagram the term 'magnetic declination' at a given place. [1 Mark]

Answer:
1. Magnetic declination is the angle between the geographic meridian and the magnetic meridian at a given place.
2. It accounts for the deviation of a magnetic compass needle from true geographic north.

Teacher's Note:
a) Declination varies from place to place on the Earth's surface.
b) Essential parameter for accurate navigation using magnetic compasses.

 

6. Define the term 'angle of dip'. What is the value of the angle of dip at the magnetic equator? What does it mean? [1 Mark]

Answer:
1. The angle of dip is the angle made by the total magnetic field of the Earth with the horizontal direction in the magnetic meridian.
2. The value of the angle of dip at the magnetic equator is 0°.
3. It means that the Earth's magnetic field is completely horizontal at the magnetic equator.

Teacher's Note:
a) Angle of dip ranges from 0° at the equator to 90° at the magnetic poles.
b) Measured using a dip circle instrument.

 

7. Two wires of equal lengths are bend in the form of two loops. One of the loop is squareshaped where as 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 reasons. [1 Mark]

Answer:
1. The circular loop will experience a greater torque.
2. Torque is given by \(\tau = NIAB \sin\theta\), where area A is maximum for a given perimeter in a circular shape compared to a square shape, yielding a larger magnetic moment.

Teacher's Note:
a) For a fixed perimeter, the circle encloses the maximum area.
b) Magnetic dipole moment is directly proportional to the enclosed area.

 

8. Explain why steel is preferred for making permanent magnets while soft iron is preferred for making electromagnets. [1 Mark]

Answer:
1. Steel has high retentivity and high coercivity, making it retain its magnetism strongly and persistently, suitable for permanent magnets.
2. Soft iron has high permeability and low retentivity, meaning it magnetizes and demagnetizes easily with low energy loss, ideal for electromagnets.

Teacher's Note:
a) Hysteresis loop area determines energy dissipation per cycle.
b) Permanent magnets require a wide hysteresis loop, while electromagnets require a narrow one.

 

9. Draw diagram to show behavior of magnetic field lines near a bar of 1)copper 2)aluminum and 3)mercury cooled at a very low temperature (4.2K). [1 Mark]

Answer:
1. For copper (diamagnetic), magnetic field lines are slightly repelled or expelled.
2. For aluminum (paramagnetic), magnetic field lines are slightly concentrated or attracted.
3. For mercury cooled to 4.2K (superconductor, perfectly diamagnetic), magnetic field lines are completely expelled (Meissner effect).

Teacher's Note:
a) Superconductors exhibit perfect diamagnetism with magnetic susceptibility equal to -1.
b) Field lines never intersect boundaries normally except for superconductors where interior B is zero.

 

10. How will the magnetic field intensity at the centre of the circular coil carrying current will change, if the current through the coil is doubled and radius of the coil is halved? [1 Mark]

Answer:
1. The formula for magnetic field at the centre is \(B = \frac{\mu_0 I}{2R}\).
2. When current \(I' = 2I\) and radius \(R' = \frac{R}{2}\), the new magnetic field is \(B' = \frac{\mu_0 (2I)}{2(R/2)} = 4 \left(\frac{\mu_0 I}{2R}\right) = 4B\).
3. Therefore, the magnetic field intensity becomes 4 times its initial value.

Teacher's Note:
a) Magnetic field is directly proportional to current and inversely proportional to radius.
b) Substitution of proportional changes avoids full recalculation of constants.

 

11. What do you mean by current sensitivity of a moving coil galvanometer? On what factors does it depend? [1 Mark]

Answer:
1. Current sensitivity is defined as the deflection produced in the galvanometer per unit current passing through it (\(I_s = \frac{\theta}{I} = \frac{NAB}{k}\)).
2. It depends on the number of turns (N), magnetic field (B), area of the coil (A), and restoring torque per unit twist (k).

Teacher's Note:
a) Higher number of turns increases sensitivity but also increases galvanometer resistance.
b) Radial magnetic field ensures uniform sensitivity independent of coil orientation.

 

12. Derive an expression for the force experienced by a current carrying straight conductor placed in a magnetic field. Under what condition is this force maximum? [1 Mark]

Answer:
1. Consider a conductor of length L, cross-sectional area A, carrying current I in a uniform magnetic field B.
2. The drift velocity of electrons is \(v_d\), and the magnetic force on one electron is \(f = ev_d B \sin\theta\).
3. Total force on N electrons in the conductor is \(F = N(ev_d B \sin\theta) = (nAL)(ev_d B \sin\theta) = I L B \sin\theta\) since \(I = nAe v_d\).
4. The force is maximum when the conductor is placed perpendicular to the magnetic field (\(\theta = 90^\circ\)), giving \(F_{max} = ILB\).

Teacher's Note:
a) Vector form of the force expression is \(F = I (L \times B)\).
b) When the conductor is parallel to the magnetic field, the magnetic force is zero.

2 MARKS

 

13. Obtain the force per unit length experienced by two parallel conductors of infinite length carrying current in the same direction. Hence define one ampere. [2 Marks]

Answer:
1. Consider two long parallel conductors \(P_1\) and \(P_2\) carrying currents \(I_1\) and \(I_2\) in the same direction, separated by a distance \(r\).
2. The magnetic field produced by conductor \(P_1\) at the location of conductor \(P_2\) is given by B₁ = \(\frac{\mu_0 I_1}{2\pi r}\).
3. The force per unit length experienced by conductor \(P_2\) due to this magnetic field is F = \(\frac{F}{L}\) = \(\frac{\mu_0 I_1 I_2}{2\pi r}\), acting attractively towards \(P_1\).
4. One ampere is defined as that constant current which, when maintained in each of two straight parallel conductors of infinite length and negligible circular cross-section, placed 1 metre apart in vacuum, produces between them a force equal to \(2 \times 10^{-7}\) newton per metre of length.

Teacher's Note:
a) The fundamental formula used is the magnetic force between two parallel current-carrying wires.
b) Students often confuse the direction of force; currents in the same direction result in an attractive force.
c) The definition of one ampere relies directly on this standard electromagnetic force relation.

 

14. A) If \(\chi\) stands for the magnetic susceptibility of a given material, identify the class of materials for which (a) \(-1 \le \chi \lt 0\), and (b) \(0 \lt \chi \lt \epsilon\) [\(\epsilon\) is a small positive number]. Write the range of relative magnetic permeability of these materials. B) Draw the pattern of the magnetic field lines when these materials are placed on a strong magnetic field. [2 Marks]

Answer:
1. (a) Diamagnetic materials; (b) Paramagnetic materials.
2. For diamagnetic materials, the range of relative magnetic permeability (\(\mu_r\)) is \(0 \le \mu_r \lt 1\), and for paramagnetic materials, it is \(\mu_r \gt 1\) (slightly greater than 1).
3. When diamagnetic materials are placed in a strong magnetic field, the field lines are expelled or repelled by the material.
4. When paramagnetic materials are placed in a strong magnetic field, the field lines are weakly attracted and become slightly concentrated inside the material.

Teacher's Note:
a) Magnetic susceptibility \(\chi\) distinguishes the magnetic behaviour of different media.
b) Remember that diamagnetic susceptibility is negative, whereas paramagnetic susceptibility is positive and small.
c) Relative permeability is related to susceptibility by the formula \(\mu_r = 1 + \chi\).

 

15. Derive an expression for the force acting on a current carrying conductor in a magnetic field. Under what conditions this force is maximum and minimum? [2 Marks]

Answer:
1. Consider a conductor of length \(L\) and cross-sectional area \(A\) carrying current \(I\) placed in a uniform magnetic field \(B\) at an angle \(\theta\).
2. The drift velocity of electrons is \(v_d\), and the magnetic force on a single moving electron is \(f = q(v_d \times B) = e v_d B \sin\theta\).
3. Total number of free electrons in the conductor is \(nAL\), where \(n\) is the electron density. Total force \(F = nAL \times e v_d B \sin\theta\).
4. Since \(I = nAe v_d\), the magnetic force becomes \(F = I L B \sin\theta\) (or in vector form, \(F = I(L \times B)\)).
5. Force is maximum when \(\theta = 90^\circ\) (\(F_{max} = ILB\)) and minimum when \(\theta = 0^\circ\) or \(180^\circ\) (\(F_{min} = 0\)).

Teacher's Note:
a) The derivation links microscopic magnetic force on moving charge carriers to macroscopic current.
b) Ensure proper vector cross-product representation for maximum clarity in examinations.

 

16. Define the term magnetic moment of current loop. Derive the expression for the magnetic moment when an electron revolves at a speed ‘v’ around an orbit of radius r in hydrogen atom. Also calculate the value of Bohr’s magnetic moment. [2 Marks]

Answer:
1. The magnetic moment of a current loop is defined as the product of the current flowing through the loop and the area vector bounded by the loop (\(M = IA\)).
2. Current due to a revolving electron of charge \(e\) with frequency \(f\) is \(I = \frac{e}{T} = \frac{ev}{2\pi r}\).
3. The area of the circular orbit is \(A = \pi r^2\), so the magnetic moment is \(M = I A = \left(\frac{ev}{2\pi r}\right)(\pi r^2) = \frac{evr}{2}\).
4. Bohr's magnetic moment is given by \(\mu_B = \frac{e h}{4\pi m}\) and its calculated value is approximately \(9.27 \times 10^{-24} \text{ A m}^2\).

Teacher's Note:
a) Magnetic moment acts as a bridge between atomic structure and magnetism.
b) Substitute fundamental constants \(e\), \(h\), and \(m\) carefully when computing Bohr magneton.

 

17. With the help of diagram explain how a galvanometer can be converted into an ammeter and a voltmeter. [2 Marks]

Answer:
1. Conversion into Ammeter: A galvanometer is converted into an ammeter by connecting a very small resistance, known as a shunt resistance (\(S\)), in parallel with it.
2. Formula for shunt is \(S = \frac{I_g G}{I - I_g}\), which protects the galvanometer and allows large currents to pass.
3. Conversion into Voltmeter: A galvanometer is converted into a voltmeter by connecting a very high resistance (\(R\)) in series with it.
4. Formula for series resistance is \(R = \frac{V}{I_g} - G\), which ensures that only a minimal current flows through the measuring device.

Teacher's Note:
a) Ammeters require a low parallel shunt to measure high currents safely.
b) Voltmeters require a high series resistance to measure potential differences accurately without drawing excessive current.

 

18. To increase the current sensitivity of a moving coil galvanometer by 50%, its resistance is increased so that the new resistance becomes twice its initial resistance. By what factor does its voltage sensitivity change? [2 Marks]

Answer:
1. Current sensitivity is given by \(I_s = \frac{NBA}{k}\). When current sensitivity increases by 50%, the new current sensitivity becomes \(I'_s = 1.5 I_s\).
2. Since resistance becomes twice, \(R' = 2R\).
3. Voltage sensitivity is given by \(V_s = \frac{I_s}{R} = \frac{NBA}{k R}\).
4. The new voltage sensitivity is \(V'_s = \frac{I'_s}{R'} = \frac{1.5 I_s}{2 R} = 0.75 V_s\). Thus, it changes by a factor of \(0.75\) or \(\frac{3}{4}\).

Teacher's Note:
a) Voltage sensitivity is defined as the deflection per unit voltage, expressed as \(V_s = I_s / R\).
b) Carefully track proportional changes in numerator and denominator.

 

3 MARKS

 

19. Write an expression for force experienced by a charged particle moving in a uniform magnetic field? With the help of labeled diagram, explain principle and working of a cyclotron. Show that cyclotron frequency does not depend upon the speed of particles. Write its two limitations. [3 Marks]

Answer:
1. The force experienced by a charged particle is given by the Lorentz magnetic force formula \(F = q(v \times B)\) or magnitude \(F = q v B \sin\theta\).
2. Principle: A charged particle can be accelerated to high energies by alternating an electric field between two hollow D-shaped metallic chambers (dees) in a strong perpendicular magnetic field.
3. Working: The magnetic field keeps the particle moving in a circular path, and each time the particle crosses the gap between the dees, the alternating electric field accelerates it because its frequency matches the cyclotron frequency.
4. Cyclotron frequency is given by \(f = \frac{qB}{2\pi m}\), which is independent of velocity \(v\) and radius \(r\).
5. Two limitations: (i) It cannot accelerate neutral particles or uncharged neutrons. (ii) At very high speeds, relativistic mass increase disrupts resonance, preventing electrons from being successfully accelerated.

Teacher's Note:
a) Always equate magnetic centripetal force to magnetic Lorentz force (\(\frac{mv^2}{r} = qvB\)) to derive frequency.
b) Mention relativistic constraints clearly as a key limitation for light particles like electrons.

 

20. State Ampere’s Circuital Law. Derive an expression for the magnetic field at a point due to straight current carrying conductor. [3 Marks]

Answer:
1. Ampere's Circuital Law states that the line integral of magnetic field \(B\) around any closed loop is equal to \(\mu_0\) times the total current \(I\) passing through the surface bounded by the loop: \(\oint B \cdot dl = \mu_0 I\).
2. Consider an infinitely long straight conductor carrying current \(I\). By symmetry, the magnetic field lines are concentric circles centred on the wire.
3. Choose a circular Amperian loop of radius \(r\) centered on the wire. The magnetic field \(B\) is tangential to every point on the loop, so \(B \cdot dl = B dl \cos 0^\circ = B dl\).
4. Applying Ampere's law: \(\oint B dl = B \oint dl = B (2\pi r) = \mu_0 I\).
5. Solving for the magnetic field gives \(B = \frac{\mu_0 I}{2\pi r}\).

Teacher's Note:
a) Ampere's law simplifies calculating magnetic fields for systems with high symmetry.
b) Ensure proper geometric justification of the Amperian loop choice.

 

21. Derive an expression for the magnetic field at a point along the axis of an air cored solenoid using a Ampere’s circuital law. [3 Marks]

Answer:
1. Consider a long solenoid with \(n\) turns per unit length carrying current \(I\).
2. Consider a rectangular Amperian loop \(abcd\) of length \(l\) where side \(ab\) is parallel to the axis inside the solenoid, side \(cd\) is far outside where the magnetic field is zero, and sides \(bc\) and \(da\) are perpendicular to the field.
3. The line integral around the closed loop is split into four parts: \(\oint B \cdot dl = \int_a^b B \cdot dl + \int_b^c B \cdot dl + \int_c^d B \cdot dl + \int_d^a B \cdot dl\).
4. The integrals along \(bc\) and \(da\) are zero because \(B\) is perpendicular to \(dl\), and the integral along \(cd\) is zero because the field outside is negligible.
5. Thus, the total integral reduces to \(\int_a^b B dl = B l\). The total current enclosed by the loop is \(I_{enclosed} = n l I\).
6. Applying Ampere's Circuital Law: \(B l = \mu_0 n l I\), which yields \(B = \mu_0 n I\).

Teacher's Note:
a) The magnetic field inside an ideal infinite solenoid is uniform and parallel to its axis.
b) The magnetic field outside an ideal solenoid is approximately zero.

22. Derive an expression for torque acting on a rectangular current carrying loop kept in a uniform magnetic field B. Indicate the direction of torque acting on the loop. [3 Marks]

Answer:
1. Consider a rectangular loop PQRS of length \(l\) and breadth \(b\) carrying current I placed in a uniform magnetic field \(\vec{B}\).
2. Let \(\theta\) be the angle between the normal to the plane of the loop and the magnetic field vector \(\vec{B}\).
3. The magnetic force on side PS is \(F_1 = IlB\) acting outwards, and on side QR is \(F_2 = IlB\) acting inwards.
4. The forces on sides PQ and RS are equal, opposite, and collinear, so they cancel out.
5. The two equal and opposite forces \(F_1\) and \(F_2\) form a couple that exerts a torque \(\tau = F \times \text{arm of the couple} = (IlB) \times (b \sin\theta) = IAB \sin\theta\), where \(A = l \times b\) is the area of the loop.
6. In vector form, the torque is given by \(\vec{\tau} = M \times \vec{B}\), where \(M = IA\) is the magnetic dipole moment.
7. The direction of the torque is perpendicular to the plane containing the magnetic moment vector and the magnetic field vector, tending to align the magnetic moment with the field.

Teacher's Note:
a) The fundamental formula used is torque equals force multiplied by perpendicular distance.
b) Students often confuse the angle of the loop with the angle of the normal to the field.
c) The magnetic dipole moment \(M = IA\) is a vector directed along the normal to the loop area.
d) This derivation forms the underlying principle for the working of electric motors and moving coil galvanometers.

 

23. With neat diagram, describe the principle, construction and working of a moving coil galvanometer. Explain the importance of radial field. [5 Marks]

[Figure: Diagram of a moving coil galvanometer showing a coil suspended in a radial magnetic field between soft iron cylindrical cores, with a spring, pointer, and scale]

Answer:
1. Principle: A current-carrying coil placed in a magnetic field experiences a magnetic torque, causing it to deflect proportional to the current.
2. Construction: It consists of a coil of many turns wound on a light non-magnetic frame, suspended between cylindrical pole pieces of a permanent magnet, with a soft iron core placed symmetrically inside.
3. Working: When an electric current I passes through the coil, a deflecting torque \(\tau = NIAB\) acts on it, where N is the number of turns, A is the area, and B is the magnetic field.
4. The deflecting torque is balanced by the restoring torque of the suspension spring \(\tau_r = k\theta\), where k is the torsional constant and \(\theta\) is the angular deflection.
5. At equilibrium, \(NIAB = k\theta\), which gives \(\theta = \left(\frac{NAB}{k}\right)I\), meaning deflection is directly proportional to current.
6. Importance of radial field: A radial magnetic field ensures that the plane of the coil is always parallel to the magnetic field lines for any orientation, making \(\sin\theta = 1\) and ensuring a linear scale where deflection is directly proportional to current.

Teacher's Note:
a) The radial magnetic field is achieved by using concave pole pieces and a soft iron cylindrical core.
b) A common mistake is omitting the soft iron core's role in producing both a strong and radial field.
c) The current sensitivity is defined as deflection per unit current, given by \(I_s = \frac{NAB}{k}\).
d) Increasing the number of turns N, area A, or magnetic field B enhances the sensitivity of the galvanometer.

 

24. State Biot Savart Law. Use this law to obtain a formula for magnetic field at the centre of a circular loop of radius R, number of turns N carrying current I. Sketch the magnetic field lines for a current loop clearly indicating the direction of the field. [5 Marks]

[Figure: Sketch of concentric circular magnetic field lines around a circular current-carrying loop, showing the direction of current and the axial magnetic field direction]

Answer:
1. Biot Savart Law: The magnetic field dB due to a current element \(Id\vec{l}\) at a position vector \(\vec{r}\) is given by \(dB = \frac{\mu_0}{4\pi} \frac{I dl \sin\theta}{r^2}\).
2. Consider a circular loop of radius R and N turns carrying current I.
3. For any small current element \(dl\) on the circular loop, the angle \(\theta\) between the current element and the position vector connecting it to the centre is \(90^\circ\).
4. The magnetic field at the centre due to a single turn element is \(dB = \frac{\mu_0}{4\pi} \frac{I dl \sin 90^\circ}{R^2} = \frac{\mu_0}{4\pi} \frac{I dl}{R^2}\).
5. Integrating around the full circular loop of perimeter \(2\pi R\), the total magnetic field for a single turn is \(B = \int dB = \frac{\mu_0 I}{4\pi R^2} \int dl = \frac{\mu_0 I}{4\pi R^2} (2\pi R) = \frac{\mu_0 I}{2R}\).
6. For a circular coil with N turns, the magnitude of the magnetic field at the centre is \(B = \frac{\mu_0 NI}{2R}\).

Teacher's Note:
a) The Biot Savart law is an inverse-square law vector field relation analogous to Coulomb's law in electrostatics.
b) Students must remember that \(\sin 90^\circ = 1\) because the tangent vector and radius vector are always perpendicular at the boundary.
c) The direction of the magnetic field at the centre is given by the right-hand thumb rule.
d) Multiplying by N accounts for the linear superposition of magnetic fields from all identical turns.

 

25. Distinguish the magnetic properties of dia, para- and ferro-magnetic substances interms of (i) Susceptibility, (ii) magnetic permeability and (iii) coercivity. Give one example of each of these materials. Draw the field lines due to an external magnetic field near a (i) diamagnetic, (ii) paramagnetic substance. [5 Marks]

[Figure: Two diagrams showing magnetic field lines being expelled slightly from a diamagnetic substance, and being slightly concentrated inside a paramagnetic substance]

Answer:
1. Comparison Table:

Property	Diamagnetic	Paramagnetic	Ferromagnetic
(i) Magnetic Susceptibility (\(\chi\))	Small and negative (-1 to 0)	Small and positive (small positive value)	Large and positive (very high)
(ii) Magnetic Permeability (\(\mu_r\))	Slightly less than 1	Slightly greater than 1	Much greater than 1
(iii) Coercivity	Not applicable / Zero	Not applicable / Zero	High (for permanent magnets) or low (for electromagnets)

2. Examples: Bismuth (Diamagnetic), Aluminium (Paramagnetic), Iron (Ferromagnetic).
3. Field lines near diamagnetic substance: The field lines are repelled or expelled slightly, bending away from the material.
4. Field lines near paramagnetic substance: The field lines are slightly attracted and crowded into the material.

Teacher's Note:
a) Magnetic susceptibility is dimensionless, measuring the degree of magnetization of a material in response to an applied magnetic field.
b) A common error is confusing relative permeability values, where diamagnetic is less than unity while paramagnetic exceeds unity.
c) Coercivity is specifically a property defining the reverse magnetic field required to reduce magnetization to zero in ferromagnets.
d) Drawing field lines correctly requires showing spacing changes to indicate field expulsion or concentration.