Magnetism-Matter. 3D visualization of a bar magnet's field lines and Earth's magnetic field.

Magnetism & Matter: Question Bank

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Magnetism and Matter Question Bank | Class 12 Physics

Magnetism & Matter: Question Bank

Practice Set: Bar Magnet, Earth’s Magnetism, and Magnetic Materials.

10 MCQs 10 Assertion-Reason Case Study
Instructions: Review the Chapter Notes before attempting. Click “Show Answer” to verify.

Part 1: Multiple Choice Questions (1 Mark)

1. The magnetic field lines inside a bar magnet:

(a) Run from North pole to South pole
(b) Run from South pole to North pole
(c) Do not exist
(d) Depend on external field

Answer: (b)
Magnetic field lines form continuous closed loops. Outside, they go N to S; inside, they go S to N.

2. A magnetic needle is kept in a non-uniform magnetic field. It experiences:

(a) A force and a torque
(b) A force but not a torque
(c) A torque but not a force
(d) Neither a force nor a torque

Answer: (a)
In a non-uniform field, the forces on the two poles (mB_1 and mB_2) are unequal, creating a net force. Since they are not collinear, there is also a torque.

3. The magnetic susceptibility \chi is negative for:

(a) Ferromagnetic materials
(b) Paramagnetic materials
(c) Diamagnetic materials
(d) Superconductors only

Answer: (c)
Diamagnetic materials repel magnetic field lines, corresponding to a small negative susceptibility.

4. According to Gauss’s law for magnetism, \oint \vec{B} \cdot d\vec{S} is equal to:

(a) \mu_0 I
(b) Zero
(c) \mu_0 q_m
(d) q_m/\epsilon_0

Answer: (b)
The net magnetic flux through any closed surface is zero because magnetic monopoles do not exist.

5. The work done in turning a magnet of magnetic moment M by an angle of 90^\circ from the magnetic meridian is n times the corresponding work done to turn it through an angle of 60^\circ. The value of n is:

(a) 2
(b) 1
(c) 0.5
(d) 1/3

Answer: (a)
W_1 = MB(1 - \cos 90^\circ) = MB.
W_2 = MB(1 - \cos 60^\circ) = MB(1 - 0.5) = 0.5 MB.
W_1 = 2 W_2 \Rightarrow n = 2.

6. The angle of dip at the magnetic equator is:

(a) 0^\circ
(b) 90^\circ
(c) 45^\circ
(d) 30^\circ

Answer: (a)
At the magnetic equator, the magnetic field lines are horizontal, so the dip angle is 0^\circ.

7. Which of the following is independent of temperature?

(a) Ferromagnetism
(b) Diamagnetism
(c) Paramagnetism
(d) None of these

Answer: (b)
Diamagnetism arises from intrinsic electronic structure and is generally temperature-independent. Paramagnetism and Ferromagnetism depend on thermal agitation (Curie’s Law).

8. The relative permeability \mu_r of a substance is slightly greater than 1. The substance is:

(a) Diamagnetic
(b) Paramagnetic
(c) Ferromagnetic
(d) Non-magnetic

Answer: (b)
For paramagnetic, \mu_r > 1 (small positive susceptibility). For ferromagnetic, \mu_r \gg 1.

9. At Curie temperature, a ferromagnetic substance becomes:

(a) Diamagnetic
(b) Paramagnetic
(c) Superconductor
(d) Non-magnetic

Answer: (b)
Above the Curie temperature, thermal agitation disrupts the alignment of domains, turning the material paramagnetic.

10. A bar magnet is cut into two equal halves perpendicular to its length. The pole strength of each piece:

(a) Becomes half
(b) Doubles
(c) Remains same
(d) Becomes zero

Answer: (c)
Cutting perpendicular to length changes the magnetic moment (M = m \times 2l \to m \times l) but the pole strength (m) remains unchanged.

Part 2: Assertion-Reason Questions

(A) Both A & R are true, R explains A.
(B) Both A & R are true, R does NOT explain A.
(C) A is true, R is false.
(D) A is false, R is true.

1. Assertion (A): Magnetic field lines are continuous and form closed loops.
Reason (R): Magnetic monopoles do not exist.

Answer: (A)
Since there are no isolated sources or sinks (monopoles), field lines must close on themselves.

2. Assertion (A): The poles of a bar magnet cannot be separated.
Reason (R): Breaking a magnet creates smaller magnets, each with a North and South pole.

Answer: (A)
This is a direct consequence of the non-existence of monopoles.

3. Assertion (A): Diamagnetic materials are repelled by magnets.
Reason (R): They tend to move from stronger to weaker parts of a magnetic field.

Answer: (A)
Repulsion implies moving away from the high-field region (pole) to the low-field region.

4. Assertion (A): A compass needle points North-South.
Reason (R): The Earth behaves like a giant bar magnet with its magnetic South pole near the geographic North pole.

Answer: (A)
The North pole of the compass is attracted to the Earth’s magnetic South pole (which is near the geographic North).

5. Assertion (A): Gauss’s law for magnetism states that the net magnetic flux through any closed surface is zero.
Reason (R): Magnetic field lines start from the North pole and terminate at the South pole.

Answer: (C)
Assertion is True. Reason is False. Magnetic lines do not “terminate”; they run S-to-N inside the magnet to form closed loops.

6. Assertion (A): Soft iron is used in making electromagnets.
Reason (R): Soft iron has low retentivity and low coercivity.

Answer: (A)
Low retentivity allows it to lose magnetism quickly when current stops; low coercivity means low hysteresis loss.

7. Assertion (A): The magnetic susceptibility of a paramagnetic substance is inversely proportional to absolute temperature.
Reason (R): This is Curie’s Law (\chi = C/T).

Answer: (A)
Correct statement and explanation.

8. Assertion (A): A superconductor exhibits perfect diamagnetism.
Reason (R): A superconductor expels all magnetic flux lines from its interior (Meissner effect).

Answer: (A)
Expulsion of flux lines corresponds to \chi = -1, perfect diamagnetism.

9. Assertion (A): Magnetic field lines can intersect.
Reason (R): At the point of intersection, there would be two directions of magnetic field, which is impossible.

Answer: (D)
Assertion is False. Reason is True (explaining why they cannot intersect).

10. Assertion (A): The angle of dip is 90^\circ at the poles.
Reason (R): At the poles, the earth’s magnetic field is vertical.

Answer: (A)
The magnetic needle points vertically downwards at the North Pole.

Part 3: Important Derivations & Theory

1. Derive an expression for the potential energy of a magnetic dipole placed in a uniform magnetic field. (3 Marks)

Key Steps:
1. Torque on dipole: \tau = mB \sin\theta.
2. Work done in rotating by d\theta: dW = \tau d\theta = mB \sin\theta d\theta.
3. Integrate from \theta_1 to \theta_2: W = \int_{\theta_1}^{\theta_2} mB \sin\theta d\theta = -mB [\cos\theta]_{\theta_1}^{\theta_2}.
4. PE is defined as work done to bring from zero energy position (\theta=90^\circ):
U(\theta) = -mB \cos\theta = -\vec{m} \cdot \vec{B}.

2. State Gauss’s Law for Magnetism. How does it differ from Gauss’s Law for Electrostatics? (2 Marks)

Statement: The net magnetic flux through any closed surface is always zero. \oint \vec{B} \cdot d\vec{S} = 0.
Difference: In electrostatics, \oint \vec{E} \cdot d\vec{S} = q/\epsilon_0, because isolated electric charges exist. In magnetism, isolated poles (monopoles) do not exist, so the integral is always zero.

3. Distinguish between Diamagnetic, Paramagnetic, and Ferromagnetic substances based on: (a) Susceptibility, (b) Permeability, (c) Effect of Temperature. (3 Marks)

[Refer to table in Notes]
1. Diamagnetic: Small -ve \chi; 0 \le \mu_r < 1; Temp independent.
2. Paramagnetic: Small +ve \chi; \mu_r > 1; \chi \propto 1/T.
3. Ferromagnetic: Large +ve \chi; \mu_r \gg 1; Becomes paramagnetic above Curie point.

Part 4: Numericals

1. A short bar magnet placed with its axis at 30^\circ with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 \times 10^{-2} J. What is the magnitude of the magnetic moment of the magnet?

Formula: \tau = mB \sin\theta.
m = \tau / (B \sin\theta) = (4.5 \times 10^{-2}) / (0.25 \times \sin 30^\circ).
m = 0.045 / (0.25 \times 0.5) = 0.045 / 0.125 = 0.36 \text{ A m}^2.

2. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 60^\circ with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.4 G. Determine the magnitude of the earth’s magnetic field at the place.

Given: Dip \delta = 60^\circ, H_E = 0.4 G.
Formula: H_E = B_E \cos\delta \Rightarrow B_E = H_E / \cos\delta.
B_E = 0.4 / \cos 60^\circ = 0.4 / 0.5 = 0.8 G.

3. An iron rod of susceptibility 599 is subjected to a magnetizing field of 1200 A/m. The permeability of the material of the rod is: (\mu_0 = 4\pi \times 10^{-7} T m A^{-1})

\mu_r = 1 + \chi = 1 + 599 = 600.
\mu = \mu_0 \mu_r = 4\pi \times 10^{-7} \times 600.
\mu = 2400\pi \times 10^{-7} = 2.4\pi \times 10^{-4} T m A^{-1}.

Part 5: Case Study

Case Study: Earth’s Magnetism
The earth acts as a huge magnet with its magnetic field extending far into space. The field is believed to arise due to electrical currents produced by convective motion of metallic fluids in the outer core (Dynamo effect). To define the magnetic field at any place on Earth, three quantities are required, known as Magnetic Elements: Magnetic Declination, Angle of Dip (Inclination), and Horizontal Component.

  1. What is Magnetic Declination?
  2. At which location on Earth is the angle of dip 90^\circ?
  3. What is the relation between the horizontal component (H_E), vertical component (Z_E), and total field (B_E)?
  4. If the dip angle is 45^\circ, compare the horizontal and vertical components.
1. The angle between the geographic meridian and the magnetic meridian at a place.
2. At the **Magnetic Poles**.
3. B_E^2 = H_E^2 + Z_E^2 and \tan\delta = Z_E / H_E.
4. Since \tan 45^\circ = 1, Z_E / H_E = 1 \Rightarrow Z_E = H_E. They are equal.

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