Magnetism & Matter

5 NEET-UG practice questions on Magnetism & Matter, part of Physics. The first 5 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

1. Question 1 · Medium

   A bar magnet of magnetic moment 4 A m² lies aligned with a uniform field of 0.05 T. How much work is needed to turn it slowly through 60° from this position?

   1. A0.3 J
   2. B0.2 J
   3. C0.1√3 J
   4. D0.1 J

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   Answer: D · 0.1 J

   Potential energy is U = -mB cosθ, so W = mB(cos 0° - cos 60°) = (4)(0.05)(1 - 1/2) = 0.1 J.

   • A: cos 60° was added rather than subtracted.
   • B: This is mB, the work for 90°, not 60°.
   • C: This is the torque mB sin 60°, not the work.

   Remember this

   Work to rotate a dipole from θ₁ to θ₂ is mB(cosθ₁ - cosθ₂), while the torque at θ is mB sinθ.

2. Question 2 · Hard

   Consider these statements about the B-H hysteresis behaviour of ferromagnetic materials. A. A transformer core material should have a narrow hysteresis loop. B. A hard magnetic material such as alnico suits permanent magnets because of its high coercivity. C. The energy dissipated per cycle of magnetisation increases with the area enclosed by the loop. D. Retentivity is the reverse magnetic intensity needed to bring B to zero. Which of these are correct?

   1. AB, C and D only
   2. BA and D only
   3. CA and B only
   4. DA, B and C only

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   Answer: D · A, B and C only

   A narrow loop keeps the heat loss per cycle small, so a transformer core needs it; A and C are correct. Alnico holds its magnetisation against reverse fields, so B is correct. D is false: the reverse intensity that brings B to zero is the coercivity, whereas retentivity is the B left at H = 0.

   • A: D confuses retentivity with coercivity, and A is also true.
   • B: D is false, and B and C are also true.
   • C: C is also true, as the loop area measures the loss per cycle.

   Remember this

   Retentivity is B remaining at H = 0; coercivity is the reverse H that reduces B to zero.

3. Question 3 · Easy

   Which pair of values correctly describes a diamagnetic substance?

   1. ARelative permeability far above 1; susceptibility large and positive
   2. BRelative permeability slightly below 1; susceptibility small and negative
   3. CRelative permeability slightly above 1; susceptibility small and negative
   4. DRelative permeability slightly above 1; susceptibility small and positive

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   Answer: B · Relative permeability slightly below 1; susceptibility small and negative

   A diamagnetic substance is weakly repelled, as its induced magnetisation opposes the field, so χ is small and negative. Since μᵣ = 1 + χ, μᵣ is slightly below 1.

   • A: These values describe a ferromagnetic substance.
   • C: A negative χ cannot give μᵣ above 1, since μᵣ = 1 + χ.
   • D: These values describe a paramagnetic substance.

   Remember this

   Diamagnetic χ is small and negative, paramagnetic χ is small and positive, ferromagnetic χ is large and positive.

4. Question 4 · Medium

   A paramagnetic sample is studied well above absolute zero. Which graph shows how its magnetic susceptibility χ varies with absolute temperature T?

   

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   1. AGraph A
   2. BGraph B
   3. CGraph D
   4. DGraph C

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   Answer: C · Graph D

   Curie's law gives χ = C/T for a paramagnet, a rectangular hyperbola. Heating disturbs the alignment of atomic dipoles, so χ falls, but it reaches zero only as T becomes very large. It stays positive at all these temperatures.

   • A: Has χ rising with heating, the reverse of what thermal motion does.
   • B: Fits a temperature-independent material, not a paramagnet.
   • D: Has χ falling linearly and vanishing at a finite temperature.

   Remember this

   Paramagnetic χ is proportional to 1/T; diamagnetic χ is nearly independent of temperature.

5. Question 5 · Medium

   Starting from its stable position, this magnet (moment 2 A m²) has been swung slowly to the angle marked in a 0.5 T field. What external work did this take?

   

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   1. A0.87 J
   2. B1.5 J
   3. C0.5 J
   4. D1.0 J

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   Answer: C · 0.5 J

   Potential energy is U = −mB cos θ, with mB = 2 × 0.5 = 1 J. So W = U(60°) − U(0°) = −0.5 − (−1) = 0.5 J. The formula W = mB(1 − cos θ) = 1 × (1 − 1/2) gives the same.

   • A: Uses the torque expression mB sin θ as the work.
   • B: Uses mB(1 + cos θ), which starts from the unstable position.
   • D: Takes mB itself as the work.

   Remember this

   Work to turn a dipole from alignment is mB(1 − cos θ); the torque is mB sin θ, a different quantity.