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NEET 2026
Exam
NEET
Year
2026
Subject
Physics
Chapter
Chapter

NEET 2026 · Physics · Chapter

A small rectangular bar magnet is suspended horizontally in a...

A small rectangular bar magnet is suspended horizontally in a uniform magnetic field of 50 μ T , as shown in the
A small rectangular bar magnet is suspended horizontally in a uniform magnetic field of 50μT50 \mu T, as shown in the figure. The moment of inertia of the magnet about its axis of suspension is 2×10−5π2 kg m2\dfrac{2 \times 10^{-5}}{\pi^2} \text{ kg m}^2. If the magnet takes 40 s40 \text{ s} to complete 10 small angular oscillations, what is the magnetic moment of the magnet?

Options

  1. A0.4 A m20.4 \text{ A m}^2
  2. B10 A m210 \text{ A m}^2
  3. C0.1 A m20.1 \text{ A m}^2
  4. D0.001 A m20.001 \text{ A m}^2

Correct answer

C. 0.1 A m20.1 \text{ A m}^2

Step-by-step solution

The time period of one oscillation is T=4010=4 sT = \dfrac{40}{10} = 4 \text{ s}.

The formula for the time period of a magnetic dipole oscillating in a uniform magnetic field is:

T=2πIMBT = 2\pi \sqrt{\dfrac{I}{MB}}

Squaring both sides and rearranging for the magnetic moment MM gives:

M=4π2IT2BM = \dfrac{4\pi^2 I}{T^2 B}

Substituting the given values I=2×10−5π2 kg m2I = \dfrac{2 \times 10^{-5}}{\pi^2} \text{ kg m}^2, B=50×10−6 TB = 50 \times 10^{-6} \text{ T}, and T=4 sT = 4 \text{ s}:

M=4π2×(2×10−5π2)42×50×10−6M = \dfrac{4\pi^2 \times \left(\dfrac{2 \times 10^{-5}}{\pi^2}\right)}{4^2 \times 50 \times 10^{-6}}

M=8×10−516×50×10−6M = \dfrac{8 \times 10^{-5}}{16 \times 50 \times 10^{-6}}

M=8×10−5800×10−6M = \dfrac{8 \times 10^{-5}}{800 \times 10^{-6}}

M=10−510−4=0.1 A m2M = \dfrac{10^{-5}}{10^{-4}} = 0.1 \text{ A m}^2

Answer: 0.1 A m20.1 \text{ A m}^2
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