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AP EAMCET · PHYSICS · Magnetic Effects of Current

A short bar magnet is placed in a uniform magnetic field of 2 T such that the axis of the magnet makes an angle of \(45^{\circ}\) with the direction of the magnetic field. If the torque acting on the magnet is \(0.36 \sqrt{2} \mathrm{Nm}\), then the moment of the magnet is

  1. A \(0.54 \mathrm{~J} \mathrm{~T}^{-1}\)
  2. B \(0.18 \mathrm{~J} \mathrm{~T}^{-1}\)
  3. C \(0.72 \mathrm{~J} \mathrm{~T}^{-1}\)
  4. D \(0.36 \mathrm{~J} \mathrm{~T}^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(0.36 \mathrm{~J} \mathrm{~T}^{-1}\)

Step-by-step Solution

Detailed explanation

\( \tau = M B \sin \theta \) \( 0.36 \sqrt{2} = M \cdot 2 \cdot \sin 45^{\circ} \) \( 0.36 \sqrt{2} = M \cdot 2 \cdot \frac{1}{\sqrt{2}} \) \( M = \frac{0.36 \sqrt{2} \cdot \sqrt{2}}{2} = \frac{0.36 \cdot 2}{2} = 0.36 \mathrm{~J} \mathrm{~T}^{-1} \)