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GUJCET · Physics · Magnetism And Matter
A short bar magnet placed with its axis at \(30^{\circ}\) and a uniform external magnetic field of 0.5 T experience torque of magnitude equal to \(4.5 \times 10^{-2} J\). Then the magnitude of the magnetic moment will be _________.
- A \(3.6 \times 10^2 J T ^{-1}\)
- B \(36 \times 10^{-2} J T ^{-1}\)
- C \(1.8 \times 10^2 J T ^{-1}\)
- D \(18 \times 10^{-2} J T ^{-1}\)
Answer & Solution
Correct Answer
(D) \(18 \times 10^{-2} J T ^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}\tau=m B \sin \theta \\ 4.5 \times 10^{-2}=m \times 0.5 \times \sin 30^{\circ} \\ 4.5 \times 10^{-2}=m \times \frac{1}{2} \times \frac{1}{2} \\ m=4.5 \times 10^{-2} \times 2 \times 2 \\ m=18 \times 10^{-2} J T ^{-1}\end{array}\)
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