AP EAMCET · PHYSICS · Magnetic Properties of Matter
A short bar magnet placed with its axis at \(45^{\circ}\) with a uniform external magnetic field of \(28.3 \times 10^{-3} \mathrm{~T}\) experiences a torque of magnitude equal to \(3.6 \times 10^{-5} \mathrm{~J}\). The magnitude of magnetic moment of the magnet is nearly
- A \(1.8 \times 10^{-3} \mathrm{~J} \mathrm{~T}^{-1}\)
- B \(1.2 \times 10^{-3} \mathrm{~J} \mathrm{~T}^{-1}\)
- C \(2.4 \times 10^{-3} \mathrm{~J} \mathrm{~T}^{-1}\)
- D \(1.6 \times 10^{-3} \mathrm{~J} \mathrm{~T}^{-1}\)
Answer & Solution
Correct Answer
(A) \(1.8 \times 10^{-3} \mathrm{~J} \mathrm{~T}^{-1}\)
Step-by-step Solution
Detailed explanation
We know that \(\tau=\mathrm{MB} \sin \theta\) \(\Rightarrow \mathrm{M}=\frac{\tau}{\mathrm{B} \sin 45^{\circ}}=\frac{3.6 \times 10^{-5}}{28.3 \times 10^{-3} \times \frac{1}{\sqrt{2}}}=1.8 \times 10^{-3} \mathrm{~J} \mathrm{~T}^{-1}\)
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