AP EAMCET · PHYSICS · Magnetic Effects of Current
A bar magnet of length \(10 \mathrm{~cm}\) and having the pole strength equal to \(10^{-3} \mathrm{~A}-\mathrm{m}\) is kept in a magnetic field having magnetic induction \(B\) equal to \(4 \pi \times 10^{-3} \mathrm{~T}\). It makes an angle of \(30^{\circ}\) with the direction of magnetic induction. The value of the torque acting on the magnet is
- A \(2 \pi \times 10^{-7} \mathrm{Nm}\)
- B \(2 \pi \times 10^{-5} \mathrm{Nm}\)
- C \(0.5 \mathrm{Nm}\)
- D \(0.5 \times 10^2 \mathrm{Nm}\)
Answer & Solution
Correct Answer
(A) \(2 \pi \times 10^{-7} \mathrm{Nm}\)
Step-by-step Solution
Detailed explanation
Given, length of bar magnet, \[ l=10 \mathrm{~cm}=10^{-1} \mathrm{~m} \] Pole strength, \(m=10^{-3} \mathrm{~A}-\mathrm{m}\) \[ \begin{aligned} & B=4 \pi \times 10^{-3} \mathrm{~T} \\ & \theta=30^{\circ} \end{aligned} \] \(\therefore\) Magnetic dipole moment,…
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