Q. A bar magnet of length 16cm has a pole strength of 500 milli amp.m. The angle at which it should be placed to the direction of external magnetic field of induction 2.5 gauss so that it may experience a torque of √3 ×10^{-5} Nm is

(a) π

(b) π/2

(c) π/3

(d) π/6

Ans: (c)

Sol:

$ \displaystyle \tau = M B sin\theta $

$ \displaystyle sin\theta = \frac{\tau}{M B} $

$ \displaystyle sin\theta = \frac{\sqrt{3}\times 10^{-5}}{500\times 10^{-3}\times 16 \times 10^{-2}\times 2.5 \times 10^{-4}} $

$ \displaystyle sin\theta = \frac{\sqrt{3}}{2} $

$ \displaystyle \theta = \frac{\pi}{3} $