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} $