When a bar magnet is placed perpendicular to a uniform a magnetic field, it is acted upon by a couple of magnitude $$1.732\times 10^{-5}Nm$$. The angle through which the magnet should be turned so that the couple acting on it becomes $$1.5\times 10^{-5}Nm$$ is

  • A
    $$60^{o}$$
  • B
    $$45^{o}$$
  • C
    $$30^{o}$$
  • D
    $$75^{o}$$

Solution

$$\theta _{1}=\pi /2$$
$$z_{1}=1.732\times 10^{-5}$$
$$z_{2}=1.5\times 10^{-5}$$
$$z=mB\sin\theta $$
$$z_{max}=mB$$
$$=1.732\times 10^{-5}$$
$$z_{2}=mB\sin\theta ^{1}$$
$$1.5\times 10^{-5}=1.732\times 10^{-5}\sin\theta ^{1}$$
$$\sin\theta ^{1}=\dfrac{\sqrt{3}}{2}$$
$$\theta ^{1}=60$$
Angle turned$$=\pi /2-\theta ^{1}$$
$$=30^o$$