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$=\pi /2-\theta ^{1}$
$=30^o$