Question

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. ${75}^o$

Answer 1

Answer: (C)

${30}^o$

${\theta }_1=\pi /2$
$z_1=1.732\times {10}^{-5}$
$z_2=1.5\times {10}^{-5}$
$z=mB\sin \theta$
$z_{max}=mB$
$=1.732\times {10}^{-5}$
$z_2=mB\sin {\theta }^1$
$1.5\times {10}^{-5}=1.732\times {10}^{-5}\sin {\theta }^1$
$\sin {\theta }^1=\frac{\sqrt{3}}{2}$
${\theta }^1=60$
Angle turned$=\pi /2-{\theta }^1$
$={30}^o$