Question

A magnetic dipole placed in two perpendicular magnetic fields $\overrightarrow{B}$ and ${\overrightarrow{B}}_{\circ }$ is in equilibrium making an angle $\theta$ with $\overrightarrow{B}$ then

• A. $B=B_{\circ }$
• B. $B\cos \theta =B_{\circ }\sin \theta$
• C. $B\sin \theta =B_{\circ }\cos \theta$
• D. $B=B_{\circ }\tan \theta$

Answer 1

The correct option is C $B\sin \theta =B_{\circ }\cos \theta$

For equilibrium of dipole net torque on the loop should be zero.
So, $\overrightarrow{Z}=\overrightarrow{M}\times \overrightarrow{B}$
${\overrightarrow{Z}}_1={\overrightarrow{Z}}_o$
$\overrightarrow{M}\times \overrightarrow{B}=\overrightarrow{M}\times {\overrightarrow{B}}_o$
$MB\sin \theta =M{B}_o\sin ({90}^o-\theta )$
$B\sin \theta =B_o\cos \theta$ $(\because \sin ({90}^o-\theta )=\cos \theta )$