Question

A bar magnet of magnetic moment $\overrightarrow{M}$, is placed in a magnetic field of induction $\overrightarrow{B}$. The torque exerted on it is :

• A. $\overrightarrow{M}\cdot \overrightarrow{B}$
• B. $-\overrightarrow{M}\cdot \overrightarrow{B}$
• C. $\overrightarrow{M}\times \overrightarrow{B}$
• D. $-\overrightarrow{B}\cdot \overrightarrow{M}$

Answer 1

Answer: (C)

$\overrightarrow{M}\times \overrightarrow{B}$

Let us suppose a bar magnet of length 2l is placed inside the magnetic field(B).

Torque(t) = Force acting $\times$ perpendicular distance(d).

Force acting $=Bm$

$\sin \theta =\frac{d}{2l}$ $T=Bm\times d$

$d=2l\sin \theta$ $T=Bm.2l\sin \theta$

$M=m\times 2l$ $T=MB\sin \theta$

$T=-\overrightarrow{B}\times \overrightarrow{M}$ or $T=\overrightarrow{M}\times \overrightarrow{B}$

Where ; $B=$ Magnetic field of induction.

$M=$ magnetic moment.

$m=$ pole strength of bar magnet.

$d=$ perpendicular distance.

$2l=$ length of bar magnet.

$T=$ torque.

Hence, the answer is $\overrightarrow{M}\times \overrightarrow{B}.$