Question

A bar magnet of magnetic moment $3.0{\text{Am}}^2$ is placed in a uniform magnetic field of $2\times {10}^{-5}\text{T}$. If each pole experiences a force of $6\times {10}^{-4}\text{N}$, the length of the magnet is

• A. $0.5\text{m}$
• B. $0.3\text{m}$
• C. $0.2\text{m}$
• D. $0.1\text{m}$

Answer 1

The correct option is D $0.1\text{m}$

As we know, magnetic moment $M$ is defined as

$M=ml.........\left(1\right)$

where, $m=$ pole strength and

$l=$ length of the magnet.

Force on each pole is given by

$F=mB$

$\Rightarrow 6\times {10}^{-4}=m\times 2\times {10}^{-5}$

$\therefore m=30\text{Am}$

From $\left(1\right)$ we get,

$l=\frac{M}{m}=\frac{3}{30}=0.1\text{m}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.