Question

A bar magnet of magnetic moment $M$ is bent in the form of an arc which makes an angle ${60}^{\circ }$ at the centre of arc. The $\%$ change in the magnetic moment is

• A. $9\%$ increase
• B. $9\%$ decrease
• C. $4.5\%$ decrease
• D. $4.5\%$ increase

Answer 1

The correct option is C $4.5\%$ decrease

Percentage change in the magnetic moment is,

$\Delta M\%=\frac{M_{new}-M}{M}\times 100.......\left(1\right)$

Let,
$m=$ pole strength of the magnet
$2l=$ length of the magnet

Then magnetic moment, $M=m\left(2l\right)$

After bending into an arc of angle $\theta$ (in $\text{radian}$) as shown in the figure, its length becomes $l_{new}$, and

$l_{new}=2l$


From the figure,

$R\theta =2l\Rightarrow R=\frac{2l}{\theta }$, and

$\sin \left(\frac{\theta }{2}\right)=\frac{l}{R}\Rightarrow l=R\sin \left(\frac{\theta }{2}\right)$

The new effective length of the bar magnet is,

$l_{new}=2l=2R\sin \left(\frac{\theta }{2}\right)$

New magnetic moment becomes,

$M_{new}=m\times l_{new}=m\times 2R\sin \left(\frac{\theta }{2}\right)$

$M_{new}=2m\times \frac{2l}{\theta }\times \sin \left(\frac{\theta }{2}\right)[\because R=2l/\theta ]$

$M_{new}=\frac{2M}{\theta }\sin \left(\frac{\theta }{2}\right)$

Given, $\theta =\frac{\pi }{3}$

$\therefore M_{new}=\frac{3M}{\pi }$

Putting the value of $M_{new}$ in eq.$\left(1\right)$ we get,

$\frac{M_{new}-M}{M}\times 100=\frac{3-\pi }{\pi }\times 100\approx -4.5$

Therefore, new magnetic moment decreases by $4.5\%$.

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

Why this question: When a bar magnet of lenght $L$ and magnetic moment $M$ is bent into an arc subtending an angle $\theta$ at it centre, then its magnetic moment becomes, $M_{new}=\frac{2M}{\theta }\sin \left(\frac{\theta }{2}\right)$ Where, $\theta \text{is in radian}$