Question

If the work done in turning a magnet of magnetic moment $M$ by an angle of ${90}^o$ from the magnetic meridian is n times the corresponding work done to turn it through an angle of ${60}^o$, then the value of $n$ is

• A. 1
• B. 2
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B 2

Given,

Magnetic moment of magnet ${\theta }_1={90}^0$

${\theta }_2={60}^0$

Work done in these process

We have, $W=-MB(\cos {\theta }_2-\cos {\theta }_1)$
So, $W_1=-MB(\cos {90}^o-\cos 0^o)=MB$
and $W_2=-MB(\cos {60}^o-\cos 0^o)=\frac{1}{2}MB$
As $W_1=n{W}_2$
$\therefore n=\frac{W_1}{W_2}=\frac{MB}{\frac{1}{2}MB}=2$

So the value of $n=2$