Question

The couple acting on a bar magnet of pole strength $2\text{Am}$ when kept in a magnetic field of intensity $10\text{A/m}$, such that axis of the magnet makes an angle ${30}^o$ with the direction of the field is $80\times {10}^{-7}\text{Nm}$. The distance between the poles of the magnet is :

• A. $8\times {10}^{-9}\text{m}$
• B. $\frac{\pi }{2}\text{m}$
• C. $63.36\text{m}$
• D. $\frac{1}{2\pi }\text{m}$

Answer 1

Answer: (A)

$8\times {10}^{-9}\text{m}$

$p=2Am$
$M=10A/B$
$z=80\times {10}^{-7}Nm$
$z=m\times BM$
$m=\text{Magnetic moment}$
$80\times {15}^{-7}=m\sin \theta$
$80\times {10}^{-7}=m\times 10\times \frac{1}{2}$
$m=1.6\times {10}^{-8}$
$m=p\times l$
$1.6\times {10}^{-8}=2\times l$
$l=8\times {10}^{-9}\text{m}$