Question

A magnet $20cm$ long with its poles concentrated at its ends is placed vertically with its north pole on the table. At a point due $20cm$ south (magnetic) of the pole, a neutral point is obtained. If $H=0.3G$, then the pole strength of the magnet is approximately :

• A. $185$ A-cm
• B. $185$ A-m
• C. $18.5$ A-cm
• D. $18.5$ A-m

Answer 1

Answer: (D)

$18.5$ A-m

$d_1=\sqrt{{0.2}^2+{0.2}^2}=0.2\sqrt{2}m{B}_H=0.3G=0.3\times {10}^{-4}T$

$B_N=\frac{{\mu }_o}{4\pi }\frac{m}{{0.2}^2}$

$B_S=\frac{{\mu }_o}{4\pi }\frac{m}{(0.2\sqrt{2}{)}^2}$

$B_H=B_N-B_S\cos {45}^o=\frac{{\mu }_{o}m}{4\pi \left({0.2}^2\right)}[1-\frac{1}{2\sqrt{2}}]$

So, $0.3\times {10}^{-4}=\frac{{10}^{-7}m}{\left({0.2}^2\right)}[1-\frac{1}{2\sqrt{2}}]$

$⟹m=18.5$ A-m