Question

A short magnet is arranged with its axis along east-west. A compass box is placed on its axial line at a distance of $20cm$ from centre. If the deflection is $45$. Find moment of magnet $(B_H=0.4\times {10}^{-4}T)$.

• A. $0.8A{M}^2$
• B. $1.6A{M}^2$
• C. $2.4A{M}^2$
• D. $3.2A{M}^2$

Answer 1

Answer: (B)

$1.6A{M}^2$

$\frac{m}{B_H}=\frac{d^3}{2}\times \tan \Theta$

$m=$moment of magnet

$d=$distance of compass box from centre of magnet=$.02m$

$\Theta =$angle of deflection=${45}^0$

$\frac{m}{0.4\times {10}^{-4}}={10}^{7\times \frac{{0.2}^3}{2}\times \tan 45}$

$m=1.6A{M}^2$