Question

If the magnetic moment of magnet is $0.8\times {10}^{-3}A{m}^2$ and the force acting on each pole in a uniform magnetic field strength of $0.48gauss$, is $0.512\times {10}^{-6}N$ . The distance between the poles of the magnet is : (in $cm$)

• A. $25$
• B. $12.5$
• C. $15$
• D. $7.5$

Answer 1

The correct option is D $7.5$

The relation between Torque and force is $\tau =\mu \ast B$

$\tau =F\ast r$

where r is distance $F\ast r=\mu \ast B$

putting values of $F=0.512\times {10}^{-6}N,B=0.48gauss,\mu =0.8\times {10}^{-3}A{m}^2$

$r=\frac{\mu \ast B}{F}=\frac{0.48\ast 0.8\times {10}^{-3}A{m}^2}{0.512\ast {10}^{-6}}=0.075m=7.5cm$