Question

A barmagnet of pole strength 2 A-m is kept in a magnetic field of induction $4\times {10}^{-5}Wb/m^2$ such that the axis of the magnet makes an angle 30? with the direction of the field. The couple acting on the magnet is found to be $80\times {10}^{-7}Mm.$ Then. the distance between the poles of the magnet is:

• A. 20 m
• B. 2 m
• C. 3 cm
• D. 20 cm

Answer 1

Answer: (D)

Given: Pole strength m = 2 A-m Magnetic field induction $B=4\times {10}^{-5}Wb/m^2$ Axis makes angle with direction $\theta ={30}^o$ of magnetic field The torque of couple acting on magnet $\tau =80\times {10}^{-17}Nm$ The torque acting on the magnet $\tau =MB\sin \theta$ $\tau =m\times 2lB\sin \theta$ $\Rightarrow$ $2l=\frac{\tau }{mB\sin \theta }$ $\Rightarrow$ $2l=\frac{80\times {10}^{-7}}{2\times 4\times {10}^{-5}\times \sin {30}^o}$ $\Rightarrow$ $2l=\frac{80\times {10}^{-7}}{2\times 4\times {10}^{-5}\times \frac{1}{2}}$ $=20\times {10}^{-2}$ = 0.2 m = 20 cm