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 2l\,B\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