A) 20 m
B) 2 m
C) 3 cm
D) 20 cm
Correct Answer: D
Solution :
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
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