Question

Two magnetic isolated north poles each of strength m A m, are placed one at each of the two vertices of an equilateral triangle of side $ \alpha $ . The resultant magnetic induction at the third vertex is: ( $ {{\mu }_{0}} $ is permeability of free space)

• A. $ \frac{{{\mu }_{0}}}{4\pi }\left( \frac{m}{{{a}^{2}}} \right) $
• B. $ \frac{{{\mu }_{0}}}{4\pi }\frac{\sqrt{2}m}{{{a}^{2}}} $
• C. $ \frac{{{\mu }_{0}}}{4\pi }\frac{\sqrt{3}\,m}{{{a}^{2}}} $
• D. $ \frac{{{\mu }_{0}}m}{2\pi {{a}^{2}}} $

Answer 1

Answer: (C)

Magnetic field at the equatorial line when isolated north pole place at B.
$ {{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{m}{{{a}^{2}}} $ Similarly, $ {{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{m}{{{a}^{2}}} $ Now, resultant magnetic field is $ {{B}_{resul\tan t}}=\sqrt{B_{1}^{2}+B_{2}^{2}+2{{B}_{1}}{{B}_{2}}\cos {{60}^{o}}} $ $ =\sqrt{{{B}^{2}}+{{B}^{2}}+2{{B}^{2}}\times \frac{1}{2}} $ $ =\sqrt{3{{B}^{2}}}=\sqrt{3B}=\sqrt{3}\frac{{{\mu }_{0}}}{4\pi }\times \frac{m}{{{a}^{2}}} $