question_answer

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)                  \[\frac{{{\mu }_{0}}m}{2\pi {{a}^{2}}}\]

Correct Answer: C

Solution :

                 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                 \[{{B}_{resul\tan t}}=\sqrt{B_{1}^{2}+B_{2}^{2}+2{{B}_{1}}{{B}_{2}}\cos {{60}^{o}}}\]                 \[=\sqrt{{{B}^{2}}+{{B}^{2}}+2{{B}^{2}}\times \frac{1}{2}}\]                 \[=\sqrt{3{{B}^{2}}}=\sqrt{3B}=\sqrt{3}\frac{{{\mu }_{0}}}{4\pi }\times \frac{m}{{{a}^{2}}}\]