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 \[{{B}_{resul\tan t}}=\sqrt{B_{1}^{2}+B_{2}^{2}+2{{B}_{1}}{{B}_{2}}\cos {{60}^{o}}}\] \[=\sqrt{{{B}^{2}}+{{B}^{2}}+2{{B}^{2}}\times \frac{1}{2}}\] \[=\sqrt{3{{B}^{2}}}=\sqrt{3B}=\sqrt{3}\frac{{{\mu }_{0}}}{4\pi }\times \frac{m}{{{a}^{2}}}\]You need to login to perform this action.
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