question_answer

If the magnetic field at P can be written as K tan \[\left( \frac{\alpha }{2} \right)\], then K

A) \[\frac{{{\mu }_{0}}I}{4\pi d}\]

B) \[\frac{{{\mu }_{0}}I}{2\pi d}\]

C) \[\frac{{{\mu }_{0}}I}{\pi d}\]

D) \[\frac{2{{\mu }_{0}}I}{\pi d}\]

Correct Answer: B

Solution :

[b] Let us compute the magnetic field due to any one segment: \[B=\frac{{{\mu }_{0}}i}{4\pi (d\,\sin \,\alpha )}\left( \cos \,{{0}^{o}}+\cos \,(180-\alpha ) \right)\] \[=\frac{{{\mu }_{0}}I}{4\pi (d\,\sin \,\alpha )}(1-\cos \alpha )\] \[=\frac{{{\mu }_{0}}I}{4\pi d}\tan \,\frac{\alpha }{2}\] Resultant field will be \[{{B}_{net}}=2B=\frac{{{\mu }_{0}}I}{2\pi d}\tan \frac{\alpha }{2}\Rightarrow K=\frac{{{\mu }_{0}}I}{2\pi d}\]