A) \[\frac{{{\mu }_{0}}i}{2\pi a}\tan \frac{\pi }{n}\]
B) \[\frac{{{\mu }_{0}}ni}{2\pi a}\tan \frac{\pi }{n}\]
C) \[\frac{2}{\pi }\frac{ni}{a}{{\mu }_{0}}\tan \frac{\pi }{n}\]
D) \[\frac{ni}{2a}{{\mu }_{0}}\tan \frac{\pi }{n}\]
Correct Answer: B
Solution :
[b] Magnetic field at the centre due to one side \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2i\sin \theta }{r}\] Where \[r=a\cos \theta \] So \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2i\sin \theta }{a\cos \theta }=\frac{{{\mu }_{0}}i}{2\pi a}\tan \theta \] Hence net magnetic field \[{{B}_{net}}=n\times \frac{{{\mu }_{0}}i}{2\pi a}\tan \frac{\pi }{n}\]You need to login to perform this action.
You will be redirected in
3 sec