A) \[4\sqrt{2}\frac{{{\mu }_{0}}}{4\pi }\frac{I}{l}\]
B) \[8\sqrt{2}\frac{{{\mu }_{0}}}{4\pi }\frac{I}{l}\]
C) \[16\sqrt{2}\frac{{{\mu }_{0}}}{4\pi }\frac{I}{l}\]
D) \[32\sqrt{2}\frac{{{\mu }_{0}}}{4\pi }\frac{I}{l}\]
Correct Answer: B
Solution :
\[B=4\left[ \frac{{{\mu }_{0}}}{4\pi }\cdot \frac{I}{a}(\sin {{\phi }_{1}}+\sin {{\phi }_{2}}) \right]\] \[=4\times \frac{{{\mu }_{0}}}{4\pi }\frac{I}{(l/2)}\left[ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \right]\] (Here,\[a=\frac{l}{2},\,\,{{\phi }_{1}}={{\phi }_{2}}={{45}^{o}}\]) \[=4\times \frac{16}{\sqrt{2}}\left[ \frac{{{\mu }_{0}}}{4\pi }\frac{I}{l} \right]\] \[=8\sqrt{2}\left( \frac{{{\mu }_{0}}}{4\pi }\frac{I}{l} \right)\]
You need to login to perform this action.
You will be redirected in
3 sec
