A) \[\frac{{{\mu }_{0}}I}{2d}\left[ 1+\frac{2}{\pi } \right]\]
B) \[\frac{{{\mu }_{0}}{{I}^{2}}}{2\pi d}\]
C) \[\frac{{{\mu }_{0}}I}{\pi d}\]
D) \[\frac{{{\mu }_{0}}I}{2d}\]
Correct Answer: A
Solution :
The bent wire can be represented as below Circular part is ABC with the centre P We have to find B atthis point, The magnetic field due to part MA is \[{{B}_{1}}=\frac{1}{2}\frac{{{\mu }_{0}}I}{2\pi \left( \frac{d}{2} \right)}=\frac{{{\mu }_{0}}I}{2\pi d}\] The same magnetic field will be produced due to part NC is, \[{{B}_{2}}=\frac{{{\mu }_{0}}I}{2\pi d}\] The field at P due circular part ABC is \[{{B}_{3}}=\frac{{{\mu }_{0}}I}{2d}\] Now, the net magnetic field\[B={{B}_{1}}+{{B}_{2}}+{{B}_{3}}=\frac{{{\mu }_{0}}I}{2d}\left[ 1+\frac{2}{\pi } \right]\]You need to login to perform this action.
You will be redirected in
3 sec