question_answer

A conductor wire carrying a current \[I\]through it is bent to form U-shape such that its straight arms is separated at a distance d. The magnetic field induction at the centre of circular part of the U-shape will be

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\[B={{B}_{1}}+{{B}_{2}}+{{B}_{3}}=\frac{{{\mu }_{0}}I}{2d}\left[ 1+\frac{2}{\pi } \right]\]