Question 49 of 276

The magnitude of magnetic induction at mid-point $O$ due to current arrangement as shown in figure will be
.

A$\dfrac{{\mu }_{0}I}{2\pi a}$B$0$C$\dfrac{{\mu }_{0}I}{4\pi a}$D$\dfrac{{\mu }_{0}I}{\pi a}$

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Answer: (D) $\dfrac{{\mu }_{0}I}{\pi a}$

The magnetic field at point $O$ due to the two semi-infinite portions of both wires lying along line $BE$ will be $0$.
Using right-hand rule, it can be determined that the magnetic field due to both the semi-infinite portions of wires $BC\text{and}ET$ which are perpendicular to $BE$ will be outwards from the plane.
The magnetic field due to one such portion will be:
$B=\dfrac{{\mu }_{0}I}{4\pi \left(\dfrac{a}{2}\right)}$
Since the magnetic field of both the portions is in same direction, the net magnetic field at $O$ will be:
${B}_{net}=2\times \dfrac{{\mu }_{0}I}{4\pi \left(\dfrac{a}{2}\right)}=\dfrac{{\mu }_{0}I}{\pi a}$

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