Question

A current I flows in a long straight wire with its cross-section having the form of a this half ring of radius R. O is the centre of its mid-section perpendicular to its length. Assume the wire extends to infinite length on either side of the mid-section. Then the magnetic induction field B at the centre O will be :

• A. $B=\frac{{\mu }_{0}i}{\pi R}$
• B. $B=\frac{{\mu }_{0}i}{2\pi R}$
• C. $B=\frac{{\mu }_{0}i}{{\pi }^{2}R}$
• D. $B=\frac{{\mu }_{0}i}{2R}$

Answer 1

The correct option is C $B=\frac{{\mu }_{0}i}{{\pi }^{2}R}$

Let semicircular ring of radius $R$ as shown in figure$,$

An elementary length $dl$ cut for finding magnetic field $.$

so$,dl=Rd\theta$

so$,$ elementary current along $dl$ is $di=i/\pi R\times Rd\theta =id\theta /\pi$

Now, you can see that elementary part of length is perpendicular upon dB

so, dB =

Now, magnetic field along axis $B=2dB$

Now$,$ put $dB$ in here$,$

So$,B=\frac{\mu ₀i}{\pi ²R}$

Hence,

option $\left(C\right)$ is correct answer.