Question

A wire of length L is bent into a semi-circle and carries a current i. The magnetic induction at the center of the semi-circle is :

• A. ${\mu }_0\pi i/2L$
• B. ${\mu }_0\pi i/4L$
• C. ${\mu }_{0}i/4\pi L$
• D. ${\mu }_{0}i/2\pi L$

Answer 1

The correct option is B ${\mu }_0\pi i/4L$

When a straight wire is bent into a semicircular loop,

Then there are two parts which can produce the magnetic feildat the center one is circular part and another is straight part due to which feild is zero.

Length $L$ is bent into semicircular loop,

Length of wire is circumference of semicirculr coil.

$L=\pi rr=\frac{L}{\pi }$

Concidering a small element $dl$ on current loop.

Magnetic feild due to small current element $Idl$ at center C

Using Biot Savorts law we have

$dB=\frac{{\mu }_0}{4\pi }.\frac{Idlsin90°}{r^2}[Idlperpendicular\therefore \theta =90°]dB=\frac{{\mu }_0}{4\pi }.\frac{Idl}{r^2}$

Net magnetic field at C, due to semi-circular loop,

$B={\int }_{semicircle}^{}\frac{{\mu }_0}{4\pi }.\frac{Idl}{r^2}B=\frac{{\mu }_0}{4\pi }.\frac{I}{r^2}{\int }_{semicircle}^{}dlB=\frac{{\mu }_0}{4\pi }.\frac{I}{r^2}.L$

But $r=\frac{L}{\pi }$

$B=\frac{{\mu }_0}{4\pi }.\frac{IL}{{\left(\frac{L}{\pi }\right)}^2}=\frac{{\mu }_0}{4\pi }.\frac{IL}{L^2}\times {\pi }^{2}B=\frac{{\mu }_{0}I\pi }{4L}$

This is the required expression

So, the correct answer is $\frac{{\mu }_{0}I\pi }{4L}$