Question

A wire PQRS shown in Fig. carries a current l. The radius of the circular part of the wire r. The magnetic field at the centre O of the circular part of the wire is given by

• A. $\frac{1}{8}\frac{{\mu }_{0}I}{r}$
• B. $\frac{1}{4}\frac{{\mu }_{0}I}{r}$
• C. $\frac{3}{8}\frac{{\mu }_{0}I}{r}$
• D. $\frac{1}{2}\frac{{\mu }_{0}I}{r}$

Answer 1

The correct option is C

$\frac{3}{8}\frac{{\mu }_{0}I}{r}$

According to Biot-Savart law, the magnetic field at a point due to current element Idl is
$dB=\frac{{\mu }_0}{4\pi }.\frac{Idlsin\theta }{r^2}$
For straight portions PQ and RS of the wire, $\theta =0$. Hence magnetic field at O due to PQ and RS will be zero. For the circular portion QR, we have (since $\theta ={90}^{\circ }$)
$dB=\frac{{\mu }_0}{4\pi }.\frac{Idl}{r^2}\therefore B=\frac{{\mu }_{0}I}{4\pi r^2}{\int }_0^{\frac{3}{4}\times 2\pi r}dl=\frac{{\mu }_{0}I}{4\pi r^2}\times \frac{3}{4}\times 2\pi r=\frac{3}{8}\frac{{\mu }_{0}I}{r}$
Hence the correct choice is (c)