Question

A wire loop $PQRSP$ formed by joining two semicircular wires of radii $R_1$ and $R_2$ carries a current $I$ as shown in figure below. The magnitude of magnetic induction at centre $C$ is

• A. $\left(\frac{{\mu }_0}{4}\right)I[\frac{1}{R_2}-\frac{1}{R_1}]$
• B. $\left(\frac{{\mu }_0}{4}\right)I[\frac{1}{R_1}-\frac{1}{R_2}]$
• C. ${\mu }_{0}I[\frac{1}{R_2}-\frac{1}{R_1}]$
• D. ${\mu }_{0}I\left(\frac{1}{R_1}\right)$

Answer 1

The correct option is B $\left(\frac{{\mu }_0}{4}\right)I[\frac{1}{R_1}-\frac{1}{R_2}]$

Magnetic field at the centre of a coil $B_c=\frac{{\mu }_{o}IN}{2r}$

where $N$ is the number of turn.

For semi-circular coil $N=0.5$

So, magnetic field at C due to semicircular coil of radius $R_1$,

$B_1=\frac{{\mu }_{o}I\left(0.5\right)}{2R_1}=\frac{{\mu }_{o}I}{4R_1}$ out of plane of paper

So, magnetic field at C due to semicircular coil of radius $R_2$,

$B_2=\frac{{\mu }_{o}I\left(0.5\right)}{2R_2}=\frac{{\mu }_{o}I}{4R_2}$ into the plane of paper

Magnetic field at C due to section SR and PQ is zero.

So net magnetic field at C $B_{net}=B_1-B_2=\frac{{\mu }_{o}I}{4}[\frac{1}{R_1}-\frac{1}{R_2}]$