Physics - Magnetism Question with Solution | TestHub

The magnetic field at the centre O due to the straight wire segments PQ and RS is zero as O lies on their axes.

The magnetic field at O due to the outer semicircular arc of radius $R_1$ is:

$B_1=\frac{{\mu }_{0}I}{4R_1}$ (directed into the plane)

The magnetic field at O due to the inner semicircular arc of radius $R_2$ is:

$B_2=\frac{{\mu }_{0}I}{4R_2}$ (directed out of the plane)

Since $R_1>R_2$, we have $B_2>B_1$. The net magnetic field at O is the difference between their magnitudes:

$B=B_2-B_1$

$B=\frac{{\mu }_{0}I}{4R_2}-\frac{{\mu }_{0}I}{4R_1}$

$B=\frac{{\mu }_{0}I}{4}\left[\frac{1}{R_2}-\frac{1}{R_1}\right]$

Answer: $\frac{{\mu }_0\mathrm{I}}{4}\left[\frac{1}{{\mathrm{R}}_2}-\frac{1}{{\mathrm{R}}_1}\right]$