Question

A wire loop is formed by joining two semi-circular wires of radii $R_{1}and{R}_2$ carrying current I as shown in the given diagram. The magnitude of net magnetic field at the centre O is

• A. $\frac{{\mu }_{0}l\pi }{4\pi R_1}$
• B. $\frac{{\mu }_{0}l\pi }{4\pi R_2}$
• C. $\frac{{\mu }_{0}l\pi }{4\pi }[\frac{1}{R_1}-\frac{1}{R_2}]$
• D. $\frac{{\mu }_{0}l\pi }{4\pi }[\frac{1}{R_1}+\frac{1}{R_2}]$

Answer 1

The correct option is D $\frac{{\mu }_{0}l\pi }{4\pi }[\frac{1}{R_1}+\frac{1}{R_2}]$

Magnetic field due to current carrying semi-circular loop at the centre, $B=\frac{{\mu }_0/\pi }{4\pi r},$ where l is the current through the loop and r is the radius of the loop.
$\therefore$ Magnetic field at the centre O due to semi-circular loop of radius $R_1,B_1=\frac{{\mu }_o/\pi }{4\pi R_1}$
Magnetic field at the centre O due to semi-circular loop of radius $R_2,B_2=\frac{{\mu }_o/\pi }{4\pi R_2}$
The magnetic field due to part of wire RS and PQ is zero.
$\therefore$ Net magnetic field at the centre, $B_{net}=B_1+B_2+B_2+0+0$
$=\frac{{\mu }_o/\pi }{4\pi }(\frac{1}{R_1}+\frac{1}{R_2})$
Hence, the correct answer is option (4).