Question

A wire loop $PQRS$ formed by joining two semi-circular wires of radii $R_1$ and $R_2$ carries a current $I$ as shown in the following diagram. The magnetic induction at the centre $O$ is

• A. $\frac{{\mu }_{0}I}{4R_1}$
• B. $\frac{{\mu }_{0}I}{4R_2}$
• C. $\frac{{\mu }_0}{4\pi }I(\frac{1}{R_1}-\frac{1}{R_2})$
• D. $\frac{{\mu }_0}{4}I(\frac{1}{R_1}+\frac{1}{R_2})$

Answer 1

Answer: (C)

$\frac{{\mu }_0}{4\pi }I(\frac{1}{R_1}-\frac{1}{R_2})$

The resultant magnetic field at O will be the sum of the magnetic fields due to the current in the two semicircles, and we can use the expression for the magnetic field at the center of a current loop to find B.

The resultant magnetic field at point O is given as B = B1+B2

The magnetic field at the center of a current loop is B = $\frac{{\mu }_{0}I}{2\pi R}$, where R is the radius of the loop.

The magnetic field at the center of half a current loop B = $\frac{1}{2}\frac{{\mu }_{0}I}{2R}=\frac{{\mu }_{0}I}{4R}$.

Therefore,

B1 = $\frac{{\mu }_{0}I}{4R1}$ and B2 = - $\frac{{\mu }_{0}I}{4R2}$.

So, B = B1+B2 = $\frac{{\mu }_{0}I}{4R1}$ - $\frac{{\mu }_{0}I}{4R2}$.

That is, $\frac{{\mu }_{0}I}{4}(\frac{1}{R1}-\frac{1}{R2})$.

Hence, the magnetic induction at the center is $\frac{{\mu }_{0}I}{4}(\frac{1}{R1}-\frac{1}{R2})$.