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. $\frac{{\mu }_0}{4}I\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
• D. $\frac{{\mu }_0}{4}I\left(\frac{1}{R_1}+\frac{1}{R_2}\right)$

Answer 1

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

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