Question

Two wire loops ABCD formed by joining two semicircular wires of radii $R_1$ and $R_2$ carries a current I as shown in figure.The magentic induction at the centre is

• A. ${\mu }_{0}I(R_2-R_1)/4R_1{R}_2$ out of the page
• B. ${\mu }_{0}I(R_2-R_1)/4R_1{R}_2$ into the page
• C. $\frac{{\mu }_{0}I{R}_1{R}_2}{4(R_2-R_1)}$ out of the page
• D. Zero

Answer 1

The correct option is A ${\mu }_{0}I(R_2-R_1)/4R_1{R}_2$ out of the page

Answer is A.
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, $\frac{{\mu }_{0}I}{4}(\frac{1}{R1}-\frac{1}{R2})={\mu }_{0}I\left(\frac{R2-R1}{4R1.R2}\right)$.
Hence, the magnetic induction at the center is ${\mu }_{0}I\left(\frac{R2-R1}{4R1.R2}\right)$ out of the page.