Question

Two concentric circular coils, each having $10$ turns with radii $0.2m$ and $0.4m$ carry currents $0.2A$ and $0.3A$ respectively in opposite direction. Magnetic field at the centre is

• A. $\left(2/3\right){\mu }_0$
• B. $\left(5/4\right){\mu }_0$
• C. $\left(1/4\right){\mu }_0$
• D. $\left(1/6\right){\mu }_0$

Answer 1

Answer: (B)

$\left(5/4\right){\mu }_0$

The magnetic field at the centre of a circular loop carrying current is given by

$B=n\frac{{\mu }_{0}I}{2R}$, where $n$ is the number of turns.

Thus, the magnetic field due to first coil is $B_1=10\times \frac{{\mu }_0\times 0.2}{2\times 0.2}=5{\mu }_0$

Magnetic field due to second coil is $B_2=10\times \frac{{\mu }_0\times 0.3}{2\times 0.4}=3.75{\mu }_0$

Net magnetic field is $B=B_1-B_2=(5-3.75){\mu }_0=\left(5/4\right){\mu }_0$