Question

Two concentric circular coils of ten turns each are situated in the same plane. Their radii are 20 and 40 cm and they carry respectively 0.2 and 0.4 ampere current in opposite direction. The magnetic field in weber/m${}^2$ at the centre is

• A. ${\mu }_0/80$
• B. $7{\mu }_0/80$
• C. $\left(5/4\right){\mu }_0$
• D. zero

Answer 1

The correct option is D zero

$\text{The magnetic field due to the circular coil at the center is given by,}$

$\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{2\pi ni}{r}$

$\text{The magnetic field due to the first circular coil is given by,}$

$B_1=\frac{{\mu }_0}{4\pi }\frac{2\pi n{i}_1}{r_1}$

$\text{The magnetic field due to the second circular coil is given by,}$

$B_2=-\frac{{\mu }_0}{4\pi }\frac{2\pi n{i}_2}{r_2}$

$\text{The resultant magnetic field is given by,}$

$B=B_1+B_2$

$B=\frac{{\mu }_0}{4\pi }\frac{2\pi n{i}_1}{r_1}-\frac{{\mu }_0}{4\pi }\frac{2\pi n{i}_2}{r_2}$

$B=\frac{{\mu }_0}{2}[\frac{n{i}_1}{r_1}-\frac{n{i}_2}{r_2}]$

$B=\frac{{\mu }_0}{2}[\frac{10\times 0.2}{0.20}-\frac{10\times 40}{0.40}]$

$B=0$