Question

Two concentric circular coils having $10$ turns each are situated in the same plane. Their radii are $20\text{cm}$ and $40\text{cm}$ and they carry $0.2\text{A}$ and $0.3\text{A}$ current in opposite direction respectively. The magnetic field $\left(\text{in}{\text{Wb m}}^{-2}\right)$ at the centre is

• A. $\frac{35}{4}{\mu }_0$
• B. $\frac{{\mu }_0}{80}$
• C. $\frac{7}{80}{\mu }_0$
• D. $\frac{5}{4}{\mu }_0$

Answer 1

The correct option is D $\frac{5}{4}{\mu }_0$

Field at the centre of a coil having $n$ turns is,

$B_c=\frac{n{\mu }_{0}i}{2r}$



${\overrightarrow{B}}_1=\frac{{\mu }_0\times 0.2\times {10}^2\times 10}{2\times 20}\odot$

${\overrightarrow{B}}_2=\frac{{\mu }_0\times 0.3\times {10}^2\times 10}{2\times 40}\otimes$

Net magnetic field at the center is,

$B={\overrightarrow{B}}_1+{\overrightarrow{B}}_2$

$=\frac{{\mu }_0\times {10}^2\times 10}{2}[\frac{0.2}{20}-\frac{0.30}{40}]$

$=50{\mu }_0\times 10[\frac{1}{100}-\frac{3}{400}]$

$=\frac{50\times 10{\mu }_0}{400}=\frac{5{\mu }_0}{4}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.