Question

A tightly wound, long solenoid is kept with its axis parallel to a large metal sheet carrying a surface current. The surface current through a width $dl$ of the sheet is $Kdl$ and the number of turns per unit length of the solenoid is $n$. The magnetic field near the centre of the solenoid is found to be zero. Find the current in the solenoid.

• A. $\frac{2n}{K}$
• B. $\frac{n}{K}$
• C. $\frac{K}{n}$
• D. $\frac{K}{2n}$

Answer 1

The correct option is D $\frac{K}{2n}$

As the magnetic field is zero at the center of the solenoid, it means the magnitude of magnetic fields due to both solenoid and metal sheet at the centre of solenoid are same and opposite in direction.

$\therefore |{\overrightarrow{B}}_{\text{plate}}|=|{\overrightarrow{B}}_{\text{solenoid}}|$

Field inside the solenoid is,

$B_{\text{solenoid}}={\mu }_{0}ni........\left(1\right)$

Using Ampere's circuital law, we can find the magnetic field due to the plate.

Now assuming a square amperian loop as shown in figure and applying Ampere's circuital law on it


$\oint {\overrightarrow{B}}_{\text{plate}}\cdot \overrightarrow{dl}={\mu }_0\left(I_{enclosed}\right)$

$B_{\text{plate}}\times 2l={\mu }_0\int Kdl[\because I=\int Kdl]$

$B_{\text{plate}}\times 2l={\mu }_{0}Kl$

$B_{\text{plate}}=\frac{{\mu }_{0}K}{2}......\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get,

$\frac{{\mu }_{0}K}{2}={\mu }_{0}ni$

$\therefore i=\frac{K}{2n}$

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