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. (a) Find the current in the solenoid. (b) If the solenoid is rotated to make its axis perpendicular to the metal sheet, what would be the magnitude of the magnetic field near its centre?

Answer 1

Given:
Number of turns per unit length of the solenoid = n
(a) Since the net magnetic field near the centre of the solenoid is 0,
$$\begin{gathered} \therefore {\overrightarrow{B}}_{\mathrm{plate}}={\overrightarrow{B}}_{\mathrm{solenoid}} \\ \mathrm{Using}\mathrm{Guass}\text{'}\mathrm{s}\mathrm{law}\mathrm{for}\mathrm{the}\mathrm{plate},\mathrm{we}\mathrm{get} \\ {B}_{\mathrm{plate}}\times 2l={\mathrm{\mu }}_{0}i={\mathrm{\mu }}_{0}kl \\ {B}_{\mathrm{plate}}=\frac{{\mathrm{\mu }}_{0}k}{2}\dots \left(1\right) \end{gathered}$$

For the solenoid,
$B_{\mathrm{solenoid}}={\mathrm{\mu }}_{0}ni$ ...(2)

From (1) and (2), we get
$i=\frac{k}{2n}$

(b) On putting the value of n in (2), we get
${\overrightarrow{B}}_{\mathrm{solenoid}}=\frac{{\mathrm{\mu }}_{0}k}{2}$
Now, ${\overrightarrow{B}}_{\mathrm{plate}}\mathrm{and}{\overrightarrow{B}}_{\mathrm{solenoid}}$ are perpendicular to each other.
Thus, the net magnetic field near the centre of the solenoid is given by
$$\begin{gathered} B_{\mathrm{net}}=\sqrt{{\left(\frac{{\mathrm{\mu }}_{0}k}{2}\right)}^2+{\left(\frac{{\mathrm{\mu }}_{0}k}{2}\right)}^2} \\ =\frac{{\mathrm{\mu }}_{0}k}{\sqrt{2}} \end{gathered}$$