A long solenoid of radius a, carrying n turns per unit length, is looped by a wire with resistance R, as shown in Fig. 7.28.

(a) If the current in the solenoid is increasing at a constant rate $\mathbf{(}\mathbf{dl}\mathbf{/}\mathbf{dt}\mathbf{=}\mathbf{k}\mathbf{)}\mathbf{,}$, what current flows in the loop, and which way (left or right) does it pass through the resistor?

(b) If the current$\mathit{l}$in the solenoid is constant but the solenoid is pulled out of the loop (toward the left, to a place far from the loop), what total charge passes through the resistor?

The radius of the solenoid is a.

The number of the turns per unit length is n.

The resistance of the wire is R.

The current in the solenoid is increasing at a constant rate,$k=\frac{dl}{dt}$

(a)

The magnetic field inside the solenoid is given by,

$\mathbf{B}\mathbf{=}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{nl}$

Here$l$is the current in the solenoid and${\mu }_0$is the magnetic permeability.

The area of the solenoid is given by,

$A={\mathrm{\pi a}}^2$

The magnetic flux inside the solenoid is given by,

$\varphi =B.A$

Substitute ${\mathrm{\pi a}}^2$for A and${\mu }_{0}nl$for B into above equation.

$$\begin{gathered} \varphi ={\mu }_{0}nl.{\mathrm{\pi a}}^2 \\ \varphi =\pi {\mu }_{0}nla \end{gathered}$$

The induced emf into the solenoid is given by,

$\mathbf{\epsilon }\mathbf{=}\mathbf{-}\frac{\mathbf{d\varphi }}{\mathbf{dt}}$

Substitute $\pi {\mu }_{0}n/a^2$for $\varphi$into above equation.

localid="1657610039816" $$\begin{gathered} \epsilon =-\frac{d}{dt}\left(\pi {\mu }_{0}n/a^2\right) \\ \epsilon =-\pi {\mu }_{0}n{a}^2\frac{dl}{dt} \end{gathered}$$

Substitute$\frac{dl}{dt}$for K into above equation.

$\epsilon =-{\mathrm{\pi \mu }}_0{\mathrm{na}}^2\mathrm{k}$

The current in the loop is given by the ohm’s law as,

$l=\frac{\epsilon }{R}$

Substitute $-\pi {\mu }_{0}n{a}^{2}k$for $\epsilon$into above equation.

$l=\frac{-\pi {\mu }_{0}n{a}^{2}k}{R}$

Hence the magnitude of the current in the loop is$\frac{\pi {\mu }_{0}n{a}^{2}k}{R}.$

The direction of the magnetic field of solenoid in toward the right. Therefore, the direction of the magnetic field of the loop should be in the opposite direction of the solenoid magnetic field.

Therefore, the direction of the current is in toward the left or counter clockwise.

(b)

When solenoid is pulled out, the magnitude of the flux remains the same but the direction becomes the opposite. Therefore, change in the flux,

$∆\varphi =2{\mu }_{0}nl{\mathrm{\pi a}}^2$

The current in term of charge is given by,

$l=\frac{dQ}{dt}$ …… (1)

The current in terms of resistance and voltage is given by,

$l=\frac{\epsilon }{R}$ …… (2)

Equate equation (1) and (2).

$\frac{dQ}{dt}=\frac{\epsilon }{R}$

Substitute$-\frac{d\varphi }{dt}$for into above equation.

$$\begin{gathered} \frac{dQ}{dt}=-\frac{1}{R}\frac{d\varphi }{dt} \\ ∆Q=-\frac{∆\varphi }{R} \end{gathered}$$

The magnitude of the charge,

$∆Q=\frac{∆\varphi }{R}$

Substitute$2{\mu }_{0}nl{\mathrm{\pi a}}^2$ forlocalid="1657611561575" $∆\varphi$into above equation.

$∆Q=\frac{2{\mu }_{0}nl{\mathrm{\pi a}}^2}{R}$

Hence the charge passing through the resistor is$\frac{2{\mu }_{0}nl{\mathrm{\pi a}}^2}{R}$