Q15P - Introduction to Electrodynamics

Question

A long solenoid with radius a and n turns per unit length carries a time-dependent current $\mathbf{l}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ in the $\hat{\mathbf{\varphi }}$ direction. Find the electric field (magnitude and direction) at a distance s from the axis (both inside and outside the solenoid), in the quasistatic approximation.

Step-by-Step Solution

Answer

The electric field at distance inside the loop is $\frac{-{\mu }_{0}ns}{2}\frac{dI}{dt}\hat{\varphi }$ and outside the loop is $\frac{-a^2{\mu }_{0}n}{2s}\frac{dI}{dt}\hat{\varphi }.$

Step 1: Write the given data from the question.

The radius of the solenoid is a.

The number of the turns per unit length is n.

The time independent current in the solenoid is l(t).

Step 2; Determine the magnetic field inside and outside the solenoid at distance s.

The magnetic field outside of the solenoid is zero.

 

The magnetic flux inside the solenoid is given by,

 $\varphi =B.A$

Here is the magnetic field and A is the area of the loop.

 

The magnetic field inside the solenoid is given by, 

 $\mathbf{B}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{=}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{nl}$

Here I is the current in the solenoid and  is the magnetic permeability.

 

Consider the loop below,

                           

Here s is the distance.

 

The change in the magnetic flux is given by,

$$\begin{gathered} -\frac{d\varphi }{dt}=\int E.dl \\ ‐\frac{d\varphi }{dt}=E\int dl ............................\left(1\right) \end{gathered}$$

Here is the small unit of the length of the loop and is the electric field.

 

Substitute for into equation (1).

 $E\int dl=-\frac{d}{dt}(B.A) ...........\left(2\right)$

When the distance s is smaller than the radius of the solenoid. Then the area of the loop is given by,

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

 

Substitute ${\mathrm{\pi s}}^2$ for A , $2\mathrm{\pi s}$ for $dl$ and ${\mu }_{0}nl$ for B into equation (2).

$$\begin{gathered} \mathit{E}\mathbf{\left(}\mathbf{2}\mathit{\pi }\mathit{s}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{d}}{\mathbf{d}\mathbf{t}}\mathbf{(}{\mathbf{\mu }}_{\mathbf{0}}\mathit{n}\mathit{l}\mathbf{.}\mathit{\pi }{\mathit{s}}^{\mathbf{2}}\mathbf{)} \\ \mathit{E}\mathbf{\left(}\mathbf{2}\mathit{\pi }\mathit{s}\mathbf{\right)}\mathbf{=}\mathbf{-}{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{nTTs}}^{\mathbf{2}}\frac{\mathbf{dl}}{\mathbf{dt}} \\ \mathbf{E}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{=}\mathbf{-}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{ns}\frac{\mathbf{dl}}{\mathbf{dt}} \\ \mathbf{E}\mathbf{=}\frac{\mathbf{-}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{ns}}{\mathbf{2}}\frac{\mathbf{dl}}{\mathbf{dt}} \end{gathered}$$

Hence the electric field at distance inside the loop is $\frac{-{\mu }_{0}ns}{2}\frac{dI}{dt}$ and direction electric field is same as the current i.e. $\hat{\varphi }$.

 

When the distance is greater than the radius of the solenoid. Then the area of the loop is given by,

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

 

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

$$\begin{gathered} E\left(2\mathrm{\pi s}\right)=-\frac{\mathrm{d}}{\mathrm{dt}}({\mathrm{\mu }}_0\mathrm{nI}.{\mathrm{\pi a}}^2) \\ \mathrm{E}\left(2\mathrm{\pi s}\right)=-{\mathrm{\mu }}_0{\mathrm{n\pi a}}^2\frac{\mathrm{dI}}{\mathrm{dt}} \\ \mathrm{E}\left(2\mathrm{\pi s}\right)=-{\mathrm{\mu }}_0{\mathrm{n\pi a}}^2\frac{\mathrm{dI}}{\mathrm{dt}} \\ \mathrm{E}=\frac{-{\mathrm{a}}^2{\mathrm{\mu }}_0\mathrm{n}}{2\mathrm{s}}\frac{\mathrm{dI}}{\mathrm{dt}} \end{gathered}$$

Hence the electric field at distance outside the loop is $\frac{-a^2{\mu }_{0}n}{2s}\frac{dI}{dt}$ and direction electric field is same as the current i.e., $\hat{\varphi }.$