Question

An alternating current $\mathbf{l}\mathbf{=}{\mathbf{l}}_{\mathbf{0}}\mathbf{cos}\left(\mathrm{wt}\right)$flows down a long straight wire, and returns along a coaxial conducting tube of radius a.

(a) In what direction does the induced electric field point (radial, circumferential, or longitudinal)?

(b) Assuming that the field goes to zero as $\mathbf{s}\mathbf{\to }\mathbf{\infty }$, find$\mathbf{E}\mathbf{=}\mathbf{(}\mathbf{s}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{.}$

Short answer

(a) The induced electrical field point is longitudinal.

(b) The expression for the electric field is$\frac{{\mu }_0}{2\mathrm{\pi }}{I}_0\omega \sin \omega tin\left(\frac{a}{s}\right).$

Step-by-step solution

Write the given data from the question.

The alternating current, $l=l_0\cos \left(\omega t\right)$

The radius of the tube is a.

Find the direction of the electrical field.

(a)

The relation between the electric field and magnetic field is given by,

$\mathbf{\nabla }\mathbf{\times }\mathbf{E}\mathbf{=}\mathbf{‐}\frac{\mathbf{\partial }\mathbf{B}}{\mathbf{\partial }\mathbf{t}}$

Here, E is the electrical field and B is the magnetic field.

It is given that the alternating current flows down and a long wire and return along a coaxial conducting tube of radius a . Therefore, the magnetic field I circumferential.

The equation analogous to the relationship between the electric and magnetic field to the magnetic field and current is given as,

$\nabla \times B={\mu }_{0}J$

Therefore, induced electrical field point is longitudinal.

Calculate the expression for E(s,t).

(b)

Consider the Ampere loop outside the coaxial conducting tube.

Here l is the length of the loop, a is the radius of the tube, l is the current flowing through straight wire,

According to the stake’s theorem,

$\int ds.(\nabla \times E)=\int E.dl$

Substitute$-\frac{dB}{dt}$for$\nabla \times E$into above equation.

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

The electric field inside the conductor,

localid="1657526376455" $$\begin{gathered} EI=-\frac{d}{dt}\int B.da \\ EI=-\frac{d}{dt}{\int }_s^a\frac{{\mu }_0}{2\mathrm{\pi }}\frac{I}{s\text{'}}Ids\text{'} \\ E=-\frac{{\mu }_0}{2\mathrm{\pi }}\frac{dI}{dt}(In{(s\text{'}))}_s^a \\ E=-\frac{{\mu }_0}{2\mathrm{\pi }}\frac{dI}{dt}in\left(\frac{a}{s}\right) \end{gathered}$$

The alternating current is given as,

$I=I_0\cos \left(\omega t\right)$

Differentiate the above equation with respect to$t.$

$\frac{dI}{dt}=-I_0\omega \sin \omega t$

Substitute $-I_0\omega \sin \omega t$ for$\frac{dI}{dt}$ into equation (1).

$$\begin{gathered} E=-\frac{{\mu }_0}{2\mathrm{\pi }}(-I_0\omega \sin \omega t)in\left(\frac{a}{s}\right) \\ E=\frac{{\mu }_0}{2\mathrm{\pi }}{I}_0\omega \sin \omega tin\left(\frac{a}{s}\right) \end{gathered}$$

Hence the expression for the electric field is$\frac{{\mu }_0}{2\mathrm{\pi }}{I}_0\omega \sin \omega tin\left(\frac{a}{s}\right).$