91影视

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

The radius of the tube is a.

(a)

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

$\mathbf{鈭�}\mathbf{脳}\mathbf{E}\mathbf{=}\mathbf{鈥�}\frac{\mathbf{鈭�}\mathbf{B}}{\mathbf{鈭�}\mathbf{t}}$

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,

$鈭�脳B={渭}_{0}J$

Therefore, induced electrical field point is longitudinal.

(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鈥檚 theorem,

$鈭�ds.(鈭�脳E)=鈭�E.dl$

Substitute$-\frac{dB}{dt}$for$鈭�脳E$into above equation.

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

The electric field inside the conductor,

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

The alternating current is given as,

$I=I_0\cos \left(蝇t\right)$

Differentiate the above equation with respect to$t.$

$\frac{dI}{dt}=-I_0蝇\sin 蝇t$

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

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

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