Answer to Question #189708 in Electricity and Magnetism for Hamza Waseem

Question

A potential difference $V$ is applied between the faces of a non-magnetic disc, generating a current $I$. The disk has a radius $r$ that is much larger than its thickness $d$.

(a) Just below the curved surface of the disc, what is the direction of the electric and the magnetic fields?

(b) What is the direction and magnitude of the Poynting vector?  

Answer 1

Given,

Applied potential difference $=V$

Current $=I$

Radius of the disc $=r$

Thickness $=d$

$r>>d$

Electric field $(E)=\frac{dV}{dx}$

Hence for this, the electric field will be $(\overrightarrow{E})=\frac{V}{d}$

$\Rightarrow \oint B. dl= \mu_o I$

$\Rightarrow B. 2\pi r = \mu_o I$

$\Rightarrow B = \frac{\mu_o I}{2 \pi r}$

Poynting vector $(\overrightarrow{S})=\frac{EB \sin\theta}{\mu_o}$

Now, substituting the angle between the electric and magnetic field be $90^\circ$

$\overrightarrow{S}=\frac{V\times \mu_o I}{d\mu_o\times 2\pi r}$

$=\frac{VI}{2\pi r d}$