Question

A long, thin superconducting wire carrying a $15A$current passes through the center of a thin, $2.0cm$-diameter ring. A uniform electric field of increasing strength also passes through the ring, parallel to the wire. The magnetic field through the ring is zero.

$a$. At what rate is the electric field strength increasing?

$b$. Is the electric field in the direction of the current or opposite to the current?

Short answer

a.Here the magnitude wise electric field is increasing at a rate of $\left|\frac{dE}{dt}\right|=5.4\times {10}^{-5}N/C.s$

b. Yes the direction of electric field is opposite to the direction of current.

Step-by-step solution

Part(a) Step 1: Given information

We have been given a wire carrying a current $15A$. A ring of diameter $2.0cm$, a uniform electric filed of increasing strength passes through the ring parallel to the wire. We need to find the rate of increasing electric field

Part(a) Step 2: Simplify 

Let $I$be the current ,$I=15A$and let $D$be the diameter of the ring and let $R$be the radius of the ring

As, $D=2.0cm$.So. $R=\frac{D}{2}=\frac{2.0}{2}=1.0cm$or $R=0.01m$

Now let $A$be the area of the ring. So $A=\mathrm{\pi }\times R^2=\mathrm{\pi }\times {(0.01)}^2$

Using general form of Ampere- Maxwell law we know that ,

$\oint \stackrel{\rightharpoonup }{B}.\stackrel{\rightharpoonup }{ds}=u_{0}I+{\epsilon }_0{u}_0\frac{d\varphi }{dt}$

now here we are given that $B=0$,

Then the term $\oint \stackrel{\rightharpoonup }{B}.\stackrel{\rightharpoonup }{ds}=0$

So $u_{0}I+{\epsilon }_0{u}_0\frac{d\varphi }{dt}=0$( where ${\epsilon }_0=8.85\times {10}^{-12}{C}^2/N.m^2$is the permittivity of free space and $u_0=4\mathrm{\pi }\times {10}^{-7}\mathrm{N}/{\mathrm{A}}^2$is the permeability of free space)

Now On rearranging terms we have,

$$\begin{gathered} \Rightarrow \frac{dE}{dt}=-\frac{I}{{\epsilon }_0\times A} \\ \Rightarrow \frac{dE}{dt}=-\frac{I}{{\epsilon }_0\times \mathrm{\pi }\times {(0.01)}^2} \\ \Rightarrow \frac{dE}{dt}=-\frac{15}{(8.85\times {10}^{-12})\times \mathrm{\pi }\times {(0.01)}^2} \\ \Rightarrow \frac{dE}{dt}=-5.4\times {10}^{15}N/C.s \end{gathered}$$

Part(b) Step 1: Given information

We need to find Is the electric field in the direction of the current or opposite to the current?

Part(b) Step 2: Simplify

Yes, the electric field is opposite to the direction of current

As we have found in the above question that $\frac{dE}{dt}$is negative which clearly indicates that electric field is in opposite direction