Question

A long, thin superconducting wire carrying a 17 A current passes through the center of a thin, 3.0-cm-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?

Answer 1

Answer:

a

 $\frac{dE}{dt} =- 2.72 *10^{15} \ N/C \cdot s$

b

The  direction of the electric field is opposite that of the current              

Explanation:

From the question we are told that

   The current is  $I = 17\ A$

   The diameter of the ring is  $d = 3.0 \ cm = 0.03 \ m$

   

Generally the  radius is mathematically represented as

       $r = \frac{d}{2}$

       $r = \frac{0.03}{2}$

       $r = 0.015 \ m$

The  cross-sectional area is mathematically represented as

       $A = \pi r^2$

=>     $A = 3.142 * (0.015^2)$

=>    $A = 7.07 *10^{-4 } \ m^ 2$

Generally  according to ampere -Maxwell equation we have that

      $\oint \= B \cdot \= ds = \mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt }$

Now given that $\= B = 0$ it implies that

     $\oint \= B \cdot \= ds = 0$

So

    $\mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt } = 0$

Where  $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^{-12 } \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

            $\mu_o$ is the permeability of free space with value  

$\mu_o = 4\pi * 10^{-7} N/A^2$

      $\phi$ is magnetic flux which is mathematically represented as

       $\phi = E * A$

Where E is the electric field strength

  So  

       $\mu_o I + \epsilon_o \mu _o \frac{ d [EA] }{dt } = 0$

=>   $\frac{dE}{dt} =- \frac{I}{\epsilon_o * A }$

=>   $\frac{dE}{dt} =- \frac{17}{8.85*10^{-12} * 7.07*10^{-4} }$

=>   $\frac{dE}{dt} =- 2.72 *10^{15} \ N/C \cdot s$

The  negative  sign shows that the  direction  of  the electric field is opposite that of the current