Question

The electric field in a region of space increases from 0 to 1000 N/C in 6.00 s. What is the magnitude of the induced magnetic field B B around a circular area with a diameter of 0.560 m oriented perpendicularly to the electric field?

Answer 1

Answer:

The magnetic field required is $B= 2.59 *10^{-15}T$

Explanation:

From the question we are told that there was a change in electric field with respect to time , this change can be mathematically represented as

               

                                $\frac{\Delta E}{\Delta t} = \frac{E_2 - E_1}{t_2- t_1 }$

from the question  $E_1 = 0 \ N/C$

                               $E_2 = 1000 \ N/C$

                                $t_1 =0s$

                                $t_2 = 6.00s$

Now substituting this into the equation

                      $\frac{\Delta E}{\Delta t} = \frac{1000-0}{6.0 - 0}$

                           $=166.67 N/C \cdot s$

Maxwell state mathematically that

          $\int\limits {B} \, dl = \mu_o \epsilon_o \int\limits {\frac{\Delta E}{\Delta t} } \, dS$

=>        ${B} \int\limits \, dl = \mu_o \epsilon_o {\frac{\Delta E}{\Delta t} } \int\limits \, dS$

          ${B} (l) = \mu_o \epsilon_o {\frac{\Delta E}{\Delta t} } (S)$

Where $l$ is the length which is mathematically represented as

                  $l = 2\pi r$

And  S is the the surface area which is mathematically represented as

                $S = \pi r^2$

And  $permeability \ of \ free \ space * permitivity \ of \ free \ space = \frac{1}{(speed \ of \ light)^2 }$

        i.e  $\mu_o * \epsilon_o = \frac{1}{c^2}$ and the value of the speed of light is $c = 3.0 *10^{8} \ m/s^2$

Now substituting these into the equation

              $B (2 \pi r ) = \frac{1}{c^2} \frac{\Delta E}{\Delta t } (\pi r^2)$

                      $B = \frac{1}{c^2} \frac{\Delta E}{\Delta t} [\frac{\pi r^2}{2 \pi r} ]$

                          $B= \frac{1}{c^2} \frac{\Delta E }{\Delta t} [\frac{r}{2} ]$

From the question we are told that the diameter $d = 0.560m$

  The radius would be $r = \frac{d}{2} = \frac{0.056}{2} =0.28m$

Substituting in to the formula for B

                 $B = \frac{1}{3.0*10^8} * (166.67) [\frac{0.28}{2} ]$

                   $B= 2.59 *10^{-15}T$

             

         

Answer 2

Answer:

The magnitude of the induced magnetic field is 2.6x10⁻¹⁶T

Explanation:

Given:

d = diameter = 0.56 m

r = radius = 0.56/2 = 0.28 m

The magnetic field is:

$B=\frac{\mu \epsilon }{2\pi r} (\pi r^{2} )\frac{dE}{dt} \\\frac{dE}{dt} =\frac{1000}{6} =166.67\frac{N}{Cs}$

Where

μ = 4πx10⁻⁷T m/A

ε = 8.854x10⁻¹²C²/N m²

Replacing:

$B=\frac{4\pi x10^{-7}*8.854x10^{-12} *\pi *(0.28)^{2}*166.67 }{2\pi *0.28} =2.6x10^{-16} T$