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 arou

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?

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Đan Thu 3 years 2021-07-20T15:08:40+00:00 2 Answers 27 views 0

Answers ( )

  1. thugiang
    0
    2021-07-20T15:10:26+00:00
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    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

                 

             

  2. Nho
    0
    2021-07-20T15:10:28+00:00
    Reply

    Answer:

    The magnitude of the induced magnetic field is 2.6×10⁻¹⁶T

    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 = 8.854×10⁻¹²C²/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

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