Answer:
The magnetic field required is [tex]B= 2.59 *10^{-15}T[/tex]
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
[tex]\frac{\Delta E}{\Delta t} = \frac{E_2 - E_1}{t_2- t_1 }[/tex]
from the question [tex]E_1 = 0 \ N/C[/tex]
[tex]E_2 = 1000 \ N/C[/tex]
[tex]t_1 =0s[/tex]
[tex]t_2 = 6.00s[/tex]
Now substituting this into the equation
[tex]\frac{\Delta E}{\Delta t} = \frac{1000-0}{6.0 - 0}[/tex]
[tex]=166.67 N/C \cdot s[/tex]
Maxwell state mathematically that
[tex]\int\limits {B} \, dl = \mu_o \epsilon_o \int\limits {\frac{\Delta E}{\Delta t} } \, dS[/tex]
=> [tex]{B} \int\limits \, dl = \mu_o \epsilon_o {\frac{\Delta E}{\Delta t} } \int\limits \, dS[/tex]
[tex]{B} (l) = \mu_o \epsilon_o {\frac{\Delta E}{\Delta t} } (S)[/tex]
Where [tex]l[/tex] is the length which is mathematically represented as
[tex]l = 2\pi r[/tex]
And S is the the surface area which is mathematically represented as
[tex]S = \pi r^2[/tex]
And [tex]permeability \ of \ free \ space * permitivity \ of \ free \ space = \frac{1}{(speed \ of \ light)^2 }[/tex]
i.e [tex]\mu_o * \epsilon_o = \frac{1}{c^2}[/tex] and the value of the speed of light is [tex]c = 3.0 *10^{8} \ m/s^2[/tex]
Now substituting these into the equation
[tex]B (2 \pi r ) = \frac{1}{c^2} \frac{\Delta E}{\Delta t } (\pi r^2)[/tex]
[tex]B = \frac{1}{c^2} \frac{\Delta E}{\Delta t} [\frac{\pi r^2}{2 \pi r} ][/tex]
[tex]B= \frac{1}{c^2} \frac{\Delta E }{\Delta t} [\frac{r}{2} ][/tex]
From the question we are told that the diameter [tex]d = 0.560m[/tex]
The radius would be [tex]r = \frac{d}{2} = \frac{0.056}{2} =0.28m[/tex]
Substituting in to the formula for B
[tex]B = \frac{1}{3.0*10^8} * (166.67) [\frac{0.28}{2} ][/tex]
[tex]B= 2.59 *10^{-15}T[/tex]
This question involves the concepts of the electric field and the magnetic field.
The induced magnetic field around a circular area with a diameter of 0.56 m is "2.59 x 10⁻¹⁶ T".
From Maxwell's equation:
[tex]B(C)=\epsilon_o \mu_o\frac{\Delta E}{\Delta T}(A)[/tex]
where,
B = induced magnetic field = ?
C = circumference of circle = πd = π(0.56 m) = 1.76 m
c = speed of light = 3 x 10⁸ m/s
[tex]\epsilon_o \mu_o = \frac{1}{c^2}[/tex] = [tex]\frac{1}{(3\ x\ 10^{8}\ m/s)^2}=1.11\ x\ 10^{-17}\ s^2/m^2[/tex]
ΔE = change in electric field = 1000 N/C
ΔT = time interval = 6 s
A = surface area = [tex]\pi \frac{d^2}{4} = \pi \frac{(0.56\ m)^2}{4}=0.246\ m^2[/tex]
Therefore,
[tex]B=\frac{(1.11\ x\ ^{-17}\ s^2/m^2)(1000\ N/C)(0.246\ m^2)}{(6\ s)(1.76\ m)}[/tex]
B = 2.59 x 10⁻¹⁶ T
Learn more about the electric field here:
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