Answer:
a
[tex]\frac{dE}{dt} =- 2.72 *10^{15} \ N/C \cdot s[/tex]
b
The direction of the electric field is opposite that of the current
Explanation:
From the question we are told that
The current is [tex]I = 17\ A[/tex]
The diameter of the ring is [tex]d = 3.0 \ cm = 0.03 \ m[/tex]
Generally the radius is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
[tex]r = \frac{0.03}{2}[/tex]
[tex]r = 0.015 \ m[/tex]
The cross-sectional area is mathematically represented as
[tex]A = \pi r^2[/tex]
=> [tex]A = 3.142 * (0.015^2)[/tex]
=> [tex]A = 7.07 *10^{-4 } \ m^ 2[/tex]
Generally according to ampere -Maxwell equation we have that
[tex]\oint \= B \cdot \= ds = \mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt }[/tex]
Now given that [tex]\= B = 0[/tex] it implies that
[tex]\oint \= B \cdot \= ds = 0[/tex]
So
[tex]\mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt } = 0[/tex]
Where [tex]\epsilon _o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85*10^{-12 } \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
[tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
[tex]\phi[/tex] is magnetic flux which is mathematically represented as
[tex]\phi = E * A[/tex]
Where E is the electric field strength
So
[tex]\mu_o I + \epsilon_o \mu _o \frac{ d [EA] }{dt } = 0[/tex]
=> [tex]\frac{dE}{dt} =- \frac{I}{\epsilon_o * A }[/tex]
=> [tex]\frac{dE}{dt} =- \frac{17}{8.85*10^{-12} * 7.07*10^{-4} }[/tex]
=> [tex]\frac{dE}{dt} =- 2.72 *10^{15} \ N/C \cdot s[/tex]
The negative sign shows that the direction of the electric field is opposite that of the current
