Given Information:
Number of turns = N = 52
Diameter of coil = d = 12 cm = 0.12 m
Time = t = 0.10 seconds
Magnetic field = B = 0.30 T
Required Information:
Induced electric field = E = ?
Answer:
Induced electric field = E = 4.68 V/m
Step-by-step explanation:
The Maxwell's third equation can be used to find out the induced electric field,
∫E.dl = -dΦ/dt
Where E is the induced electric field, dl is the circumference of the loop and dΦ/dt is the rate of change of magnetic flux and is given by
Φ = NABcos(θ)
Where N is the number of turns, A is the area of coil and B is the magnetic field and cos(θ) = 1
Φ = NAB
∫E.dl = -dΦ/dt
E(2πr) = -d(NAB)/dt
E =1/(2πr)*-d(NAB)/dt
E =NA/(2πr)*-dB/dt
Area is given by
A = πr²
E =Nπr²/(2πr)*-dB/dt
E =Nr/2*-dB/dt
The magnetic field reduce from 0.30 to zero in 0.10 seconds
E =Nr/2*-(0.30 - 0)/(0 - 0.10)
E =Nr/2*-(0.30)/(-0.10)
E = Nr/2*-(-3)
The radius r is given by
r = d/2 = 0.12/2 = 0.06 m
E = (52*0.06)/2*(3)
E = 1.56*3
E = 1.56*3
E = 4.68 V/m
Therefore, the induced electric field in the coil is 4.68 V/m