Question

Calculate the induced electric field (in V/m) in a 52-turn coil with a diameter of 12 cm that is placed in a spatially uniform magnetic field of magnitude 0.30 T so that the face of the coil and the magnetic field are perpendicular. This magnetic field is reduced to zero in 0.10 seconds. Assume that the magnetic field is cylindrically symmetric with respect to the central axis of the coil. (Enter the magnitude.)

Answer 1

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

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