Question 2 A 0.200 mm diameter copper wire makes up a 98.0 turns, 6.90 cm diamet

Question

Question 2 A 0.200 mm diameter copper wire makes up a 98.0 turns, 6.90 cm diameter coil. There is a magnetic field parallel to the axis of the coil. If the induced current in the coil is 8.20 A, what is the rate of change of the magnetic field? 2.52x104 T/s 6.26×10-6 T/s O257 T/s 0.746 Ts Question 3 A 6.80 cm diameter solenoid has a magnetic field inside of 1.10 T. The magnetic field changes at a rate of 4.30 T/s. What is the electric field strength inside the solenoid at a point on the axis? O 0 V/m 0.146 V/m 0.0187 V/m 0 2.15 V/m Click "Submit Answer" below to submit all 3 of your answers.

Explanation / Answer

Q2.

EMF is given by:

EMF = i*R = N*d(phi)/dt

phi = B*A

i*R = N*A*dB/dt

dB/dt = i*R/(N*A)

i = 8.20 Amp

N = 98 turns

A = area of coil = pi*0.069^2/4 = 3.74*10^-3 m^2

R = resistance = rho*L/A1

rho = resistivity of copper = 1.7*10^-8

A1 = surface area of wire = pi*(0.1*10^-3)^2 = 3.14*10^-8 m^2

L = length of wire = N*2*pi*r = 98*pi*0.069 = 21.24 m

R = 1.7*10^-8*21.24/(3.14*10^-8) = 11.499 ohm

Using given values:

dB/dt = 8.20*11.499/(98*3.74*10^-3)

dB/dt = 257.2 T/sec

Correct option is C.

3.

Electric field is given by

E = (r/2)*dB/dt

for a point on the axis, r = 0

So, E = 0 V/m

Correct option is A.

Please Upvote.

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