A long, thin solenoid has 700 turns per meter and radius 2.50 cm. The current in

Question

A long, thin solenoid has 700 turns per meter and radius
2.50 cm. The current in the solenoid is increasing at a uniform rate of 39.0 A/s .

Part A

What is the magnitude of the induced electric field at a point near the center of the solenoid?
Express your answer with the appropriate units.

Part B

What is the magnitude of the induced electric field at a point 0.500 cm from the axis of the solenoid?

Express your answer with the appropriate units

Part C

What is the magnitude of the induced electric field at a point 1.00 cm from the axis of the solenoid?

Express your answer with the appropriate units.

Explanation / Answer

Given that -

N = 700 turns/m, r = 2.50 cm = 0.025 m, I = 39.0 A

Write the Maxwell-Faraday equation,
? E ? ds = - ? ?B/?t ? dS

B = (?o)(n)(I)
?B/?t = (?o)(n)(dI/dt)
?B/?t = (1.257e-6 N/A²)(700 turns/m)(+39.0 A/s)
?B/?t = 3.432e-2 T/s

E[2?r] = -(?o)(n)(dI/dt)(?r²)
E = -(?o)(n)(dI/dt)(r) / 2

Part A -

At the center of the solenoid, r = 0

So, E = 0

So, magnitude of electric field = 0

Part B -

r = 0.50 cm = 0.005 m

||E|| = (1.257e-6 N/A²)(700 turns/m)(+39.0 A/s)(0.005 m) / 2
||E|| = 8.58 x 10^-5 N/C

Part C -

r = 1 cm = 0.01

So,
||E|| = (1.257e-6 N/A²)(700 turns/m)(+39 A/s)(0.01 m) / 2
||E|| = 3.43 x 10^-4 N/C

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