A solenoid of radius 2.80 cm has 350 turns and a length of 17.0 cm.

(b) Find the rate at which current must change through it to produce an emf of 72.0 µV.

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+4 votes

Best answer

The formula for the inductance of a solenoid is:

Inductance = [(permeability of free space)*(number of turns)^2 * Area] / length

permeability of free space = (4*pi)*10^-7 T*m/A

number of turns = 350

Area = pi*r^2, r = 2.8 cm = 0.028 m, so Area = pi*(0.028 m)^2

length = 17 cm = 0.17 m

Plugging back in to the formula for inductance of solenoid we get:

Inductance = [((4*pi)*10^-7 T*m/A)*(350)^2 * pi*(0.028 m)^2] / 0.17 m

Inductance = 0.00223 H = 2.23 mH

Part B:

The formula for self induced emf is:

emf = - inductance * (change in current / change in time)

emf = 72 µV = 72*10^-6 V

inductance = 0.00223 H

plugging back into emf formula we get:

72*10^-6 V= - 0.00223 H * (change in current / change in time)

And so

- (change in current / change in time) = (72*10^-6 V) / (0.00223 H)

(change in current / change in time) = 0.0322 A/s = 32.29 mA/s

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