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

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

(a) Find its inductance.
 mH

(b) Find the rate at which current must change through it to produce an emf of 72.0 µV.
 mA/s
asked Mar 6, 2012 in Physics by Joey33 ~Expert~ (1,216 points)
    

1 Answer

+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

answered Mar 6, 2012 by kirby ~Expert~ (3,020 points)
selected Mar 7, 2012 by Joey33

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