A square, single-turn wire loop = 1.00 cm on a side is placed inside a solenoid that has a circular cross section of radius r = 3.00 cm, as shown in the end view of the figure below. The solenoid is 15.0 cm long and wound with 92 turns of wire.

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A square, single-turn wire loop script l = 1.00 cm on a side is placed inside a solenoid that has a circular cross section of radiusr = 3.00 cm, as shown in the end view of the figure below. The solenoid is 15.0 cm long and wound with 92 turns of wire.

image

(a) If the current in the solenoid is 4.00 A, what is the magnetic flux through the square loop?
 T · m2

(b) If the current in the solenoid is reduced to zero in 4.00 s, what is the magnitude of the average induced emf in the square loop?
 V
asked Feb 28, 2012 in Physics by Joey33 ~Expert~ (1,216 points)
    

1 Answer

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Best answer

To find the magnetic flux you must first find the magnetic field of the solenoid.

The magnetic field of a solenoid is:

magnetic field of a solenoid = (permeability of free space) * (# turns)/(length) * current

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

# turns = 92

current = 4 A

length = 15 cm = 0.15 m

pluging these values back in to the equation above we get:

 

magnetic field of a solenoid = ((4*pi )*10^-7 T*m/A) * (92)/(0.15 m) * 4 A

magnetic field of a solenoid = 0.0030829496 T

 

The magnetic flux is given by:

magnetic flux = magnetic field * area

area of square = length^2 = (1 cm)^2 = (0.1 m)^2

magnetic flux = (0.0030829496 T) * (0.1 m)^2 = 3.08 *10^-7 T*m^2

 

Part B:

Emf = - change in magnetic flux / change in time

t = 4s

Emf = 3.08 *10^-7 T*m^2 / 4 s  =     7.7 * 10^-8 V

answered Mar 3, 2012 by pokemonmaster ~Expert~ (3,856 points)
selected Mar 3, 2012 by Joey33
Thanks! Great Answer.

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