A square, single-turn wire loop = 1.00 cm on a side is placed inside a solenoid

Question

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 23.0 cm long and wound with 124 turns of wire.

(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

Explanation / 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 = 124

current = 4 A

length = 23 cm = 0.23 m

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



magnetic field of a solenoid = ((4*pi )*10^-7 T*m/A) * (124)/(0.23 m) * 4 A

magnetic field of a solenoid = 2.71*10-3 T



The magnetic flux is given by:

magnetic flux = magnetic field * area

area of square = length^2 = (1 cm)^2 = (0.01 m)^2

magnetic flux = (2.71*10-3 T) * (0.01 m)^2 = 2.71 *10^-7 T*m^2

Part B:

Emf = - change in magnetic flux / change in time

t = 4s

Emf = 2.71 *10^-7 T*m^2 / 4 s = 6.775 * 10^-8 V

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