(a) If the current in the solenoid is 2.00 A, what is the magnetic flux through

Question

(a) If the current in the solenoid is 2.00 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

A square, single-turn wire loop (a) If the current in the solenoid is 2.00 A, what is the magnetic flux through the square loop? T ½ 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 l = 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 21.0 cm long and wound with 92 turns of wire.

Explanation / Answer

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 = 2 A

length = 21 cm = 0.21 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)/(0.21 m) * 2 A

magnetic field of a solenoid = 0.001100495 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 = (0.001100495 T) * (0.1 m)^2 = 1.100 *10^-7 T*m^2

Part B:

Emf = - change in magnetic flux / change in time

t = 4s

Emf = 1.100 *10^-7 T*m^2 / 4 s = 2.75 * 10^-8 V

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