A solenoid has 480 turns, a length of 12 cm, and a radius of 5.0 cm. The current

Question

A solenoid has 480 turns, a length of 12 cm, and a radius of 5.0 cm. The current in the solenoid increases from 0 A to 4.0 A rate in a time of 40 ms. a) What is the final magnetic field strength in the solenoid? b) What is the magnitude of the induced emf while the current is changing? (Assume that the magnetic field is parallel to the axis of the solenoid.) A Solenoid has 520 turns and a cross-sectional area of 2.0 times 10^-3 m^2. A magnetic field from an external source in the solenoid has a time dependence shown in the plot below. Assume the magnetic field is parallel to the axis of the solenoid. What is the induced emf in the coil at the following times a) t = 1.0 ms, b) t = 3.5 ms, c) t = 5.0 ms?

Explanation / Answer

1.

a)

Magnetic field strength inside the solenoid is

B=uoNI/L =(4pi*10-7)*480*4/0.12

B=0.0201 T

b)

Inductance of the solenoid

L=uoN2A/l =(4pi*10-7)(4802)(pi*0.052)/0.12 =0.01895 H

Induced emf

E=L(dI/dt) =0.01895*(4-0)/(40*10-3) =1.89 Volts

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