Question

A solenoid with a cross-sectional area of 1.55×10−3 m2 is 0.730 m long and has 965 turns per meter. Find the induced emf in this solenoid if the current in it is increased from 0 to 2.10 A in 30.3 ms . The answer "172 mV" is incorrect.

Answer 1

here,

E = d(BAN)/dt

where B is the magnetic flux density,

A is the cross sectional area,

N is the total number of turns

and t is the time.

The magnitude of the induced emf is directly proportional to the rate of change of flux linkage which is represented by what is on the right hand side of the equation.

Initial B = u0 * n * I0

where u is the permeability of free space,

n is the number of turns per metre and

I is the current

Initial B= (4.0 * pi * 10^-7)(965)(0)
= 0T

Final B= μ1nI1= (4.0 * pi * 10^-7)(965)(2.1)

B = 2.54 * 10^-3 T

Thus,

E = (2.54 * 10^-3) (1.55 * 10^-3) * (965 * 0.73)/ 0.0303

E = 0.0915 V = 91.5 mV