Along solenoid of length 9.10 × 10-2 m and cross-sectional area 5.0 × 10-5 m2 contains 6500 turns per meter of length. determine the magnitude of the emf induced in the solenoid when the current in the solenoid changes from 0 to 1.5 a during the time interval from 0 to 0.20 s.

1.8 mV

Explanation:

The total number of turns in the solenoid is

where

n = 6500 is the number of turns per meter of length

is the length of the solenoid

The flux linkage through the solenoid is given by

where

B is the strength of the magnetic field

is the cross-sectional area of the solenoid

The strength of the field in the solenoid is given by

where I is the current.

At the beginning, I = 0, so the field is

And the flux linkage is

Later, the current is I = 1.5 A, so the field in the solenoid is

So, the flux linkage is

So, the change in flux linkage is

And therefore, the emf induced in the solenoid is (in magnitude)