Question

A long solenoid has a diameter 12.7 cm. When a current i is passed through its winding, a uniform magnetic field B = 38.2 mT is produced in its interior. By decreasing i, the field is caused to decrease at the rate 6.71 mT/s. Calculate the magnitude of the induced electric field

(a) 1.28 cm from the axis of the solenoid.

V/m

(b) 7.79 cm from the axis of the solenoid

V/m

Answer 1

here,

radius of solenoid, r = 12.7 /2 = 6.35 cm = 0.0635 m

change in field, db/dt = 6.71 mt/s = 0.00671 T/s

Since :
induced electric field, E = indueced emf / 2*pi*d

induced electric field, E = (n * (db/dt)*(pi*r^2) ) / 2*pi*d

Where,
d is distance to point of field
r i radius of solenoid
n is no of coils

part a:
d = 1.28cm = 0.0128 m

induced electric field, E = (1 * (0.00671)*(pi*0.0635^2) ) / (2*pi*0.0128)

induced electric field, E = 0.001057 V/m

part b:
d = 7.79cm = 0.0779 m

induced electric field, E = (1 * (0.00671)*(pi*0.0635^2) ) / (2*pi*0.0779)

induced electric field, E = 0.0017366 V/m