Question

A long solenoid has a diameter of 11.1 cm. When a current i exists in its windings, a uniform magnetic field of magnitude B = 42.9 mT is produced in its interior. By decreasing i, the field is caused to decrease at the rate of 6.65 mT/s. Calculate the magnitude of the induced electric field (a) 4.34 cm and (b) 7.46 cm from the axis of the solenoid.

Answer 1

`E(1) = 1.44* 10^(-4) v/m`

`E(2) = 1.34* 10^(-4) v/m`

Step-by-step explanation:

Given data:

diameter of solenoid is 11.1 cm

B = 42.9 mT

dR = 6.65 mT/s

d(1) = 4.34 cm

d(2) = 7.46 cm

we know that

electric field due to solenoid is given as

E = 1/2 dB/dt (r)

`E(1)= (0.0434)/(2) * (6.65* 10^(-3))`

`E(1) = 1.44* 10^(-4) v/m`

E = 1/2 dB/dt (R^2)/r

`E(2)= (0.055^2)/(2* 0.0746) * (6.65* 10^(-3))`

`E(2) = 1.34* 10^(-4) v/m`