Question

A solenoid has a radius Rs = 14.0 cm, length L = 3.50 m, and Ns = 6500 turns. The current in the solenoid decreases at the rate of 79.0 A/s. A circular coil with a single turn and radius rc = 20.0 cm encircles the solenoid with its plane perpendicular to the axis of the solenoid. Determine the magnitude of the average induced electric field in the coil.

Answer 1

Answer:

E = 58.7 V/m

Explanation:

As we know that flux linked with the coil is given as

$\phi = NBA$

here we have

$A = \pi R_s^2$

$B = \mu_o N i/L$

now we have

$\phi = N(\mu_o N i/L)(\pi R_s^2)$

now the induced EMF is rate of change in magnetic flux

$EMF = \frac{d\phi}{dt} = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L$

now for induced electric field in the coil is linked with the EMF as

$\int E. dL = EMF$

$E(2\pi r_c) = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L$

$E = \frac{\mu_o N^2 R_s^2 \frac{di}{dt}}{2 r_c L}$

$E = \frac{(4\pi \times 10^{-7})(6500^2)(0.14^2)(79)}{2(0.20)(3.50)}$

$E = 58.7 V/m$