The magnetic field inside a 5 cm diameter superconducting solenoid varies sinusoidally between 10t and 14t at a frequency of 10 hz.
a) what is the maximum electric field strength at a point 2 cm from the solenoid axis?
b) what is the value of b at the instant e reaches its maximum value?

a) E(max)= 7.5398 v/m

b) B= 2 Tesla.

Explanation:

first we need to know the  law of faraday   look the attachment to see the formula.

if r<R where R= radius of supercondunting solenoid

E(2πr) =   -d * B(πr²) / dt

E= -π/2*dB/dt

B(t)= 2+12sin(20πt)

answer for A question:

dB/dt= 12(20π) cos(20πt)

E= -r/2 (12*20π) cos(20πt)

E= -120πr cos(20πt)

max electric field strenght at:

r=2cm

r=0.02m

E(max)= (120π)*0.02)

E(max)= 7.5398 v/m

answer for question B:

value of B at the instant E reaches its maximum value

when

20πt = π, 3π, 5π,....

B= 2+12sinπ

B= 2 Tesla.

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm