A very long, straight solenoid with a cross-sectional area of 2.15 cm^2 is wound

Question

A very long, straight solenoid with a cross-sectional area of 2.15 cm^2 is wound with 90.6 turns of wire per centimeter. Starting at t = 0, the current in the solenoid is increasing according to i(t) = (0.169 A/s^2)t^2. A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2 A ? Express your answer with the appropriate units.

Explanation / Answer

given that

number of turns/m n = 90.6 /cm = 9060 / m

A = 2.15 cm^2 = 2.15*10^(-4) m^2

current (i)t =(0.169 A/s^2)*t^2

we know that

megnetic field for solenoid

B = uo*n*i

B = 4*pi*10^(-7) * 9060 * (0.169)*t^2

B = 1,92*10^(-3)*t^2 T

dB/dt = 1,92*10^(-3) * 2*t T = 3.84*10^(-3)*t T

we know that

e = rate of change of flux

given that ,a secondery winding of 5 turns encircle the solenoid

so , e = -N*A*dB/dt

e = -5 * 2.15 * 10^(-4) * 3.84*10^(-3) *t

e = -4.128*10^(-6) *t V

If t is the time at which the current = 3.2A

3.2 = 0.169*t^2

t = 4.35 s

so magnitude of emf induced

|emf| = 4.128*10^(-6) *t = 4.128*10^(-6) * 4.35

|emf| = 17.96*10^(-6) V

e = 17.96 uV

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