600-turn solenoid, 29 cm long, has a diameter of 2.9 cm . A 13-turn coil is woun

Question

600-turn solenoid, 29 cm long, has a diameter of 2.9 cm . A 13-turn coil is wound tightly around the center of the solenoid.

Part A

If the current in the solenoid increases uniformly from 0 to 4.7 A in 0.70 s , what will be the induced emf in the short coil during this time?

600-turn solenoid, 29 cm long, has a diameter of 2.9 cm . A 13-turn coil is wound tightly around the center of the solenoid.

Part A

If the current in the solenoid increases uniformly from 0 to 4.7 A in 0.70 s , what will be the induced emf in the short coil during this time?

Explanation / Answer

Vcoil = -Ldi/dt, where L is the self
inductance of the coil.

Similarly, if the time varying
magnetic flux of one coil links another coil, then it will
induce a voltage in the 2nd coil equal to V’coil = -MdI/dt
where I is the current in the 1st coil and M is the mutual
inductance of the two coils.

The magnetic field of the solenoid(away from the edges) is
given by:

B = oNI/L. The flux through the short coil is
= BA, where the area of the short coil is A.

Then, by Faraday’s law of induction:

emf: V12 = -N2 d/dt = -N2 AdB/dt = -(N2)A o(N1)(dI/dt)/L
where,
N1 = 600, N2 = 13, L = 0.29 m, d = 0.029 m,

A = d²/4 = (0.029²)/4 =6.60x10^-4 m²

o = 4x10^-7 weber/A-m, dI/dt = 4.7/0.75 = 6.267 A/s

V12 = -(4x10^-7)(600)(13)(6.6x10^-4)(6.267)/0.2...

V12= 2.027x10^-4 volts

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