Question

A 450-turn solenoid, 24 cm long, has a diameter of 2.2 cm . A 10-turn coil is wound tightly around the center of the solenoid. If the current in the solenoid increases uniformly from 0 to 2.9 A in 0.57 s , what will be the induced emf in the short coil during this time?

Please explain each step. Thanks!

Answer 1

here

If a current I in a coil is changing at a rate dI/dt, then a
voltage across the coil will be induced in the direction to
oppose the change, 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 = μoNI/L. The flux through the short coil is
φ = BA, where the area of the short coil is A.

Then, by Faraday’s law of induction:

emf: V12 = - N2 dφ / dt = - N2 A dB / dt = -(N2) A μo (N1) (dI/dt) /L
where,
N1 = 450 , N2 = 10 , L = 0.24 m, d = 0.022 m,
A = pie * d^2 / 4 = pie * (0.022^2) / 4 = 3.8 * 10^-4 m²

μo = 4*pie * 10^-7 weber /A-m, dI / dt = 2.9 / 0.57 = 5.087 A/s

V12 = -(4 * pie * 10^-7) * (450 ) * (10 ) * (3.8 * 10^-4) * (5.087 ) / 0.24

V12 = 4.55 *10^-5 volts

V12= 4.56x10^-5 volts

the induced emf in the short coil during this time is 4.56 * 10^-5 volts