In: Physics
A 410-turn solenoid, 21 cm long, has a diameter of 4.0 cm . A 12-turn coil is wound tightly around the center of the solenoid.
If the current in the solenoid increases uniformly from 0 to 5.0 A in 0.51 s , what will be the induced emf in the short coil during this time?
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 = ?oNI/L. The flux through the short coil is
phi = BA, where the area of the short coil is A.
Then, by Faraday’s law of induction:
emf: V12 = -N2 d(phi)/dt = -N2 AdB/dt = -(N2)A ?o(N1)(dI/dt)/L
where,
N1 = 410, N2 = 12, L = 0.21 m, d = 0.04 m,
A = pid²/4 = 1.2566 x 10^-3 m²
?o = 4?x10^-7 weber/A-m, dI/dt = 5/0.51 = 9.80 A/s
V12 = -(4?x10^-7)(410)(12)(1.2566x10^-3)(9.8)/0.21
V12= -3.7 x 10^-5 volts
[Drop the – sign if only magnitude is needed]