Question

A 100-turn coil of radius 1.6 cm and resistance 5.0 \( \Omega \) is co-axial with a solenoid of 250 turns/cm and radius 1.8 cm. The solenoid current drops from 1.5 A to zero in 25 ms. Calculate the current induced in the coil in this duration. (Take \( \pi^2 = 10 \))

Hint:

Induced current depends on the rate of change of magnetic flux and the resistance of the coil.

Answer 1

The induced emf \( \varepsilon \) is given by Faraday’s law of induction: \[ \varepsilon = -N \frac{d\phi}{dt}. \] Here, \( N \) is the number of turns in the coil, and \( \phi \) is the magnetic flux. The flux is related to the magnetic field \( B \) and the area \( A \) of the coil. The magnetic flux changes due to the change in current in the solenoid, causing an induced emf. The induced emf is: \[ \varepsilon = -N \mu_0 n \pi r^2 \frac{dI}{dt}. \] Given values: \( N = 100 \),
\( n = 250 \, \text{turns/cm} \),
\( r = 1.6 \, \text{cm} \),
\( dI = 1.5 \, \text{A} \),
\( dt = 25 \, \text{ms} \).
Substituting the values: \[ \varepsilon = 0.1536 \, \text{V}. \] The current induced in the coil is: \[ I = \frac{\varepsilon}{R} = \frac{0.1536}{5} = 0.03 \, \text{A}. \]