Question

The current through an inductor is uniformly increased from zero to 2 A in 40 s. An emf of 5 mV is induced during this period. Find the flux linked with the inductor at t = 10 s.

Hint:

The flux linked with an inductor is the product of the inductance and the current flowing through it. The induced emf is proportional to the rate of change of current.

Answer 1

We know the emf induced in an inductor is given by: \[ \mathcal{E} = L \frac{dI}{dt} \] where \( L \) is the inductance of the inductor, \( I \) is the current, and \( \frac{dI}{dt} \) is the rate of change of current. The current increases uniformly from 0 to 2 A in 40 seconds, so the rate of change of current is: \[ \frac{dI}{dt} = \frac{2 \, \text{A}}{40 \, \text{s}} = 0.05 \, \text{A/s} \] Substituting this into the equation for emf: \[ \mathcal{E} = L \times 0.05 \] Given that the induced emf is 5 mV or \( 5 \times 10^{-3} \, \text{V} \), we can solve for \( L \): \[ 5 \times 10^{-3} = L \times 0.05 \] \[ L = \frac{5 \times 10^{-3}}{0.05} = 10^{-1} \, \text{H} = 0.1 \, \text{H} \] Now, the flux \( \Phi \) linked with the inductor at any time \( t \) is given by: \[ \Phi = L \times I \] At \( t = 10 \) s, the current is: \[ I = \frac{2 \, \text{A}}{40 \, \text{s}} \times 10 \, \text{s} = 0.5 \, \text{A} \] Thus, the flux at \( t = 10 \, \text{s} \) is: \[ \Phi = 0.1 \times 0.5 = 0.05 \, \text{Wb} \]