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

The electromotive force (emf) induced in an inductor is defined by the equation: \[ \mathcal{E} = L \frac{dI}{dt} \] Here, \( L \) represents the inductance of the inductor, \( I \) denotes the current, and \( \frac{dI}{dt} \) signifies the rate of change of current. Given that the current rises linearly from 0 A to 2 A over a period of 40 seconds, the rate of current change is calculated as: \[ \frac{dI}{dt} = \frac{2 \, \text{A}}{40 \, \text{s}} = 0.05 \, \text{A/s} \] Substituting this value into the emf equation yields: \[ \mathcal{E} = L \times 0.05 \] With a known induced emf of 5 mV, equivalent to \( 5 \times 10^{-3} \, \text{V} \), we can determine the inductance \( L \): \[ 5 \times 10^{-3} = L \times 0.05 \] \[ L = \frac{5 \times 10^{-3}}{0.05} = 10^{-1} \, \text{H} = 0.1 \, \text{H} \] Subsequently, the magnetic flux \( \Phi \) linked with the inductor at any given time \( t \) is expressed as: \[ \Phi = L \times I \] At \( t = 10 \) seconds, the current is determined by: \[ I = \frac{2 \, \text{A}}{40 \, \text{s}} \times 10 \, \text{s} = 0.5 \, \text{A} \] Therefore, the flux at \( t = 10 \, \text{s} \) is: \[ \Phi = 0.1 \times 0.5 = 0.05 \, \text{Wb} \]