Question

The current in a solenoid decreases steadily from 6 mA to 2 mA in 50 ms. If an average emf of 0.4 V is induced, find the self-inductance of the solenoid.

Hint:

Use \( \mathcal{E} = L \cdot \frac{dI}{dt} \) to calculate self-inductance when emf and rate of change of current are given. Ensure units are in SI (A, s, V).

Answer 1

Given: \[ I_1 = 6\,\text{mA} = 6 \times 10^{-3}\,\text{A}, \quad I_2 = 2\,\text{mA} = 2 \times 10^{-3}\,\text{A} \] \[ \Delta t = 50\,\text{ms} = 50 \times 10^{-3}\,\text{s}, \quad \mathcal{E} = 0.4\,\text{V} \] Formula: \[ \mathcal{E} = L \cdot \frac{|\Delta I|}{\Delta t} \Rightarrow L = \frac{\mathcal{E} \cdot \Delta t}{|\Delta I|} \] \[ \Delta I = I_1 - I_2 = (6 - 2)\times 10^{-3} = 4 \times 10^{-3}\,\text{A} \] \[ L = \frac{0.4 \times 50 \times 10^{-3}}{4 \times 10^{-3}} = \frac{20 \times 10^{-3}}{4 \times 10^{-3}} = 5\,\text{H} \] \[ \boxed{L = 5\,\text{henry}} \]