Question

The current following through an inductance coil of self inductance 6 mH at different time instants is as shown. The emf induced between t = 20s and t = 40s is nearly

• A. 2 × 10^-2 V
• B. 3 × 10^-4 V
• C. 4 × 10^-3 V
• D. 30 × 10² V

Answer 1

The Correct Option is B

The equations describe the induced electromotive force (emf) in a circuit:

\[ |e| = L \frac{dI}{dt} \]

The magnitude of the emf (\(|e|\)) is given by the product of inductance (\(L\)) and the rate of change of current (\(\frac{dI}{dt}\)).

\[ |e| = 6 \times 10^{-3} \left[ \frac{I_2 - I_1}{t_2 - t_1} \right] \]

Substituting the values, the emf is calculated using the discrete current change over time.

\[ |e| = 6 \times 10^{-3} \left[ \frac{1}{40 - 20} \right] \]

Plugging in the specific time interval (\(t_1 = 20\), \(t_2 = 40\)) and assuming the current change (\(\Delta I = 1\)).

\[ e = 3 \times 10^{-4} \text{ V} \]

The final result for the induced emf is \(3 \times 10^{-4}\) volts.

Answer 2

The self-inductance of the coil is given as $L = 6 \, \text{mH} = 6 \times 10^{-3} \, \text{H}$.

The induced emf in an inductor is given by Faraday's Law of induction:

$e = -L \frac{dI}{dt}$

From the graph, we need to find the rate of change of current $\frac{dI}{dt}$ between $t = 20 \, \text{s}$ and $t = 40 \, \text{s}$.

At $t = 20 \, \text{s}$, the current $I_1 = 4 \, \text{A}$.
At $t = 40 \, \text{s}$, the current $I_2 = 3 \, \text{A}$.
The change in current $\Delta I = I_2 - I_1 = 3 - 4 = -1 \, \text{A}$.
The change in time $\Delta t = 40 - 20 = 20 \, \text{s}$.
The rate of change of current $\frac{dI}{dt} \approx \frac{\Delta I}{\Delta t} = \frac{-1 \, \text{A}}{20 \, \text{s}} = -0.05 \, \text{A/s}$.

Now, we calculate the induced emf:

$e = -L \frac{dI}{dt} = -(6 \times 10^{-3} \, \text{H}) \times (-0.05 \, \text{A/s})$

$e = 6 \times 10^{-3} \times 0.05 \, \text{V}$

$e = 0.3 \times 10^{-3} \, \text{V} = 3 \times 10^{-4} \, \text{V}$.