Question

What back emf is induced in a coil of self-inductance 0.008 H when the current in the coil is changing at the rate of 110 A/s?

• A. 0.88V
• B. 0.78V
• C. 0.98V
• D. None

Hint:

The induced back emf in a coil is proportional to the rate of change of current and the self-inductance of the coil.

Answer 1

The Correct Option is A

To determine the back electromotive force (emf) induced in a coil, we use the formula involving self-inductance (\(L\)) and the rate of change of current (\(\frac{di}{dt}\)). The formula for the induced emf (\(E\)) is given by:

\[E = -L \cdot \frac{di}{dt}\]

Where:

• \(E\) is the induced emf in volts (V).
• \(L\) is the self-inductance in henrys (H).
• \(\frac{di}{dt}\) is the rate of change of current in amperes per second (A/s).

For this problem, we have:

• \(L = 0.008 \, H\)
• \(\frac{di}{dt} = 110 \, A/s\)

Substitute these values into the formula:

\[E = -0.008 \times 110\]

\[E = -0.88 \, V\]

The negative sign indicates the direction of the induced emf opposes the change in current (as per Lenz's law), but since we are interested in the magnitude of the back emf, we take:

Magnitude of induced emf, \(E = 0.88 \, V\)

Therefore, the back emf induced in the coil is 0.88 V.

Answer 2

The induced emf \( \mathcal{E} \) is given by: \[ \mathcal{E} = L \frac{dI}{dt} \] Where:
- \( L = 0.008 \, \text{H} \),
- \( \frac{dI}{dt} = 110 \, \text{A/s} \). Substituting the values: \[ \mathcal{E} = 0.008 \times 110 = 0.88 \, \text{V} \]