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: \[ \mathcal{E} = 0.008 \times 110 = 0.88 \, \text{V} \]

Answer 3

This question is about the back emf induced in a coil due to a changing current, using the self-inductance of the coil. The induced emf is given by \( E = L \dfrac{dI}{dt} \), where \( L = 0.008 \, \text{H} \) and \( \dfrac{dI}{dt} = 110 \, \text{A/s} \). Let's check each option against this relation.

1. 0.88 V: Multiplying gives \( 0.008 \times 110 = 0.88 \). This matches the product of the self-inductance and the rate of change of current exactly.
2. 0.78 V: For this to be correct, the rate of change of current would need to be \( 0.78/0.008 = 97.5 \, \text{A/s} \), not the given 110 A/s, so this option does not fit the data.
3. 0.98 V: This would require a rate of change of \( 0.98/0.008 = 122.5 \, \text{A/s} \), again different from the 110 A/s stated in the question, so this option is also inconsistent.
4. None: Since one of the listed values does exactly satisfy \( E = L \, dI/dt \), this option cannot be correct, because a valid answer is present among the choices.

Only the value 0.88 V agrees with the direct product of the given self-inductance and rate of change of current, so it is the back emf induced in the coil.

Therefore, the correct answer is 0.88 V.