Question

What is the phenomenon of mutual induction? What is meant by 1 henry mutual inductance? If a current of 4 A is reduced to zero in 10 \(\mu\)s in the primary coil of a transformer, then 40 kV of induced e.m.f. is produced in the secondary coil. Find out the mutual inductance between the primary and the secondary coils.

Hint:

The mutual inductance \( M \) measures the ability of two coils to induce e.m.f. in each other. It depends on the rate of change of current in one coil and the induced voltage in the other.

Answer 1

Mutual Induction is the phenomenon in which a change in current in one coil induces a voltage in another coil that is placed nearby. The amount of voltage induced in the secondary coil depends on the rate of change of current in the primary coil. Mutual induction occurs when two coils are magnetically coupled, and the changing magnetic field from the primary coil induces an electromotive force (e.m.f.) in the secondary coil.
The mutual inductance \( M \) between two coils is defined as the ratio of the induced e.m.f. in one coil to the rate of change of current in the other coil. The formula for mutual inductance is:
\[ M = \frac{\text{Induced e.m.f.}}{\frac{dI}{dt}} \] Where:
- \( M \) is the mutual inductance in henries (H),
- Induced e.m.f. is the voltage induced in the secondary coil,
- \( \frac{dI}{dt} \) is the rate of change of current in the primary coil.
Given:
- Induced e.m.f. = 40 kV = \( 40 \times 10^3 \, \text{V} \),
- Initial current \( I = 4 \, \text{A} \),
- Final current = 0 (current reduces to zero),
- Time interval \( dt = 10 \, \mu \text{s} = 10 \times 10^{-6} \, \text{s} \).
First, calculate the rate of change of current:
\[ \frac{dI}{dt} = \frac{I - 0}{dt} = \frac{4}{10 \times 10^{-6}} = 4 \times 10^5 \, \text{A/s} \] Now, substitute the values into the formula for mutual inductance:
\[ M = \frac{40 \times 10^3}{4 \times 10^5} = 0.1 \, \text{H} \] Thus, the mutual inductance between the primary and secondary coils is \( M = 0.1 \, \text{H} \).