When the primary current in the spark-coil of a car changes from 4A to zero in `10muS`, an emf of 40,000 V is induced in the secondary. The mutual inductance between the primary and the secondary windings of the spark-coil will be -

Options

A. 1H
B. 0.1H
C. 10 H
D. zero

Explanation

To find the mutual inductance between the primary and secondary windings of the spark coil, we can follow these steps:

### Step 1: Understand the relationship between induced EMF and mutual inductance
The induced electromotive force (EMF) in the secondary winding can be expressed using the formula:
\[ E = M \frac{di}{dt} \]
where:
- \(E\) is the induced EMF,
- \(M\) is the mutual inductance,
- \(\frac{di}{dt}\) is the rate of change of current in the primary winding.

### Step 2: Identify the given values
From the problem, we have:
- The change in current (\(di\)) is from 4 A to 0 A, so:
\[ di = 4 - 0 = 4 \, \text{A} \]
- The time interval (\(dt\)) is given as \(10 \, \mu s = 10 \times 10^{-6} \, s\).
- The induced EMF (\(E\)) is 40,000 V.

### Step 3: Calculate the rate of change of current
The rate of change of current (\(\frac{di}{dt}\)) can be calculated as:
\[ \frac{di}{dt} = \frac{di}{dt} = \frac{4 \, \text{A}}{10 \times 10^{-6} \, \text{s}} = \frac{4}{10 \times 10^{-6}} = 4 \times 10^{5} \, \text{A/s} \]

### Step 4: Rearrange the formula to find mutual inductance
Now, we can rearrange the formula to solve for mutual inductance \(M\):
\[ M = \frac{E}{\frac{di}{dt}} \]

### Step 5: Substitute the values into the formula
Substituting the known values into the equation:
\[ M = \frac{40000 \, \text{V}}{4 \times 10^{5} \, \text{A/s}} = \frac{40000}{400000} = 0.1 \, \text{H} \]

### Step 6: Conclusion
Thus, the mutual inductance between the primary and secondary windings of the spark coil is:
\[ M = 0.1 \, \text{H} \]