If a current of `3.0` amperes flowing in the primary coil is reduced to zero in `0.001` second, then the induced e.m.f. in the secondary coil is `15000 "volts"`. The mutual inductance between the two coils is

Options

A. (a)`0.5`henry
B. (b)5 henry
C. (c)`1.5`henry
D. (d)10 henry

Answer

To find the mutual inductance between the two coils, we can use the formula for induced electromotive force (e.m.f.) in terms of mutual inductance. The induced e.m.f. (ε) is given by the equation:

\[ \epsilon = -M \frac{dI}{dt} \]

Where:
- \( \epsilon \) is the induced e.m.f. (in volts),
- \( M \) is the mutual inductance (in henries),
- \( \frac{dI}{dt} \) is the rate of change of current (in amperes per second).

### Step 1: Identify the given values
- Current change (\( \Delta I \)): The current changes from \( 3.0 \, \text{A} \) to \( 0 \, \text{A} \), so:
\[ \Delta I = 0 - 3.0 = -3.0 \, \text{A} \]
- Time interval (\( dt \)): The time taken for this change is \( 0.001 \, \text{s} \) or \( 1 \times 10^{-3} \, \text{s} \).
- Induced e.m.f. (\( \epsilon \)): Given as \( 15000 \, \text{V} \) or \( 15 \times 10^{3} \, \text{V} \).

### Step 2: Calculate the rate of change of current
The rate of change of current (\( \frac{dI}{dt} \)) can be calculated as:
\[ \frac{dI}{dt} = \frac{\Delta I}{dt} = \frac{-3.0 \, \text{A}}{1 \times 10^{-3} \, \text{s}} = -3000 \, \text{A/s} \]

### Step 3: Substitute values into the e.m.f. equation
Now, substituting the values into the e.m.f. equation:
\[ 15000 = -M \left(-3000\right) \]
This simplifies to:
\[ 15000 = M \times 3000 \]

### Step 4: Solve for mutual inductance (M)
To find \( M \), rearrange the equation:
\[ M = \frac{15000}{3000} = 5 \, \text{H} \]

### Conclusion
The mutual inductance between the two coils is \( 5 \, \text{H} \).