A varying current at the rate of `3A//s` in a coil generates an e.m.f. of `8mV` in a nearby coil. The mutual inductance of the two coils is

Options

A. 2.66 mH
B. `2.66xx10^(-3)`mH
C. 2.66 H
D. 0.266 H

Explanation

To find the mutual inductance of the two coils, we can use the formula that relates the induced electromotive force (e.m.f.) in one coil to the rate of change of current in the other coil. The formula is given by:

\[ \text{e.m.f.} = -M \frac{dI}{dt} \]

Where:
- \( M \) is the mutual inductance,
- \( \frac{dI}{dt} \) is the rate of change of current in the primary coil,
- e.m.f. is the induced electromotive force in the secondary coil.

### Step 1: Identify the given values
From the problem, we have:
- e.m.f. in the secondary coil, \( \text{e.m.f.} = 8 \text{ mV} = 8 \times 10^{-3} \text{ V} \)
- Rate of change of current in the primary coil, \( \frac{dI}{dt} = 3 \text{ A/s} \)

### Step 2: Rearrange the formula to solve for mutual inductance \( M \)
We can rearrange the formula to find \( M \):

\[ M = -\frac{\text{e.m.f.}}{\frac{dI}{dt}} \]

Since we are interested in the magnitude, we can ignore the negative sign:

\[ M = \frac{\text{e.m.f.}}{\frac{dI}{dt}} \]

### Step 3: Substitute the values into the equation
Now, substitute the values we identified:

\[ M = \frac{8 \times 10^{-3} \text{ V}}{3 \text{ A/s}} \]

### Step 4: Calculate the mutual inductance
Perform the calculation:

\[ M = \frac{8 \times 10^{-3}}{3} = 2.6667 \times 10^{-3} \text{ H} \]

### Step 5: Convert to milliHenry
Since \( 1 \text{ H} = 1000 \text{ mH} \):

\[ M = 2.6667 \text{ mH} \approx 2.67 \text{ mH} \]

### Conclusion
Thus, the mutual inductance of the two coils is approximately:

\[ M \approx 2.67 \text{ mH} \]

### Final Answer
The mutual inductance of the two coils is \( 2.67 \text{ mH} \).

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