Two coils P and Q are kept near each other. When no current flows through coil P and current increases in coil Q at the rate `10 A//s`, the emf in coil P is 15 mV. When coil Q carries no current and current of `1.8 A` flows through coil P, the magnetic flux linked with the coil Q is

Options

A. `1.4 mWb`
B. `2.2 mWb`
C. 2.7 mWb
D. 2.9 mWb

Answer

To solve the problem, we need to follow these steps:

### Step 1: Understand the given data
- We have two coils, P and Q.
- When no current flows through coil P, the current in coil Q increases at a rate of \( \frac{dI}{dt} = 10 \, \text{A/s} \).
- The induced emf in coil P is given as \( E = 15 \, \text{mV} = 15 \times 10^{-3} \, \text{V} \).
- When coil Q carries no current, coil P carries a current of \( I_P = 1.8 \, \text{A} \).

### Step 2: Calculate the mutual inductance (M)
The mutual inductance can be calculated using the formula for induced emf:
\[ E = -M \frac{dI_Q}{dt} \]
Rearranging the formula gives us:
\[ M = -\frac{E}{\frac{dI_Q}{dt}} \]
Substituting the values:
\[ M = -\frac{15 \times 10^{-3}}{10} = -1.5 \times 10^{-3} \, \text{H} \]
(Note: The negative sign indicates the direction of induced emf according to Lenz's law, but we will consider the magnitude for further calculations.)

### Step 3: Calculate the magnetic flux linked with coil Q
The magnetic flux \( \Phi \) linked with coil Q can be calculated using the formula:
\[ \Phi = M \cdot I_P \]
Substituting the values we found:
\[ \Phi = (1.5 \times 10^{-3}) \cdot (1.8) = 2.7 \times 10^{-3} \, \text{Wb} \]
Converting this to milliWebers:
\[ \Phi = 2.7 \, \text{mWb} \]

### Final Answer
The magnetic flux linked with coil Q is \( \Phi = 2.7 \, \text{mWb} \).

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