The coefficient of mutual inductance of two circuits `A` and `B` is `3 mH` and their respective resistanaces are `10` and `4 Omega`. How much current should change in `0.02 s` in circuit `A`, so that the induced current in `B` should be `0.0060 A`?

Options

A. `0.24 A`
B. `1.6 A`
C. `0.18 A`
D. `0.16 A`

Explanation

To solve the problem step by step, we will use the relationship between mutual inductance, induced EMF, and the change in current.

### Step 1: Understand the given values
- Coefficient of mutual inductance, \( M = 3 \, \text{mH} = 3 \times 10^{-3} \, \text{H} \)
- Resistance of circuit \( A \), \( R_A = 10 \, \Omega \)
- Resistance of circuit \( B \), \( R_B = 4 \, \Omega \)
- Induced current in circuit \( B \), \( I_B = 0.0060 \, \text{A} = 6 \times 10^{-3} \, \text{A} \)
- Time interval, \( dt = 0.02 \, \text{s} \)

### Step 2: Calculate the induced EMF in circuit B
The induced EMF (\( E \)) in circuit \( B \) can be calculated using Ohm's law:
\[ E = I_B \times R_B \]
Substituting the values:
\[ E = (6 \times 10^{-3} \, \text{A}) \times (4 \, \Omega) = 24 \times 10^{-3} \, \text{V} = 0.024 \, \text{V} \]

### Step 3: Relate induced EMF to the change in current in circuit A
The induced EMF can also be expressed in terms of mutual inductance and the rate of change of current in circuit \( A \):
\[ E = M \frac{dI}{dt} \]
Where \( dI \) is the change in current in circuit \( A \).

### Step 4: Rearranging the equation to find \( dI \)
From the equation above, we can rearrange to find \( dI \):
\[ dI = \frac{E \cdot dt}{M} \]

### Step 5: Substitute the known values
Now, substituting the values we have:
\[ dI = \frac{(24 \times 10^{-3} \, \text{V}) \cdot (0.02 \, \text{s})}{3 \times 10^{-3} \, \text{H}} \]

### Step 6: Calculate \( dI \)
Calculating the above expression:
\[ dI = \frac{(24 \times 10^{-3}) \cdot (0.02)}{3 \times 10^{-3}} = \frac{0.00048}{0.003} = 0.16 \, \text{A} \]

### Final Answer
The change in current in circuit \( A \) should be \( dI = 0.16 \, \text{A} \).

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