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. `0.16 A`
C. `0.18 A`
D. `0.16 A`

Explanation

To solve the problem, we need to find how much current should change in circuit A in order to induce a current of 0.006 A in circuit B. We will use the formula for mutual inductance and the relationship between induced EMF, current, and resistance.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Coefficient of mutual inductance (M) = 3 mH = \(3 \times 10^{-3} \, \text{H}\)
- Resistance of circuit A (R_A) = 10 Ω
- Resistance of circuit B (R_B) = 4 Ω
- Induced current in circuit B (I_B) = 0.006 A

2. **Calculate the Induced EMF in Circuit B:**
The induced EMF (E_B) in circuit B can be calculated using Ohm's Law:
\[ E_B = I_B \times R_B \]
Substituting the values:
\[ E_B = 0.006 \, \text{A} \times 4 \, \Omega = 0.024 \, \text{V} = 24 \times 10^{-3} \, \text{V} \]

3. **Relate Induced EMF to Change in Current in Circuit A:**
The induced EMF in circuit B can also be expressed using the formula:
\[ E_B = -M \frac{dI_A}{dt} \]
Where \(dI_A\) is the change in current in circuit A and \(dt\) is the time interval (0.02 s).

4. **Rearranging the Equation:**
Rearranging the above equation to find \(dI_A\):
\[ dI_A = -\frac{E_B \cdot dt}{M} \]

5. **Substituting the Known Values:**
Now substitute the values we have:
\[ dI_A = -\frac{24 \times 10^{-3} \, \text{V} \times 0.02 \, \text{s}}{3 \times 10^{-3} \, \text{H}} \]
\[ dI_A = -\frac{0.00048 \, \text{V} \cdot \text{s}}{3 \times 10^{-3}} = -0.16 \, \text{A} \]

6. **Final Result:**
The change in current in circuit A should be \(0.16 \, \text{A}\).

### Summary:
The current in circuit A should change by \(0.16 \, \text{A}\) in order to induce a current of \(0.006 \, \text{A}\) in circuit B.