Two circular loops of radius 6 cm and 8 cm are kept in Y-Z plane with their centres at (0, 0, 0) and (14 cm, 0, 0). Current in the first loop is 1 A and is in an anticlockwise direction, when seen from the second loop. Calculate the magnitude and direction of current in the second loop such that net magnetic field at point (8 cm, 0, 0) is zero.

Answer

To solve the problem, we need to find the magnitude and direction of the current in the second loop such that the net magnetic field at the point (8 cm, 0, 0) is zero. We will use the formula for the magnetic field due to a circular loop and apply the right-hand rule to determine the direction of the currents.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Radius of the first loop \( R_1 = 6 \, \text{cm} = 0.06 \, \text{m} \)
- Radius of the second loop \( R_2 = 8 \, \text{cm} = 0.08 \, \text{m} \)
- Current in the first loop \( I_1 = 1 \, \text{A} \) (anticlockwise when viewed from the second loop)
- Distance from the center of the first loop to the point (8 cm, 0, 0) is \( x_1 = 8 \, \text{cm} = 0.08 \, \text{m} \)
- Distance from the center of the second loop to the point (8 cm, 0, 0) is \( x_2 = 14 \, \text{cm} - 8 \, \text{cm} = 6 \, \text{cm} = 0.06 \, \text{m} \)

2. **Magnetic Field Due to the First Loop:**
The magnetic field \( B_1 \) at a distance \( x_1 \) from the center of a circular loop is given by:
\[ B_1 = \frac{\mu_0 I_1 R_1^2}{2 (x_1^2 + R_1^2)^{3/2}} \]
Substituting the values:
\[ B_1 = \frac{\mu_0 \cdot 1 \cdot (0.06)^2}{2 \left((0.08)^2 + (0.06)^2\right)^{3/2}} \]

3. **Magnetic Field Due to the Second Loop:**
The magnetic field \( B_2 \) at a distance \( x_2 \) from the center of the second loop is given by:
\[ B_2 = \frac{\mu_0 I_2 R_2^2}{2 (x_2^2 + R_2^2)^{3/2}} \]
We need to find \( I_2 \) such that \( B_1 + B_2 = 0 \) (i.e., \( B_2 = -B_1 \)).

4. **Setting the Magnetic Fields Equal:**
\[ \frac{\mu_0 \cdot 1 \cdot (0.06)^2}{2 \left((0.08)^2 + (0.06)^2\right)^{3/2}} = \frac{\mu_0 I_2 (0.08)^2}{2 \left((0.06)^2 + (0.08)^2\right)^{3/2}} \]
Canceling \( \mu_0 \) and \( 2 \) from both sides:
\[ I_2 \cdot (0.08)^2 = (0.06)^2 \]

5. **Solving for \( I_2 \):**
\[ I_2 = \frac{(0.06)^2}{(0.08)^2} = \frac{36}{64} = \frac{9}{16} \, \text{A} \]

6. **Direction of Current in the Second Loop:**
Since the current in the first loop is anticlockwise when viewed from the second loop, the current in the second loop must be clockwise to produce a magnetic field in the opposite direction to \( B_1 \).

### Final Answer:
- Magnitude of current in the second loop \( I_2 = \frac{9}{16} \, \text{A} \)
- Direction: Clockwise