Two circular loops A and B of radii R and 4R are kept concentrically in the same plane. The loops carry currents `I_(1)` and `I_(2)` respectively. If the magnetic field at a point on the common axis of the loops at a distance 2R from their common centre is zero, the ratio `(I_(1))/(I_(2))` is _________.

Explanation

To solve the problem, we need to find the ratio of the currents \( \frac{I_1}{I_2} \) in two concentric circular loops A and B, where loop A has a radius \( R \) and loop B has a radius \( 4R \). The magnetic field at a point on the common axis of the loops at a distance \( 2R \) from their common center is zero.

### Step-by-Step Solution:

1. **Magnetic Field Due to Loop A**:
The magnetic field \( B_1 \) at a distance \( x \) from the center of a circular loop of radius \( R \) carrying current \( I_1 \) is given by the formula:
\[ B_1 = \frac{\mu_0 I_1 R^2}{2(R^2 + x^2)^{3/2}} \]
Here, we will substitute \( x = 2R \):
\[ B_1 = \frac{\mu_0 I_1 R^2}{2(R^2 + (2R)^2)^{3/2}} = \frac{\mu_0 I_1 R^2}{2(R^2 + 4R^2)^{3/2}} = \frac{\mu_0 I_1 R^2}{2(5R^2)^{3/2}} = \frac{\mu_0 I_1 R^2}{2 \cdot 5^{3/2} R^3} = \frac{\mu_0 I_1}{10\sqrt{5} R} \]

2. **Magnetic Field Due to Loop B**:
The magnetic field \( B_2 \) at the same distance \( x = 2R \) from the center of loop B (which has a radius \( 4R \) and carries current \( I_2 \)) is given by:
\[ B_2 = \frac{\mu_0 I_2 (4R)^2}{2((4R)^2 + (2R)^2)^{3/2}} = \frac{\mu_0 I_2 \cdot 16R^2}{2(16R^2 + 4R^2)^{3/2}} = \frac{\mu_0 I_2 \cdot 16R^2}{2(20R^2)^{3/2}} = \frac{\mu_0 I_2 \cdot 16R^2}{2 \cdot 20^{3/2} R^3} = \frac{\mu_0 I_2 \cdot 16}{2 \cdot 20\sqrt{20}} = \frac{\mu_0 I_2 \cdot 8}{20\sqrt{20}} = \frac{2\mu_0 I_2}{5\sqrt{20}} \]

3. **Setting the Magnetic Fields Equal**:
Since the magnetic field at the point is zero, we have:
\[ B_1 - B_2 = 0 \implies B_1 = B_2 \]
Substituting the expressions we derived:
\[ \frac{\mu_0 I_1}{10\sqrt{5} R} = \frac{2\mu_0 I_2}{5\sqrt{20}} \]

4. **Simplifying the Equation**:
Cancel \( \mu_0 \) and \( R \) from both sides:
\[ \frac{I_1}{10\sqrt{5}} = \frac{2 I_2}{5\sqrt{20}} \]
Rearranging gives:
\[ I_1 = \frac{2 I_2}{5\sqrt{20}} \cdot 10\sqrt{5} \]

5. **Calculating the Ratio**:
\[ I_1 = \frac{2 \cdot 10 \cdot \sqrt{5}}{5 \cdot \sqrt{20}} I_2 \]
Since \( \sqrt{20} = 2\sqrt{5} \):
\[ I_1 = \frac{20\sqrt{5}}{10\sqrt{5}} I_2 = 2 I_2 \]
Thus, the ratio \( \frac{I_1}{I_2} \) is:
\[ \frac{I_1}{I_2} = 2 \]

### Final Answer:
The ratio \( \frac{I_1}{I_2} \) is \( 2 \).

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