Two coplanar circular coils of equal radius carrying currents `i_(1),i_(2)` in opposite directions are at a large distance `'d'`. The distance from the first coil where the resultant magnetic induction is zero is

Options

A. `d/1+i_(2)/i_(1)`
B. `d/1+sqrt(i_(2)/i_(1))`
C. `d/1+(i_(2)/i_(1))^(1/3)`
D. `d/1+(i_(2)/i_(1))^(2)`

Answer

To find the distance from the first coil where the resultant magnetic induction is zero, we can follow these steps:

### Step-by-Step Solution:

1. **Visualize the Setup**:
- We have two coplanar circular coils, Coil 1 and Coil 2, separated by a distance 'd'.
- Let the current in Coil 1 be \( i_1 \) and in Coil 2 be \( i_2 \). The currents are in opposite directions.

2. **Define the Point of Interest**:
- Let point P be at a distance \( x \) from Coil 1 and \( d - x \) from Coil 2.

3. **Magnetic Field Due to a Circular Coil**:
- The magnetic field \( B \) at a point on the axis of a circular coil carrying current \( i \) is given by:
\[ B = \frac{\mu_0 i a^2}{2 (x^2 + a^2)^{3/2}} \]
- Here, \( a \) is the radius of the coil.

4. **Approximation**:
- Since \( d \) is much greater than \( a \), we can approximate \( x^2 + a^2 \) as \( x^2 \) when \( x \) is large compared to \( a \).
- Thus, the magnetic field at point P due to Coil 1 becomes:
\[ B_1 = \frac{\mu_0 i_1 a^2}{2 x^3} \]
- Similarly, for Coil 2, the magnetic field at point P is:
\[ B_2 = \frac{\mu_0 i_2 a^2}{2 (d - x)^3} \]

5. **Setting the Magnetic Fields Equal**:
- For the resultant magnetic induction to be zero at point P, the magnitudes of the magnetic fields due to both coils must be equal:
\[ B_1 = B_2 \]
- This leads to:
\[ \frac{\mu_0 i_1 a^2}{2 x^3} = \frac{\mu_0 i_2 a^2}{2 (d - x)^3} \]
- The \( \mu_0 \) and \( a^2 \) terms cancel out:
\[ i_1 (d - x)^3 = i_2 x^3 \]

6. **Rearranging the Equation**:
- Expanding and rearranging gives:
\[ i_1 (d - x)^3 = i_2 x^3 \]
- Taking the cube root of both sides:
\[ (d - x) = \left(\frac{i_2}{i_1}\right)^{1/3} x \]

7. **Solving for x**:
- Rearranging gives:
\[ d = x + \left(\frac{i_2}{i_1}\right)^{1/3} x \]
- Factoring out \( x \):
\[ d = x \left(1 + \left(\frac{i_2}{i_1}\right)^{1/3}\right) \]
- Thus, solving for \( x \):
\[ x = \frac{d}{1 + \left(\frac{i_2}{i_1}\right)^{1/3}} \]

### Final Answer:
The distance from the first coil where the resultant magnetic induction is zero is:
\[ x = \frac{d}{1 + \left(\frac{i_2}{i_1}\right)^{1/3}} \]