Two long straight horizontal parallel wires one above the other are separated by a distance `'2a'` . If the wires carry equal currents in opposite directions, the magnitude of the magnitude induction in the plane of the wires at a distance `'a'` above the upper wire is

Options

A. `(mu_(0))/(2pia)`
B. `(mu_(0))/(2pia)+(mu_(0)i)/(4pia)`
C. `(mu_(0)i)/(2pia)-(mu_(i))/(4pia)`
D. `(mu_(0)i)/(3pia)`

Answer

To solve the problem of finding the magnetic induction at a point above the upper wire due to two long straight parallel wires carrying equal currents in opposite directions, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Configuration**:
- We have two long straight parallel wires, one above the other, separated by a distance of `2a`.
- Let the upper wire be Wire 1 and the lower wire be Wire 2.
- The distance from the upper wire to the point of interest (point P) is `a`.

2. **Determine the Magnetic Field due to Wire 1**:
- The magnetic field (B1) at a distance `a` above the upper wire (Wire 1) can be calculated using the formula for the magnetic field around a long straight current-carrying wire:
\[ B_1 = \frac{\mu_0 I}{2 \pi a} \]
- Here, `I` is the current flowing through Wire 1, and `μ₀` is the permeability of free space.

3. **Determine the Magnetic Field due to Wire 2**:
- The distance from Wire 2 to point P is `3a` (since the total distance between the wires is `2a` and point P is `a` above Wire 1).
- The magnetic field (B2) at point P due to Wire 2 is given by:
\[ B_2 = \frac{\mu_0 I}{2 \pi (3a)} \]

4. **Direction of the Magnetic Fields**:
- Since the currents in the two wires are in opposite directions, the magnetic fields will also be in opposite directions.
- Assume that the current in Wire 1 is flowing in the positive direction (creating a magnetic field in one direction), while the current in Wire 2 flows in the opposite direction (creating a magnetic field in the opposite direction).

5. **Calculate the Resultant Magnetic Field**:
- The resultant magnetic field (B) at point P is the difference between B1 and B2:
\[ B = B_1 - B_2 \]
- Substitute the expressions for B1 and B2:
\[ B = \frac{\mu_0 I}{2 \pi a} - \frac{\mu_0 I}{2 \pi (3a)} \]

6. **Simplify the Expression**:
- Factor out the common terms:
\[ B = \frac{\mu_0 I}{2 \pi a} \left(1 - \frac{1}{3}\right) \]
- Simplifying the term in parentheses:
\[ 1 - \frac{1}{3} = \frac{2}{3} \]
- Thus, we have:
\[ B = \frac{\mu_0 I}{2 \pi a} \cdot \frac{2}{3} = \frac{\mu_0 I}{3 \pi a} \]

7. **Final Result**:
- The magnitude of the magnetic induction at a distance `a` above the upper wire is:
\[ B = \frac{\mu_0 I}{3 \pi a} \]

### Conclusion:
The final answer is:
\[ \text{Magnitude of magnetic induction} = \frac{\mu_0 I}{3 \pi a} \]