Two very long, straight, parallel wires carry steady currents `I` and `-I` respectively.The distance between the wires is `d`.At a certain instant of time, a point charge `q` is at a point equidistant from the two wires,in the plane of wires.Its instantaneous velocity `vecv` is perpendicular to this plane.The magnitude of the force due to the magnetic field acting on the charge at this instant is:

Options

A. `(mu_(0)Iqv)/(2pid)`
B. `(mu_(0)Iqv)/(pid)`
C. `(2mu_(0)Iqv)/(pid)`
D. zero

Answer

To solve the problem, we need to analyze the situation involving two long, straight, parallel wires carrying currents and the magnetic field they produce, as well as the force experienced by a point charge in that magnetic field.

### Step-by-Step Solution:

1. **Understand the Configuration**:
- We have two long, straight, parallel wires. Wire 1 carries a current \( I \) upwards, and Wire 2 carries a current \( -I \) (downwards).
- The distance between the two wires is \( d \).
- A point charge \( q \) is placed at a point equidistant from both wires, which we can denote as point A.
- The instantaneous velocity \( \vec{v} \) of the charge is perpendicular to the plane formed by the two wires.

2. **Determine the Magnetic Fields**:
- The magnetic field \( \vec{B} \) produced by a long straight wire at a distance \( r \) from it is given by the formula:
\[ B = \frac{\mu_0 I}{2\pi r} \]
- At point A, which is equidistant from both wires, the distance from each wire to point A is \( \frac{d}{2} \).

3. **Calculate the Magnetic Fields at Point A**:
- For Wire 1 (current \( I \)):
\[ B_1 = \frac{\mu_0 I}{2\pi \frac{d}{2}} = \frac{\mu_0 I}{\pi d} \]
- For Wire 2 (current \( -I \)):
\[ B_2 = \frac{\mu_0 (-I)}{2\pi \frac{d}{2}} = -\frac{\mu_0 I}{\pi d} \]
- The direction of \( B_1 \) (due to Wire 1) is out of the plane, and the direction of \( B_2 \) (due to Wire 2) is into the plane.

4. **Net Magnetic Field at Point A**:
- Since \( B_1 \) and \( B_2 \) are equal in magnitude but opposite in direction, they will cancel each other out:
\[ B_{\text{net}} = B_1 + B_2 = \frac{\mu_0 I}{\pi d} - \frac{\mu_0 I}{\pi d} = 0 \]

5. **Calculate the Magnetic Force on the Charge**:
- The magnetic force \( \vec{F} \) on a charge moving in a magnetic field is given by:
\[ \vec{F} = q \vec{v} \times \vec{B} \]
- Since the net magnetic field \( B_{\text{net}} = 0 \), the force on the charge will also be zero:
\[ F = q v B_{\text{net}} = q v \cdot 0 = 0 \]

6. **Conclusion**:
- The magnitude of the force due to the magnetic field acting on the charge at this instant is:
\[ F = 0 \]