Two very long, straight , parallel wires carry steady currents ` I &-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 wires , in the plane of the wires. Its instantaneous vel,ocity ` v` 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_0Iqv)/(2pid)`
B. `(mu_0Iqv)/(pid)`
C. `(2mu_0Iqv)/(pid)`
D. `0`

Explanation

To solve the problem, we need to find the magnitude of the force acting on a point charge \( q \) due to the magnetic field created by two long parallel wires carrying currents \( I \) and \( -I \) respectively. The charge is located at a point equidistant from the wires, and its velocity is perpendicular to the plane of the wires.

### Step-by-Step Solution:

1. **Identify the Configuration**:
- We have two long parallel wires. One wire carries a current \( I \) in the positive x-direction, and the other wire carries a current \( -I \) in the negative x-direction.
- The distance between the wires is \( d \).
- The point charge \( q \) is located at the midpoint between the two wires, which is at a distance \( \frac{d}{2} \) from each wire.

2. **Calculate the Magnetic Field at the Charge's Location**:
- The magnetic field \( B \) due to a long straight current-carrying wire at a distance \( r \) is given by:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
- For the wire carrying current \( I \) at a distance \( \frac{d}{2} \):
\[ B_1 = \frac{\mu_0 I}{2 \pi \left(\frac{d}{2}\right)} = \frac{\mu_0 I}{\pi d} \]
- For the wire carrying current \( -I \) at the same distance \( \frac{d}{2} \):
\[ B_2 = \frac{\mu_0 (-I)}{2 \pi \left(\frac{d}{2}\right)} = -\frac{\mu_0 I}{\pi d} \]
- The net magnetic field \( B \) at the midpoint (where the charge is located) is:
\[ B = B_1 + B_2 = \frac{\mu_0 I}{\pi d} - \frac{\mu_0 I}{\pi d} = 0 \]

3. **Determine the Force on the Charge**:
- The magnetic force \( F \) on a charge \( q \) moving with velocity \( v \) in a magnetic field \( B \) is given by:
\[ F = q v B \sin(\theta) \]
- Here, \( \theta \) is the angle between the velocity vector \( v \) and the magnetic field \( B \). Since \( B = 0 \), the force becomes:
\[ F = q v \cdot 0 \cdot \sin(\theta) = 0 \]

4. **Conclusion**:
- The magnitude of the force acting on the charge \( q \) due to the magnetic field created by the two wires is \( 0 \).

### Final Answer:
The magnitude of the force due to the magnetic field acting on the charge at this instant is \( 0 \).