Two very long straight parallel wires carry steady currents `i` and `2i` in opposite directions. 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 the wires. Its instantaneous velocity `vecv` is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on t he charge at this instant is

Options

A. `(mu_(0)iqv)/(2 pid)`
B. `(mu_(0)iqv)/( pid)`
C. `(3mu_(0)iqv)/( 2pid)`
D. zero

Answer

To solve the problem, we need to determine the magnetic force acting on a point charge \( q \) that is located equidistant from two long parallel wires carrying currents \( i \) and \( 2i \) in opposite directions. The charge is moving with a velocity \( \vec{v} \) that is perpendicular to the plane of the wires.

### Step-by-Step Solution:

1. **Identify the Magnetic Fields**:
- The magnetic field \( B \) created by a long straight wire carrying current \( I \) at a distance \( r \) from the wire is given by the formula:
\[ B = \frac{\mu_0 I}{2\pi r} \]
- For the two wires, the magnetic fields at the point where the charge is located (which is equidistant from both wires) will be calculated.

2. **Determine the Direction of Magnetic Fields**:
- For the wire carrying current \( i \) (let's say it is on the left), the magnetic field at the point will be directed into the plane (using the right-hand rule).
- For the wire carrying current \( 2i \) (on the right), the magnetic field at the same point will also be directed into the plane.
- Since both currents are in opposite directions, the magnetic fields at the point will add up.

3. **Calculate the Total Magnetic Field**:
- Let the distance from each wire to the point charge be \( r = \frac{d}{2} \).
- The magnetic field due to the wire with current \( i \):
\[ B_1 = \frac{\mu_0 i}{2\pi \frac{d}{2}} = \frac{\mu_0 i}{\pi d} \]
- The magnetic field due to the wire with current \( 2i \):
\[ B_2 = \frac{\mu_0 (2i)}{2\pi \frac{d}{2}} = \frac{\mu_0 (2i)}{\pi d} \]
- The total magnetic field \( B \) at the point is:
\[ B = B_1 + B_2 = \frac{\mu_0 i}{\pi d} + \frac{\mu_0 (2i)}{\pi d} = \frac{3\mu_0 i}{\pi d} \]

4. **Calculate the Magnetic Force on the Charge**:
- The magnetic force \( \vec{F} \) on a charge \( q \) moving with velocity \( \vec{v} \) in a magnetic field \( \vec{B} \) is given by:
\[ \vec{F} = q \vec{v} \times \vec{B} \]
- Since the velocity \( \vec{v} \) is perpendicular to the plane of the wires and the magnetic field \( \vec{B} \) is directed into the plane, the angle \( \theta \) between \( \vec{v} \) and \( \vec{B} \) is \( 90^\circ \).

5. **Magnitude of the Force**:
- The magnitude of the force can be calculated as:
\[ F = q v B \sin(90^\circ) = q v B \]
- Substituting for \( B \):
\[ F = q v \left(\frac{3\mu_0 i}{\pi d}\right) \]

### Final Answer:
The magnitude of the force due to the magnetic field acting on the charge \( q \) at this instant is:
\[ F = \frac{3\mu_0 q i v}{\pi d} \]