Question

Two very long straight parallel wires carry currents $ i $ and $ 2i $ in opposite directions. The distance between the wires is $ r $. 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 $ v $ is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is

• A. zero
• B. \( \frac{\mu_0 i}{2\pi r} \cdot qv \)
• C. \( \frac{\mu_0 2i}{2\pi r} \cdot qv \)
• D. \( \frac{\mu_0 i}{2\pi r} \cdot 2qv \)

Hint:

When the magnetic fields from two sources are equal in magnitude but opposite in direction, they cancel each other out, leading to zero net magnetic field and hence zero magnetic force on the charge.

Answer 1

The Correct Option is A

Consider the magnetic field created by each of the two wires at the position of the point charge \( q \). The magnetic field due to a long straight current-carrying wire at a distance \( r \) from the wire is given by: \[ B = \frac{\mu_0 I}{2\pi r} \] where:
- \( \mu_0 \) is the permeability of free space,
- \( I \) is the current in the wire,
- \( r \) is the distance from the wire. In this problem, there are two wires, and they carry currents \( i \) and \( 2i \) in opposite directions. Since the point charge \( q \) is equidistant from both wires, the magnetic fields due to each wire will be in opposite directions, and they will cancel each other out.
Thus, the total magnetic field at the location of the point charge is zero. Since the magnetic field is zero, the magnetic force on the point charge is also zero. The force due to a magnetic field on a moving charge is given by: \[ F = q(\vec{v} \times \vec{B}) \] Since \( B = 0 \), the force acting on the charge is: \[ F = 0 \] Therefore, the correct answer is: \[ \text{(A) zero} \]