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

To determine the magnitude of the force due to the magnetic field acting on the point charge \( q \), we start by analyzing the magnetic fields created by the two wires. The magnetic field \( B \) at a distance \( d \) from a long straight wire carrying current \( I \) is given by the formula:

\( B = \frac{\mu_0 I}{2\pi d} \)

where \( \mu_0 \) is the permeability of free space.

Step 1: Calculate the magnetic field created by each wire at the position of the charge.

• The magnetic field due to the wire carrying current \( i \) at distance \( r/2 \) (since the charge is equidistant):
  \( B_1 = \frac{\mu_0 i}{2\pi (r/2)} = \frac{\mu_0 i}{\pi r} \)
• The magnetic field due to the wire carrying current \( 2i \) at the same distance \( r/2 \):
  \( B_2 = \frac{\mu_0 2i}{2\pi (r/2)} = \frac{2\mu_0 i}{\pi r} \)

Both fields have opposite directions since currents are in opposite directions.

Step 2: Determine the net magnetic field \( B_{\text{net}} \) acting at the charge's location.

Since the magnetic fields due to the two wires are equal in magnitude and opposite in direction, they cancel each other out:
\( B_{\text{net}} = B_2 - B_1 = 0 \)

Step 3: Calculate the magnetic force on the charge.

The magnetic force \( F \) on a charge moving with velocity \( v \) in a magnetic field \( B \) is given by:
\( F = qvB \sin \theta \)
where \( \theta \) is the angle between the velocity \( v \) and the magnetic field vector.

In this case, since the net magnetic field is zero, the force is:
\( F = 0 \)

Conclusion: The magnitude of the force due to the magnetic field acting on the charge at this instant is zero.

Answer 2

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: \[ \text{(A) zero} \]