Question

Two very long, straight, parallel wires carry steady currents \(I\) and \(2I\) 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 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. \( \frac{\mu_0 I q v}{d} \)
• B. zero
• C. \( \frac{3\mu_0 I q v}{d} \)
• D. \( \frac{\mu_0 I q v}{2d} \)

Hint:

At midpoint of two parallel wires with same current direction, magnetic fields cancel → net field zero → force zero.

Answer 1

The Correct Option is B

Step 1: Magnetic field due to straight wire.
Magnetic field due to a long straight wire is circular around the wire.
\[ B = \frac{\mu_0 I}{2\pi r} \]

Step 2: Fields at midpoint.
At the midpoint, both wires produce magnetic fields.

Step 3: Direction of magnetic fields.
Using right-hand rule, magnetic fields due to both wires at the midpoint are in opposite directions.

Step 4: Net magnetic field.
Because currents are in same direction, their magnetic fields cancel at midpoint.
\[ B_{\text{net}} = 0 \]

Step 5: Magnetic force formula.
\[ F = qvB \sin\theta \]

Step 6: Substitute \(B = 0\).
\[ F = 0 \]

Step 7: Final conclusion.
\[ \boxed{0} \] Hence, correct answer is option (B).