Question:
Two very long, straight, parallel wires carry steady currents I and -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 two wires, in the plane of the wires. Its instantaneous velocity is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is
Two very long, straight, parallel wires carry steady currents I and -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 two wires, in the plane of the wires. Its instantaneous velocity is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is
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For opposite currents, magnetic fields at the mid-point add up instead of canceling.
Updated On: Mar 19, 2026
- (mu₀ I q v)/(2π d)
- (mu₀ I q v)/(π d)
- (2mu₀ I q v)/(π d)
- (mu₀ I q v)/(π d)
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The Correct Option is C
Solution and Explanation
Step 1: Magnetic field due to one long straight wire at distance r:
B = (mu₀ I)/(2π r)
Here, the point is equidistant from both wires:
r = (d)/(2)
Step 2: Magnetic field due to each wire:
B₁ = B₂ = (mu₀ I)/(π d)
Step 3: Since currents are opposite, magnetic fields are in the same direction and add up:
Bₙet = (2mu₀ I)/(π d)
Step 4: Magnetic force on moving charge:
F = qvB
F = qv · (2mu₀ I)/(π d)
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