Question:
Two 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. The instantaneous velocity v is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge is
Two 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. The instantaneous velocity v is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge is
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Equal and opposite magnetic fields cancel at symmetric points.
Updated On: Mar 20, 2026
- \(\dfrac{\mu_0 I q v}{2\pi d}\)
- \(\dfrac{\mu_0 I q v}{\pi d}\)
- \(\dfrac{2\mu_0 I q v}{\pi d}\)
- 0
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The Correct Option is D
Solution and Explanation
Step 1: Magnetic fields due to the two wires at the midpoint are equal in magnitude.
Step 2: Directions of magnetic fields are opposite.
Step 3: Net magnetic field at the point is zero.
Step 4: Hence magnetic force F = qvB = 0.
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