Two very long, straight , parallel wires carry steady currents ` I &-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 wires , in the plane of the wires. Its instantaneous vel,ocity ` 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
`(mu_(0)Iqv)/(2 pid)`
B
`(mu_(0)Iqv)/(pid)`
C
`( 2mu_(0)Iqv)/( pid)`
D
`0`
Text Solution
Verified by Experts
The correct Answer is:
D
Net magnetic field due to the wires will be downward as shown below in the figure . Since angle between ` vec(v) and vec(B)` is ` 180@`, Therefore , magnetic force ` vec(F)_(m) = q( vec(v) xx vec(B)) = 0 `
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