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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_0Iqv)/(2pid)`
B. `(mu_0Iqv)/(pid)`
C. `(2mu_0Iqv)/(pid)`
D. `0`

1 Answer

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Best answer
Correct Answer - D
(d) Net magnetic field due to the wires
will be downwards as shown below in
Fig. Since angle between `barv and barB is 180^@`, therefor magnetic force
`vecF_m=q(vecvxxvecB)=0`
image

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