Two very long straight parallel wires carry steady currents I and –I. 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

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\frac{μ_{0} I qv}{2 π d}

\frac{μ_{0} I qv}{π d}

\frac{2 μ_{0} I qv}{π d}

Zero

answer is D.

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Detailed Solution

B = \frac{μ_{0}}{4 π} . \frac{2 i}{\frac{r}{2}} + \frac{μ_{0}}{4 π} . \frac{2 (2 i)}{\frac{r}{2}}

$= \frac{μ_{0}}{4 π} . \frac{12 i}{r}$ acting on perpendicular to the plane of wire inwards.

Now, B and V are acting in the same direction i.e, $q = 0 °$

Force on charged particle is $F = qvBsinq = qvB \times 0 = 0$

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