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 v is perpendicular to the plane of wires. The magnitude of the force due to the magnetic field acting on the charge at this instant is

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 v is perpendicular to the plane of wires. The magnitude of the force due to the magnetic field acting on the charge at this instant is (A) (μ0 I q v/2 π d) (B) (2 μ0 I q v/π d) (C) (μ0 I q v/π d) (D) zero

Tardigrade Admin · Accepted Answer

Both vecv and vecB are perpendicular to plane of parallel wires. ∴ vecF=q( vecv × vecB)=0

$\frac{μ _{0} I q v}{2 π d}$

$\frac{2 μ _{0} I q v}{π d}$

$\frac{μ _{0} I q v}{π d}$

zero

Both $v$ and $B$ are perpendicular to plane of parallel wires.
$∴ F = q (v \times B) = 0$

2πdμ0​Iqv​

πd2μ0​Iqv​

πdμ0​Iqv​

zero

Both v and B are perpendicular to plane of parallel wires. ∴F=q(v×B)=0

$\frac{μ _{0} I q v}{2 π d}$

$\frac{2 μ _{0} I q v}{π d}$

$\frac{μ _{0} I q v}{π d}$

Both $v$ and $B$ are perpendicular to plane of parallel wires.
$∴ F = q (v \times B) = 0$