Magnetics Ques 114
- 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 $\mathbf{v}$ is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is
(1998, 2M)
(a) $\frac{\mu _0 I q v}{2 \pi d}$
(b) $\frac{\mu _0 I q v}{\pi d}$
(c) $\frac{2 \mu _0 I q v}{\pi d}$
(d) zero
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Answer:
Correct Answer: 114.(d)
Solution:
Formula:
Magnetic Force Acting On A Current Carrying Wire:
- Net magnetic field due to both the wires will be downward as shown in the figure.
Since, angle between $\mathbf{v}$ and $\mathbf{B}$ is $180^{\circ}$.
Therefore, magnetic force
$ \mathbf{F} _m=q(\mathbf{v} \times \mathbf{B})=0 $
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