A) \[\frac{{{\mu }_{0}}Iqv}{2\pi d}\]
B) \[\frac{{{\mu }_{0}}Iqv}{\pi d}\]
C) \[\frac{2{{\mu }_{0}}Iqv}{\pi d}\]
D) zero
Correct Answer: D
Solution :
The magnetic field due to both of the wires will downwards. \[\vec{F}=q\,(\vec{v}\times \vec{B})=0\] Since \[\vec{v}\] is parallel to \[\vec{B}\].You need to login to perform this action.
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