Q. An electron moving with velocity $v$ along the axis approaches a circular current carrying loop as shown in the figure. The magnitude of magnetic force on electron at this instant is-

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A

$\frac{\left(\mu \right)_{0}}{4}\frac{e v i R^{2} x}{\left(\right. x^{2} + R^{2} ⁡ \left(\left.\right)^{3 / 2}}$

B

$\left(\mu \right)_{0}\frac{e v i R^{2} x}{\left(\right. x^{2} + R^{2} ⁡ \left(\left.\right)^{3 / 2}}$

C

$\frac{\left(\mu \right)_{0}}{4 \pi }\frac{e v i R^{2} x}{\left(\right. x^{2} + R^{2} ⁡ \left(\left.\right)^{3 / 2}}$

D

none of these

Solution:

The magnetic force on a moving charge in electric field is given by
$\overset{ \rightarrow }{F}=q\left(\overset{ \rightarrow }{v} \times \overset{ \rightarrow }{B}\right)$
Where, $\overset{ \rightarrow }{v}$ is the velocity vector and $\overset{ \rightarrow }{B}$ is the magnetic field.
The direction of the magnetic field due to the current-carrying circular loop is along its axis.
As the direction of electron velocity is along magnetic field lines, the magnetic force $F=0$ .

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