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An electron moving with a velocity v alo...

An electron moving with a velocity `v` along the positive `x`-axis approaches a circular current carrying loop as shown in the fig. the magnitude of magnetic force on electron at this instant is

A

`(mu_(0))/(Z)("eviR"^(2)X)/((X^(2)+R^(2))^(3//2))`

B

`mu_(0) ("eviR"^(2)X)/((X^(2)+R^(2))^(3//2))`

C

`(mu_(0))/(4pi) ("eviR"^(2)X)/((X^(2)+R^(2))^(3//2))`

D

0

Text Solution

Verified by Experts

The correct Answer is:
D
Direction of electron velocity is along magnetic lines of force. Hence F = 0
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