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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) eviR^(2))/(2(x^(2)+R^(2))^(3//2))`B. `mu_(0)=(eviR^(2)x)/((x^(2)+R^(2))^(3//2))`C. `(mu_(0) eviR^(2)x)/(4pi(x^(2)+R^(2))^(3//2))`D. 0 |
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Answer» Correct Answer - D `F=q[v(+hati)]xxB(-hati)]=0` Because B as well as v is are along axis of circular CN. |
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