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In the given fig. x and y are two long straight parallel conductors each carrying a current of 2A. The force on each conductor is F newtons. When the current in each is changed to IA and reversed in direction, the force on each is now,

 

Option: 1

\frac{F}{4} and unchanged in direction


Option: 2

\frac{F}{2} and reversed in direction


Option: 3

\frac{F}{2} and unchanged in direction


Option: 4

\frac{F}{4}


Answers (1)

best_answer

from, \frac{\mu_{0} L_1^{\prime} L_2^{\prime}}{2 \pi d}=F
When the current in the same direction there is an attraction force.

\begin{aligned} F^{\prime} & =\frac{\mu_0 i_1 \mid 2 i_2 / 2}{2 \pi d} \\ & =\frac{F}{4} \end{aligned}

Posted by

manish painkra

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