Two long conductors, separated by a distance d carry current <math> <mrow> <msub> <mrow> <mi>I</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> </math> and <math> <mrow> <msub> <mrow> <mi>I</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </math> in the same direction. They exert a force F on each other. Now the current in one of them is increased to two times and its directions is reversed. The distance is also increased to 3 d. The new value of the force between them is

F ∝ i 1 i 2 a Since one of the current increase two timesand distance increases three times, so force become 2 3 <!- [if gte mso 9]><xml> </xml><! [endif]-> times. Also due to the reversal of direction of current force becomes negative.

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physics ncert exemplar solution class 12th chapter one Force between Two Parallel Currents Physics Moving Charges and Magnetism Physics Class 12th

Two long conductors, separated by a distance d carry current $I_{1}$ and $I_{2}$ in the same direction. They exert a force F on each other. Now the current in one of them is increased to two times and its directions is reversed. The distance is also increased to 3 d. The new value of the force between them is

Option 1 -

-2F

Option 2 -

F/3

Option 3 -

2F/3

Option 4 -

-F/3

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1 Answer

Answered by

Raj Pandey | Contributor-Level 9

a month ago

Correct Option - 3

Detailed Solution:

$F \propto \frac{i_{1} i_{2}}{a}$ Since one of the current increase two times

and distance increases three times, so force become $\frac{2}{3}$ times. Also due to the reversal of direction of current force becomes negative.

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