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 - Physics MCQ AIEEE 2004 2026 | ExamTest.live

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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 increased to two times and its direction reversed. The distance is also increased to 3 d . The new value of the force between them is:

Solution & Step-by-step Explanation

The force per unit length between two parallel wires is: F = \frac{μ _{0} I _{1} I _{2}}{2 π d} Initially, the force is attractive (F).Case 2: I_{1}^{'} = - 2 I_{1} (reversed and doubled), d^{'} = 3 d . F^{'} = \frac{μ _{0} ( - 2 I _{1} ) I _{2}}{2 π ( 3 d )} = - \frac{2}{3} (\frac{μ _{0} I _{1} I _{2}}{2 π d}) = - \frac{2 F}{3} The negative sign indicates the force is now repulsive.

Try it yourself before checking the explanation above.

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 increased to two times and its direction reversed. The distance is also increased to 3 d . The new value of the force between them is:

- 2 F

F /3

- 2 F /3

- F /3

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