Question

Two long conductors, separated by a distance r carry current I1 and I2 in the same direction. They exert a force F on each other. Now, the current in one them is increased to 3 times and the direction is reversed. The distance is also increased to 5r. What is the new value of the force between them?

(a) \(\frac {3F}{5}\)

(b) \(\frac {-3F}{5}\)

(c) \(\frac {5F}{3}\)

(d) \(\frac {-5F}{3}\)

The question was posed to me in examination.

Asked question is from Ampere and Forces between Two Parallel Currents topic in portion Moving Charges and Magnetism of Physics – Class 12

Answer 1

The correct answer is (b) \(\frac {-3F}{5}\)

The explanation is: First case:

F=\(\frac {\mu_o}{2\pi } \big [ \frac {I_1 I_2}{r} \big ] \)l

Second case:

F^‘=\(\frac {\mu_o}{2\pi } \big [ \frac {(-3I_1)(I_2)}{5r} \big ] \)l

\(\frac {F^{‘}}{F} = \frac {\frac {\mu_o}{2\pi } \big [ \frac {(-3I_1)(I_2)}{5r} \big ] l}{\frac {\mu_o}{2\pi } \big [ \frac {I_1 I_2}{r} \big ] l}\)

\(\frac {F^{‘}}{F}= -\frac {3}{5}\)

Therefore, F^‘=\((-\frac {3}{5})\)F=\(\underline{\frac {-3F}{5}}\)