MHT CET · Physics · Magnetic Effects of Current
Two long conductors, separated by a distance 'd' carry currents ' \(I_1\) ' and ' \(I_2\) ' in the same directions. They exert a force ' \(F\) ' on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to ' \(3 \mathrm{~d}\). The new value of the force between them is
- A -2F
- B -F
- C \(-\frac{2 \mathrm{~F}}{3}\)
- D \(\frac{F}{3}\)
Answer & Solution
Correct Answer
(C) \(-\frac{2 \mathrm{~F}}{3}\)
Step-by-step Solution
Detailed explanation
Force on each conductor is given by
\(
\mathrm{F}=\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{I}_1 \mathrm{I}_2}{\mathrm{~d}} \ell
\)
This force will be attractive.
If the direction of current is reversed in one conductor, the force will become repulsive.
\(
\therefore \mathrm{F}^{\prime}=-\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{I}_1 \mathrm{I}_2}{3 \mathrm{~d}} \cdot \ell=-\frac{2}{3} \mathrm{~F}
\)
\(
\mathrm{F}=\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{I}_1 \mathrm{I}_2}{\mathrm{~d}} \ell
\)
This force will be attractive.
If the direction of current is reversed in one conductor, the force will become repulsive.
\(
\therefore \mathrm{F}^{\prime}=-\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{I}_1 \mathrm{I}_2}{3 \mathrm{~d}} \cdot \ell=-\frac{2}{3} \mathrm{~F}
\)
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