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 direction. They exert a force ' \(\mathrm{F}\) ' on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance between them is also increased to \(3 \mathrm{~d}\). The new value of force between them is
- A \(-2 \mathrm{~F}\)
- B \(\frac{\mathrm{F}}{3}\)
- C \(\frac{-2 \mathrm{~F}}{3}\)
- D \(\frac{-\mathrm{F}}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{-2 \mathrm{~F}}{3}\)
Step-by-step Solution
Detailed explanation
The force per unit length of the conductors is given as:
\(\mathrm{F}=\frac{\mu_4 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \mathrm{d}}\)
When the value and direction of current in the first conductor and the distance between the conductors are changed,
\(\begin{aligned}
\therefore \quad \mathrm{F}_2 & =\frac{-\mu_0 2 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \times 3 \mathrm{~d}} \\
\therefore \quad & \frac{\mathrm{F}_2}{\mathrm{~F}}=\frac{-2}{3} \\
\mathrm{~F}_2 & =-\frac{2 \mathrm{~F}}{3}
\end{aligned}\)
\(\mathrm{F}=\frac{\mu_4 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \mathrm{d}}\)
When the value and direction of current in the first conductor and the distance between the conductors are changed,
\(\begin{aligned}
\therefore \quad \mathrm{F}_2 & =\frac{-\mu_0 2 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \times 3 \mathrm{~d}} \\
\therefore \quad & \frac{\mathrm{F}_2}{\mathrm{~F}}=\frac{-2}{3} \\
\mathrm{~F}_2 & =-\frac{2 \mathrm{~F}}{3}
\end{aligned}\)
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