MCQ
10 A current is flowing in two straight parallel wires in the same direction. Force of attraction between them is $1 \times 10^{-3} N$. If the current is doubled in both the wires the force will be
  • A
    $1 \times 10^{-3} N$
  • B
    $2 \times 10^{-3} N$
  • C
    $4 \times 10^{-3} N$
  • D
    $0.25 \times 10^{-3} N$

Answer

(c) : In case of straight parallel wires, force$F=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{r}$
If current is doubled $F^{\prime}=\frac{\mu_0}{4 \pi} \frac{2 \times 2 I_1 \times 2 I_2}{r}=4 F$$\therefore \quad F^{\prime}=4 \times 1 \times 10^{-3}=4 \times 10^{-3} N$

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