MHT CET · Physics · Current Electricity
10 A current is flowing in two straight parallel wires in the same direction. Force of attraction between them is \(1 \times 10^{-3} \mathrm{~N}\). If the current is doubled in both the wires the force will be
- A \(1 \times 10^{-3} \mathrm{~N}\)
- B \(2 \times 10^{-3} \mathrm{~N}\)
- C \(4 \times 10^{-3} \mathrm{~N}\)
- D \(0.25 \times 10^{-3} \mathrm{~N}\)
Answer & Solution
Correct Answer
(C) \(4 \times 10^{-3} \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{F}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{I}_1 \mathrm{I}_2}{\mathrm{r}} l=10^{-3} \mathrm{~N}\)
When current in both the wires is doubled, then
\(\mathrm{F}^{\prime}=\frac{\mu_0}{4 \pi} \frac{2\left(2 \mathrm{I}_1 \times 2 \mathrm{I}_2\right)}{\mathrm{r}} j=4 \times 10^{-3} \mathrm{~N}\)
When current in both the wires is doubled, then
\(\mathrm{F}^{\prime}=\frac{\mu_0}{4 \pi} \frac{2\left(2 \mathrm{I}_1 \times 2 \mathrm{I}_2\right)}{\mathrm{r}} j=4 \times 10^{-3} \mathrm{~N}\)
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