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Two infinite line-charges parallel to each other are moving with a constant velocity $v$ in the same direction as shown in the figure. The separation between two line-charges is $d$. The magnetic attraction balances the electric repulsion when, [c = speed of light in free space]
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The correct answer is:
$v=c$
$\mathrm{F}_{\mathrm{E}}=\mathrm{q}_2 \mathrm{E}_1=\lambda_2 \ell \frac{2 \mathrm{k} \lambda_1}{\mathrm{~d}}$
$\mathrm{~F}_{\mathrm{B}}=\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 \ell}{2 \pi \mathrm{d}}=\frac{\mu_0 \lambda_1 \mathrm{~V} \lambda_2 \mathrm{~V}}{2 \pi \mathrm{d}} \times \ell$
$\mathrm{F}_{\mathrm{E}}=\mathrm{F}_{\mathrm{B}}$
$\Rightarrow \frac{2 \mathrm{k} \lambda_1 \lambda_2 \ell}{\mathrm{d}}=\frac{\mu_0 \lambda_1 \lambda_2 \mathrm{~V}^2 \ell}{2 \pi \mathrm{d}}$
$\Rightarrow \frac{2 \lambda_1 \lambda_2 \ell}{4 \pi \varepsilon_0}=\frac{\mu_0 \lambda_1 \lambda_2 \mathrm{~V}^2 \ell}{2 \pi}$
$\mathrm{V}^2=\frac{1}{\mu_0 \varepsilon_0}=\mathrm{c}^2 \Rightarrow \mathrm{V}=\mathrm{c}$
$\mathrm{~F}_{\mathrm{B}}=\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 \ell}{2 \pi \mathrm{d}}=\frac{\mu_0 \lambda_1 \mathrm{~V} \lambda_2 \mathrm{~V}}{2 \pi \mathrm{d}} \times \ell$
$\mathrm{F}_{\mathrm{E}}=\mathrm{F}_{\mathrm{B}}$
$\Rightarrow \frac{2 \mathrm{k} \lambda_1 \lambda_2 \ell}{\mathrm{d}}=\frac{\mu_0 \lambda_1 \lambda_2 \mathrm{~V}^2 \ell}{2 \pi \mathrm{d}}$
$\Rightarrow \frac{2 \lambda_1 \lambda_2 \ell}{4 \pi \varepsilon_0}=\frac{\mu_0 \lambda_1 \lambda_2 \mathrm{~V}^2 \ell}{2 \pi}$
$\mathrm{V}^2=\frac{1}{\mu_0 \varepsilon_0}=\mathrm{c}^2 \Rightarrow \mathrm{V}=\mathrm{c}$
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