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Three parallel wires a, b and c carrying currents $i_a, i_b$ and $i_c$ as shown in the figure are placed next to each other.
The magnitude force on a length $l$ of the wire a, if $\mathrm{d}_2=2 \mathrm{~d}_1, i_b=i_a$ and $i_c=4 i_a$ is
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The magnitude force on a length $l$ of the wire a, if $\mathrm{d}_2=2 \mathrm{~d}_1, i_b=i_a$ and $i_c=4 i_a$ is
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Verified Answer
The correct answer is:
$\frac{\mu_0}{6 \pi \mathrm{d}_1} i_a^2 l$
$\begin{aligned} & \left(\frac{\mathrm{F}}{\ell}\right)_{\mathrm{a}}=\frac{\mu_0}{2 \pi} \frac{\mathrm{i}_{\mathrm{a}} \mathrm{i}_{\mathrm{b}}}{\mathrm{d}_1}-\frac{\mu_0}{2 \pi} \frac{\mathrm{i}_{\mathrm{a}} \mathrm{i}_{\mathrm{c}}}{\left(\mathrm{d}_2+\mathrm{d}_1\right)} \\ & =\frac{\mu_0}{2 \pi} \frac{\mathrm{i}_{\mathrm{a}}^2}{\mathrm{~d}_1}-\frac{\mu_0}{2 \pi} \frac{4 \mathrm{i}_{\mathrm{a}}^2}{3 \mathrm{~d}_1} \\ & =\frac{\mu_0}{2 \pi} \frac{\mathrm{i}_{\mathrm{a}}^2}{\mathrm{~d}_1}\left[1-\frac{4}{3}\right] \\ & =\frac{\mu_0}{2 \pi} \frac{\mathrm{i}_{\mathrm{a}}^2}{\mathrm{~d}_1} \times \frac{-1}{3},- \text { ve sign indicate repulsion } \\ & \text { So, } \mathrm{F}=\left|\left(\frac{\mathrm{F}}{\ell}\right)_{\mathrm{a}}\right| \times \ell=\frac{\mu_0 \mathrm{i}_{\mathrm{a}}^2}{6 \pi \mathrm{d}_1}\end{aligned}$
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