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A long straight wire carrying a current of $25 \mathrm{~A}$ rests on the table. Another wire PQ of length $1 \mathrm{~m}$ and mass $2.5 \mathrm{~g}$ carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will wire $\mathrm{PQ}$ rise? $\left(\mu_0=4 \pi \times 10^{-7}\right.$ SI unit $)$
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Verified Answer
The correct answer is:
5 mm
Given $\mathrm{I}_1=\mathrm{I}_2=25 \mathrm{~A}, l=1 \mathrm{~m}$,
$\begin{aligned}
& \mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{h}} \\
& \mathrm{F}=\mathrm{BI} / \sin \theta=\mathrm{BI} l
\end{aligned}$
Force applied on $\mathrm{PQ}=$ Weight of the smaller current carrying wire
$\begin{aligned}
& \quad \text { i.e, } \mathrm{mg}=\frac{\mu_0 \mathrm{I}^2 l}{2 \pi \mathrm{h}} \\
& \therefore \quad \mathrm{h}=\frac{4 \pi \times 10^{-7} \times 250 \times 25 \times 1}{2 \pi \times 2.5 \times 10^{-3} \times 9.8}=5 \mathrm{~mm}
\end{aligned}$
$\begin{aligned}
& \mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{h}} \\
& \mathrm{F}=\mathrm{BI} / \sin \theta=\mathrm{BI} l
\end{aligned}$
Force applied on $\mathrm{PQ}=$ Weight of the smaller current carrying wire
$\begin{aligned}
& \quad \text { i.e, } \mathrm{mg}=\frac{\mu_0 \mathrm{I}^2 l}{2 \pi \mathrm{h}} \\
& \therefore \quad \mathrm{h}=\frac{4 \pi \times 10^{-7} \times 250 \times 25 \times 1}{2 \pi \times 2.5 \times 10^{-3} \times 9.8}=5 \mathrm{~mm}
\end{aligned}$
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