A long straight wire carrying current of $25$ $\mathrm{A}$ rests on a table as shown in figure. Another wire $\mathrm{PQ}$ of length $1$ $\mathrm{m}$, mass $2.5$ $\mathrm{g}$ carries the same current but in the opposite direction. The wire $\mathrm{PQ}$ is free to slide up and down. To what height will $\mathrm{PQ}$ rise ?
A long straight wire carrying current of $25$ $\mathrm{A}$ rests on a table as shown in figure. Another wire $\mathrm{PQ}$ of length $1$ $\mathrm{m}$, mass $2.5$ $\mathrm{g}$ carries the same current but in the opposite direction. The wire $\mathrm{PQ}$ is free to slide up and down. To what height will $\mathrm{PQ}$ rise ?
Force on $PQ$ wire which is held in air by $25$ $A$ current carrying wire which balanced with weight force of same wire$ PQ.$
Magnetic field at $h$ distance due to $25$ $A$ current carrying wire,
$\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \pi h}$
Small wire place in this magnetic field, magnetic force on that wire is given by,
$\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{I}} l \times \overrightarrow{\mathrm{B}}$ $\therefore \mathrm{F}=\mathrm{BI} / \sin \theta$ $\therefore \mathrm{F}=\mathrm{BI} l$ $\therefore \mathrm{F}=\mathrm{I} l\left(\frac{\mu_{0} \mathrm{I}}{2 \pi h}\right)$ $\therefore \mathrm{F}=\left(\frac{\mu_{0} \mathrm{I}^{2} l}{2 \pi h}\right)$
Currents are in opposite direction from both of wires so force on wire PQ is in upward direc tion.
Wire is displace by ' $h$ ' till weight force $m g$ balances by this magnetic force,
$\therefore m g=\frac{\mu_{0} \mathrm{I}^{2} l}{2 \pi h}$ $h=\frac{\mu_{0} \mathrm{I}^{2} l}{2 \pi m g}$ $h=\frac{\left(4 \pi \times 10^{-7}\right) \times(25)^{2} \times 1}{(2 \pi) \times\left(2.5 \times 10^{-3}\right)(9.8)}$ $h=51 \times 10^{-4}$ $h=0.51 \mathrm{~cm}$
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