Question

A wire of length $L$ , mass $m$ and carrying current $i$ is suspended from point ${}^{\prime }{O}^{\prime }$ as shown. An another infinitely long wire carrying the same current $i$ is at a distance $L$ below the lower end of the wire. Given $i=2\text{A},L=1\text{m}$ and $m=0.1\text{kg}$. $[\ln 2=0.693]$ At what distance $r$ from the suspended wire, the new wire (having the same current) should be palced to keep it stationary.

• A. $2.9\text{m}$
• B. $1.9\text{m}$
• C. $1.3\text{m}$
• D. $2.4\text{m}$

Answer 1

The correct option is C $1.3\text{m}$

Let us consider a infinitesimal length element in suspended wire which is at a distance $x$ from infinitely long wire.


Torque acting on the suspended wire be ${\tau }_0=\underset{L}{\overset{2L}{\int }}(\frac{{\mu }_{o}i}{2\pi x})\left(i\right)(2L-x)dx$

$\Rightarrow {\tau }_0=\frac{{\mu }_o{i}^2}{2\pi }[2\ln x-x]{|}_L^{2L}$

$\Rightarrow {\tau }_0=\frac{0.386{\mu }_0{i}^{2}L}{2\pi }=31\times {10}^{-7}\text{Nm}$

Torque about $O$ due to the new wire is ${\tau }^{\prime }=(\frac{{\mu }_{0}i}{2\pi r})\times Li\times \frac{L}{2}=\frac{{\mu }_0{L}^2{i}^2}{2\pi \left(2r\right)}$

For the suspended wire to be stationary, ${\tau }_0={\tau }^{\prime }$

$\therefore \frac{L}{2r}=0.386$

$\Rightarrow r=\frac{L}{2\times 0.386}=1.3\text{m}$

Hence, option (c) is the correct answer.