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 $m=0.1\text{kg}$. $[\ln 2=0.693]$ Find the angular acceleration of the wire just after it is released from the position shown.

• A. $6.2\times {10}^{-8}{\text{rad/s}}^2$
• B. $2.1\times {10}^{-4}{\text{rad/s}}^2$
• C. $4.5\times {10}^{-5}{\text{rad/s}}^2$
• D. $9.3\times {10}^{-6}{\text{rad/s}}^2$

Answer 1

The correct option is D $9.3\times {10}^{-6}{\text{rad/s}}^2$

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 ${\tau }_0=\underset{L}{\overset{2L}{\int }}(\frac{{\mu }_{o}i}{2\pi x})\left(i\right)(2L-x)dx$

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

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

Moment of inertia of wire $I_0=\frac{m{L}^2}{3}=\frac{1}{30}{\text{kgm}}^2$

Thus, angular acceleration ${\alpha }_0=\frac{{\tau }_0}{I_0}=9.3\times {10}^{-6}{\text{rad/s}}^2$

Hence, option (d) is the correct answer.