Question

A long wire carrying a current of 20A is fixed. Another long wire AB is to be balanced mid-air a distance 2cm. Find the magnitude of the current in AB. Also find the angular frequency of SHM. If AB is slightly disturbed from this position. Given the mass per unit length of the wires is ${10}^{-6}kg/m$

• A. $20A,10rad{s}^{-1}$
• B. $10A,5rad{s}^{-1}$
• C. $\frac{1}{20}A,10\sqrt{5}rad{s}^{-1}$
• D. $\frac{1}{10}A,\sqrt{5}rad{s}^{-1}$

Answer 1

The correct option is C

$\frac{1}{20}A,10\sqrt{5}rad{s}^{-1}$

$\frac{m}{l}={10}^{-6}kg{m}^{-1}{F}_l=\frac{mg}{l}$
$\Rightarrow \frac{{\mu }_0{i}_1{i}_2}{2\pi d}=\frac{m}{l}g$
$2\times {10}^{-7}\times \frac{20\times i}{2\times {10}^{-2}}={10}^{-6}\times 10$
$i=\frac{1}{20}A$
$\frac{{\mu }_0{i}_1{i}_2}{2\pi (d-y)}-\frac{m}{l}g=\frac{m}{l}a$
$\frac{{\mu }_0{i}_1{i}_2}{2\pi d}{(1-\frac{y}{d})}^{-1}-\frac{mg}{l}=\frac{ma}{l}$
$f_l(1+\frac{y}{d})-\frac{mg}{l}=\frac{ma}{l}$
$F_l+\frac{f_{l}y}{d}-\frac{mg}{d}-\frac{ma}{l}$
$\Rightarrow a=\left(\frac{d_l}{d\times m/l}\right)\times y=\frac{10}{2\times {10}^{-2}}y=500y$
$W=\sqrt{500}=10\sqrt{5}rad{s}^{-1}$