get app

Go To Your Study Dashboard

Two wires A & B are carrying currents $I_{1}$ & $I_{2}$ as shown in the figure.

The separation between them is d.A third wire C carrying a current

I is to be kept parallel to them at a distance x from A such that the

net force acting on it is zero. The possible values of x are:

Option 1)
$x=(\frac{I_{1}}{I_{1}-I_{2}})d$ and $x=(\frac{I_{2}}{I_{1}+I_{2}})d$

Option 2)
$x=(\frac{I_{2}}{I_{1}+I_{2}})d$ and $x=(\frac{I_{2}}{I_{1}-I_{2}})d$

Option 3)
$x=(\frac{I_{1}}{I_{1}+I_{2}})d$ and $x=(\frac{I_{2}}{I_{1}-I_{2}})d$

Option 4)
$x=\pm (\frac{I_{1}}{I_{1}-I_{2}})d$

Answers (1)

S solutionqc

Force between two parallel current carrying conductors -

$F=\frac{\mu }{4\pi } \frac{2I_{1 I_{2}}}{a} l$

$\frac{F}{l}=\frac{\mu o}{4\pi } \frac{2I_{1 I_{2}}}{a}$

- wherein

I₁ and I₂ current carrying two parallel wires

a-seperation between two wires

when $I_2>I_1$ ( that is wire 3 is on left side of both 1&2)

For equilibrium,

$F_{31}=F_{32}$

$\frac{\mu _oI_1Il}{2\pi x}=\frac{\mu _oI_2Il}{2\pi (d+x)}$

$\frac{d+x}{x}=\frac{I_2}{I_1}$

=> $x=(\frac{I_1}{I_2-I_1})d$ ........................(1)

when $I_1>I_2$ ( that is wire 3 is on right side of both 1&2)

$F_{31}=F_{32}$

$\frac{\mu _oI_1Il}{2\pi x}=\frac{\mu _oI_2Il}{2\pi (x-d)}$

$\frac{x-d}{x}=\frac{I_2}{I_1}$

=> $x=(\frac{-I_1}{I_2-I_1})d$ .....................(2)

from (1) & (2)

$x=\pm (\frac{I_1d}{I_1-I_2})$

Option 1)

$x=(\frac{I_{1}}{I_{1}-I_{2}})d$ and $x=(\frac{I_{2}}{I_{1}+I_{2}})d$

Option 2)

$x=(\frac{I_{2}}{I_{1}+I_{2}})d$ and $x=(\frac{I_{2}}{I_{1}-I_{2}})d$

Option 3)

$x=(\frac{I_{1}}{I_{1}+I_{2}})d$ and $x=(\frac{I_{2}}{I_{1}-I_{2}})d$

Option 4)

$x=\pm (\frac{I_{1}}{I_{1}-I_{2}})d$

Similar Questions

Preparation Products

Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-

Buy Now

Rank Booster JEE Main 2020

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 9999/- ₹ 4999/-

Buy Now

Test Series JEE Main Sept 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 1999/-

Buy Now

Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-

Buy Now

Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-

Buy Now

Exams

Articles

Questions