Two wires A and B are carrying currents| EMF |JEE MAIN 2019

_ JEE MAIN _2019_

S2_100419

Question

Two wires A and B are carrying currents $\mathrm{I}_1$ and $\mathrm{I}_2$ as shown in the figure. The separation between them is d . A third wire C carrying a current 1 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 :

Select the correct option:

A

$x= \pm \frac{I_1 d}{\left(I_1-I_2\right)}$

B

$x=\frac{I_1}{\left(I_1+I_2\right)} d$ and $x=\left(\frac{I_2}{I_1-I_2}\right) d$

C

$x=\left(\frac{l_2}{\left(l_1+l_2\right)}\right) d$ and $x=\left(\frac{l_2}{l_1+l_2}\right) d$

D

$x=\left(\frac{l_1}{\left(l_1-l_2\right)}\right) d$ and $x=\left(\frac{l_2}{l_1+l_2}\right) d$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Physics

Easy

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