JEE Main 2019 — Physics solution

JEE Main2019Physics

Two wires A and B are carrying currents I₁ and I₂ 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 :

A
$x = \pm \frac{I _{1} d}{( I _{1} - I _{2} )}$
B
$x = (\frac{I _{1}}{I _{1} + I _{2}}) d and x = (\frac{I _{2}}{I _{1} - I _{2}}) d$
C
$x = (\frac{I _{2}}{I _{1} + I _{2}}) d and x = (\frac{I _{2}}{I _{1} - I _{2}}) d$
D
$x = (\frac{I _{1}}{I _{1} - I _{2}}) d and x = (\frac{I _{2}}{I _{1} - I _{2}}) d$

Solution

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