Question

Two wires A and B are carrying currents ${{I}_{1}}$ and ${{I}_{2}}$ as shown in the figure. The separation between them is $d$. A third wire Carrying a current $I$ is kept parallel to them at a distance $x$ from A such that net force acting on it is zero. The possible values of $x$ are:

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

Answer 1

Hint In the given question, we have been provided with two current-carrying wires at a known separation and a third current-carrying wire is placed at an unknown distance somewhere between the two wires. We know that any current-carrying wire produces a magnetic field and the field thus produced will exert a force on the other current-carrying wires in its vicinity. We have been told that the force acting on the wire in the middle is zero and hence all we need to do is express the force on the wire in terms of the current and the distances between the wires and we will have our answer. Let’s see the detailed solution.
Formula Used: $F=\dfrac{{{\mu }_{0}}}{2\pi }\dfrac{I'I}{x}$

Complete step by step answer:
As discussed above, the force exerted on the wire C due to the wire A must be equal to the force exerted on the wire C due to the wire B.
Now we know that the force exerted on unit length of a wire carrying current I’ placed at a distance x from another wire carrying current I is given as $F=\dfrac{{{\mu }_{0}}}{2\pi }\dfrac{I'I}{x}$
Using this formula, we can say that
The force acting on unit length of the wire C due to wire A ${{F}_{1}}'=\dfrac{{{\mu }_{0}}}{2\pi }\dfrac{{{I}_{1}}I}{x}$
Force acting on the entire length of the wire C due to the wire A will be ${{F}_{1}}=\dfrac{{{\mu }_{0}}}{2\pi }\dfrac{{{I}_{1}}I}{x}d$
Similarly, the force acting on unit length of the wire C due to the wire B ${{F}_{2}}'=\dfrac{{{\mu }_{0}}}{2\pi }\dfrac{{{I}_{2}}I}{\left( d-x \right)}$
Now, force acting on the entire length of the wire C due to the wire B would be ${{F}_{2}}=\dfrac{{{\mu }_{0}}}{2\pi }\dfrac{{{I}_{2}}I}{\left( d-x \right)}d$
Since the net force on the wire C is zero, we can equate the two force calculated above, that is

& {{F}_{1}}={{F}_{2}} \\\ & \Rightarrow \dfrac{{{\mu }_{0}}}{2\pi }\dfrac{{{I}_{1}}I}{x}d=\dfrac{{{\mu }_{0}}}{2\pi }\dfrac{{{I}_{2}}I}{\left( d-x \right)}d \\\ & \Rightarrow \dfrac{{{I}_{1}}}{x}=\dfrac{{{I}_{2}}}{d-x} \\\ & \Rightarrow \dfrac{d-x}{x}=\dfrac{{{I}_{2}}}{{{I}_{1}}} \\\ & \Rightarrow \dfrac{d}{x}-1=\dfrac{{{I}_{2}}}{{{I}_{1}}} \\\ & \Rightarrow \dfrac{d}{x}=1+\dfrac{{{I}_{2}}}{{{I}_{1}}}=\dfrac{{{I}_{1}}+{{I}_{2}}}{{{I}_{1}}} \\\ \end{aligned}$$ Inverting the equation obtained above and simplifying further, we get $$\begin{aligned} & \dfrac{x}{d}=\dfrac{{{I}_{1}}}{{{I}_{1}}+{{I}_{2}}} \\\ & \Rightarrow x=\left( \dfrac{{{I}_{1}}}{{{I}_{1}}+{{I}_{2}}} \right)d \\\ \end{aligned}$$ **Hence we can say that none of the options give the correct answer to the given question.** **Note** In the solution above, we have first stated the force acting on a unit length of the wire and then used the unitary method concept to find the force on the entire length of the wire. In this case, the length of wire C remains unchanged hence we could have skipped the step where we found the force on the entire length. But we have shown it because you might come across questions where the length is being altered and hence calculation of the force on the total wire is crucial for obtaining the correct solution.