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JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2019
  • Medium
  • MCQ

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 \(C\) 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

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Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

Net force on wire carrying current \(I\) per unit length is

\(\frac{\mu_{0} I_{1} I}{2 \pi x}+\frac{\mu_{0} I_{2} I}{2 \pi(d-x)}=0\)

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

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

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