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JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
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

- A \(x = \left( {\frac{{{I_1}}}{{{I_1} - {I_2}}}} \right)d\) and \(\,x = \frac{{{I_2}}}{{\left( {{I_1} + {I_2}} \right)}}d\)
- B \(\,\,x = \pm \frac{{{I_1}d}}{{\left( {{I_1} - {I_2}} \right)}}\)
- C \(x = \left( {\frac{{{I_2}}}{{{I_1} + {I_2}}}} \right)d\) and \(\,x = \frac{{{I_2}}}{{\left( {{I_1} - {I_2}} \right)}}d\)
- D \(x = \left( {\frac{{{I_1}}}{{{I_1} + {I_2}}}} \right)d\) and \(\,x = \frac{{{I_2}}}{{\left( {{I_1} - {I_2}} \right)}}d\)
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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