JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A wire \(X\) of length \(50\; cm\) carrying a current of \(2\,A\) is placed parallel to a long wire \(Y\) of length \(5\,m\). The wire \(Y\) carries a current of \(3\,A\). The distance between two wires is \(5\,cm\) and currents flow in the same direction. The force acting on the wire \(Y\) is.

- A \(1.2 \times 10^{-5} \;N\) directed towards wire \(X\).
- B \(1.2 \times 10^{-4}\;N\) directed away from wire \(X\).
- C \(1.2 \times 10^{-4}\; N\) directed towards wire \(X\).
- D \(2.4 \times 10^{-5} \;N\) directed towards wire \(X\).
Answer & Solution
Correct Answer
(A) \(1.2 \times 10^{-5} \;N\) directed towards wire \(X\).
Step-by-step Solution
Detailed explanation
Force of interaction \(=I_{1} \ell_{1} B_{12}\) \(=\frac{\mu_{0} I _{1} I _{2}}{2 \pi r } \ell_{1}\) \(=\frac{4 \pi \times 10^{-7} \times 6 \times 5}{2 \pi \times 5 \times 10^{-2}}\) \(=1.2 \times 10^{-5} \text { towards } X\)
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