ExamBro
ExamBro
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.

  1. A \(1.2 \times 10^{-5} \;N\) directed towards wire \(X\).
  2. B \(1.2 \times 10^{-4}\;N\) directed away from wire \(X\).
  3. C \(1.2 \times 10^{-4}\; N\) directed towards wire \(X\).
  4. D \(2.4 \times 10^{-5} \;N\) directed towards wire \(X\).
Verified Solution

Answer & Solution

Correct Answer

(A) \(1.2 \times 10^{-5} \;N\) directed towards wire \(X\).

Step-by-step Solution

Detailed explanation

Force of interaction \(=I_{1} \ell_{1} B_{12}\) \(=\frac{\mu_{0} I _{1} I _{2}}{2 \pi r } \ell_{1}\) \(=\frac{4 \pi \times 10^{-7} \times 6 \times 5}{2 \pi \times 5 \times 10^{-2}}\) \(=1.2 \times 10^{-5} \text { towards } X\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app