JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
Two parallel, long wires are kept \(0.20\,m\) apart in vacuum, each carrying current of \(x\) in the same direction. If the force of attraction per meter of each wire is \(2 \times 10^{-6}\,N\), then the value of \(x\) is approximately
- A \(1\)
- B \(2.4\)
- C \(1.4\)
- D \(2\)
Answer & Solution
Correct Answer
(C) \(1.4\)
Step-by-step Solution
Detailed explanation
Force per unit length \(=\frac{\mu_{0} i_{1} i_{2}}{2 \pi d}\) \(=\frac{\mu_{0} \cdot x ^{2}}{2 \pi \times 0.2}\) \(F =2 \times 10^{-6}=\frac{4 \pi \times 10^{-7} \times x ^{2}}{2 \pi \times 0.2}\) \(\Rightarrow 10^{-6}=10^{-7} \frac{ x ^{2}}{0.2}\)…
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