JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
Two long straight wires \(P\) and \(Q\) carrying equal current \(10\,A\) each were kept parallel to each other at \(5\,cm\) distance. Magnitude of magnetic force experienced by \(10\,cm\) length of wire \(P\) is \(F_1\). If distance between wires is halved and currents on them are doubled, force \(F_2\) on \(10\,cm\) length of wire \(P\) will be :
- A \(8 F_1\)
- B \(10 F_1\)
- C \(F_1 / 8\)
- D \(F_1 / 10\)
Answer & Solution
Correct Answer
(A) \(8 F_1\)
Step-by-step Solution
Detailed explanation
Force per unit length between two parallel straight \(\text { wires }=\frac{\mu_0 i_1 i_2}{2 \pi d }\) \(\frac{ F _1}{ F _2}=\frac{\frac{\mu_0(10)^2}{2 \pi(5\,cm )}}{\frac{\mu_0(20)^2}{2 \pi\left(\frac{5 cm }{2}\right)}}=\frac{1}{8}\) \(\Rightarrow F_2=8 F_1\)
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