MHT CET · Physics · Magnetic Effects of Current
A long straight wire carrying a current of \(25 \mathrm{~A}\) rests on the table. Another wire PQ of length \(1 \mathrm{~m}\) and mass \(2.5 \mathrm{~g}\) carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will wire \(\mathrm{PQ}\) rise? \(\left(\mu_0=4 \pi \times 10^{-7}\right.\) SI unit \()\)
- A 3 mm
- B 4 mm
- C 5 mm
- D 8 mm
Answer & Solution
Correct Answer
(C) 5 mm
Step-by-step Solution
Detailed explanation
Given \(\mathrm{I}_1=\mathrm{I}_2=25 \mathrm{~A}, l=1 \mathrm{~m}\),
\(\begin{aligned}
& \mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{h}} \\
& \mathrm{F}=\mathrm{BI} / \sin \theta=\mathrm{BI} l
\end{aligned}\)
Force applied on \(\mathrm{PQ}=\) Weight of the smaller current carrying wire
\(\begin{aligned}
& \quad \text { i.e, } \mathrm{mg}=\frac{\mu_0 \mathrm{I}^2 l}{2 \pi \mathrm{h}} \\
& \therefore \quad \mathrm{h}=\frac{4 \pi \times 10^{-7} \times 250 \times 25 \times 1}{2 \pi \times 2.5 \times 10^{-3} \times 9.8}=5 \mathrm{~mm}
\end{aligned}\)
\(\begin{aligned}
& \mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{h}} \\
& \mathrm{F}=\mathrm{BI} / \sin \theta=\mathrm{BI} l
\end{aligned}\)
Force applied on \(\mathrm{PQ}=\) Weight of the smaller current carrying wire
\(\begin{aligned}
& \quad \text { i.e, } \mathrm{mg}=\frac{\mu_0 \mathrm{I}^2 l}{2 \pi \mathrm{h}} \\
& \therefore \quad \mathrm{h}=\frac{4 \pi \times 10^{-7} \times 250 \times 25 \times 1}{2 \pi \times 2.5 \times 10^{-3} \times 9.8}=5 \mathrm{~mm}
\end{aligned}\)
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