A long straight wire carrying a current of $25 A$ rests on the table. Another wire PQ of length $1 m$ and mass $2.5 g$ carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will wire $PQ$ rise? $(μ_{0} = 4 π \times 1 0^{- 7}$ SI unit $)$

Question

TestHub · Accepted Answer

Given I 1 ​ = I 2 ​ = 25 A , l = 1 m , ​ B = 2 π h μ 0 ​ I ​ F = BI / sin θ = BI l ​ Force applied on PQ = Weight of the smaller current carrying wire ​ i.e, mg = 2 π h μ 0 ​ I 2 l ​ ∴ h = 2 π × 2.5 × 1 0 − 3 × 9.8 4 π × 1 0 − 7 × 250 × 25 × 1 ​ = 5 mm ​

Physics - Magnetism Question with Solution | TestHub

Options:

Doubts & Discussion