A long horizontal rigidly supported wire carries a current is = 96 A. Directly above it and parallel to it at a distance, another wire of 0.144 N weight per metre carrying a current ib = 24 A, in a direction opposite to that of ia. If the upper wire is to float in air due to magnetic repulsion, then its distance (in mm) from the lower wire is:

Question

A long horizontal rigidly supported wire carries a current is = 96 A. Directly above it and parallel to it at a distance, another wire of 0.144 N weight per metre carrying a current ib = 24 A, in a direction opposite to that of ia. If the upper wire is to float in air due to magnetic repulsion, then its distance (in mm) from the lower wire is: (A) 9 . 6 (B) 4 . 8 (C) 3 . 2 (D) 1 . 6

Tardigrade Admin · Accepted Answer

(F/l)=(μ 0/4π )(2iaib/r) ∴ 0.144=(10-7× 2× 96× 24/r) ∴ r=(10-7× 2× 96× 24/0.144) =3.2× 10-3m =3.2 mm

9 . 6

4 . 8

3 . 2

1 . 6

lF​=4πμ0​​r2ia​ib​​ ∴ 0.144=r10−7×2×96×24​ ∴ r=0.14410−7×2×96×24​ =3.2×10−3m =3.2mm

$\frac{F}{l} = \frac{μ _{0}}{4 π} \frac{2 i _{a} i _{b}}{r}$ $∴$ $0.144 = \frac{10 ^{- 7} \times 2 \times 96 \times 24}{r}$ $∴$ $r = \frac{10 ^{- 7} \times 2 \times 96 \times 24}{0.144}$ $= 3.2 \times 10^{- 3} m$ $= 3.2 mm$