Question

A long straight wire is fixed horizontally and carries a current of $50A.$ A second wire having linear mass density $1\cdot 0\times {10}^{-4}kg{m}^{-1}$ is placed parallel to and directly above this wire at a separation of $5\cdot 0mm$. What current should this second wire carry such that the magnetic repulsion can balance its weight ?

Answer 1

$\frac{{\mu }_0{i}_1{i}_2}{2\pi d}=mg$ (For a portion of wire of length $1m$)

$\Rightarrow \frac{{\mu }_0\times 50\times i_2}{2\pi \times 5\times {10}^{-3}}=1\times {10}^{-4}\times 9.8$

$\Rightarrow \frac{4\pi \times {10}^{-7}\times 5\times i_2}{2\pi \times 5\times {10}^{-3}}=9.8\times {10}^{-4}$

$\Rightarrow 2\times i_2\times {10}^{-3}=9.3\times {10}^{-3}\times {10}^{-1}$

$\Rightarrow i_2=\frac{9.8}{2}\times {10}^{-1}=0.49A$