Physics - Magnetism Question with Solution | TestHub

The magnetic force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $h$ is given by:

$f=\frac{{\mu }_0{I}_1{I}_2}{2\pi h}$

Given $I_1=I$ and $I_2=2I$, the upward attractive force per unit length on wire A due to wire B is:

$f=\frac{{\mu }_0(I)(2I)}{2\pi h}=\frac{{\mu }_0{I}^2}{\pi h}$

The downward gravitational force per unit length on wire A is its weight per unit length:

$W=\lambda g$

For wire A not to rise from the floor, the upward magnetic force must be less than or equal to the downward gravitational force:

$f\le W$

$\frac{{\mu }_0{I}^2}{\pi h}\le \lambda g$

$h\ge \frac{{\mu }_0{I}^2}{\pi \lambda g}$

Therefore, the minimum magnitude of $h$ is $\frac{{\mu }_0{I}^2}{\pi \lambda g}$.

Answer: $\frac{{\mu }_0{\text{I}}^2}{\pi \lambda g}$