Question

Two horizontal parallel straight conductors, each $20\, cm$ long, are arranged one vertically above the other and carry equal currents in opposite directions. The lower conductor is fixed while the other is free to move in guides remaining parallel to the lower. If the upper conductor weights $1.0 \,g$, what is the approximate current (in A) that will maintain the conductors at a distance $0.50\, cm$ apart?

Answer 1

In equilibrium magnetic force $F_{m}$ will balance weight $m g$.

So $m g=F_{m} \Rightarrow m g=\frac{\mu_{0} i^{2} l}{2 \pi d}$
$\Rightarrow =\sqrt{\frac{2 \pi m g d}{\mu_{0} l}}$
$=\sqrt{\frac{2 \pi \times 2.0 \times 10^{-3} \times 10 \times 0.50 \times 10^{-2}}{4 \pi \times 10^{-7} \times 20 \times 10^{-2}}} $
$=\sqrt{2500}=50 A$