Question

Two horizontal parallel straight conductors, each $20\text{cm}$ 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 vertical plane remaining parallel to the lower. If the upper conductor weights $1.20\text{g}$, what is the approximate current that will maintain the conductors at a distance $0.75\text{cm}$ apart. (Given : $g=9.8{\text{m/s}}^2$)

• A. $27\text{A}$
• B. $47\text{A}$
• C. $35\text{A}$
• D. $15\text{A}$

Answer 1

The correct option is B $47\text{A}$

The direction of magnetic force on the upper wire will be upward due to repulsion.

In equilibrium, magnetic force $F_m$ will balance weight $mg$.

So, $mg=F_m\Rightarrow mg=\frac{{\mu }_0{i}^{2}l}{2\pi d}$

$\Rightarrow i=\sqrt{\frac{2\pi mgd}{{\mu }_{0}l}}$

$\Rightarrow i=\sqrt{\frac{2\pi \times 1.2\times {10}^{-3}\times 9.8\times 0.75\times {10}^{-2}}{4\pi \times {10}^{-7}\times 20\times {10}^{-2}}}=\sqrt{2205}$

$i=47\text{A}$