Question

Three long, straight and parallel wires are arranged as shown in figure. The force experienced by $10cm$ length of wire $Q$ is

• A. $1.4\times {10}^{-4}N$ toward the right
• B. $1.4\times {10}^{-4}N$ toward the left
• C. $2.6\times {10}^{-4}N$ toward the right
• D. $2.6\times {10}^{-4}N$ toward the left

Answer 1

The correct option is A $1.4\times {10}^{-4}N$ toward the right

Magnetic field produced by wire R at Q: $B_1=\frac{{\mu }_0{i}_1}{2\pi r_1}=\frac{{\mu }_0\times 20}{2\pi \times (2\times {10}^{-2})}$

Magnetic field produced by wire P at Q: $B_2=\frac{{\mu }_0{i}_2}{2\pi r_2}=\frac{{\mu }_0\times 30}{2\pi \times (10\times {10}^{-2})}$

$B_{net}=B_1-B_2=\frac{{\mu }_0\times 7}{2\pi \times {10}^{-2}}$

Force on 10 cm of wire at Q: $F_Q=i_{Q}l{B}_{net}$

$F=\frac{4\pi \times {10}^{-7}\times 20\times 10\times 10\times {10}^{-2}}{2\pi \times 10\times {10}^{-2}}$

$=\frac{4\pi \times {10}^{-7}\times 10\times {10}^{-2}}{2\pi \times {10}^{-2}}[100-30]$

$=20\times {10}^{-7}\times 70=1400\times {10}^{-7}$

$=1.4\times {10}^{-4}N$ toward right.