Question

Three long, straight parallel wires carrying current are arranged as shown in figure. The net magnetic force experienced by unit length of wire C is

• A. $8\times {10}^{-4}N/m$
• B. $4\times {10}^{-4}N/m$
• C. $12\times {10}^{-4}N/m$
• D. $20\times {10}^{-4}N/m$

Answer 1

The correct option is A $8\times {10}^{-4}N/m$

The force between two parallel current carrying wires having current $i_{1}and{i}_2$ per unit length placed at a separation r is $\frac{{\mu }_0{i}_1{i}_2}{2\pi r}$

Force per unit length on wire C due to wire $A=\frac{{\mu }_0\times \left(30\right)\times \left(10\right)}{2\pi \times (5\times {10}^{-2})}$
Force per unit length on wire C due to wire $B=\frac{{\mu }_0\times \left(20\right)\times \left(10\right)}{2\pi \times (10\times {10}^{-2})}$
$\therefore$ Net force per unit length on wire $C=\frac{{\mu }_0\times \left(30\right)\times \left(10\right)}{2\pi \times (5\times {10}^{-2})}-\frac{{\mu }_0\times \left(20\right)\times \left(10\right)}{2\pi \times (10\times {10}^{-2})}$
$=\frac{{\mu }_0}{2\pi }(\frac{300}{5\times {10}^{-2}}-\frac{200}{10\times {10}^2})$
$=8\times {10}^{-4}N/m$
Hence, the correct answer is option (1).