Question

Two parallel wires P and Q carry electric currents of $10A$ and $2A$ respectively in mutually opposite directions. The distance between the wires is $10cm$. If the wire P is of infinite length and wire Q is $2m$ long, then the force acting Q will be

• A. $4\times {10}^{-5}N$
• B. $8\times {10}^{-5}N$
• C. $6\times {10}^{-5}N$
• D. $14\times {10}^{-5}N$

Answer 1

The correct option is B $8\times {10}^{-5}N$

Force on a current carrying conductor is $F=\int i(\overrightarrow{dl}\times \overrightarrow{B})$
Magnetic field due to one wire at the location of the second wire ( other one ) is $B_1=\frac{{\mu }_0{i}_1}{2\pi d}$
Force on the 2nd wire due to $B_1$ is $F_{21}=i_2{l}_2{B}_1=i_2{l}_2\frac{{\mu }_0{i}_1}{2\pi d}=\frac{{\mu }_0{i}_1{i}_2{l}_2}{2\pi d}$
Force per unit length for wire $l_2$ is $F_{perunitlength}=\frac{F_{21}}{l_2}$
$B_1$ is in $-\hat{k}$ direction.
$l_2$ is in $\hat{j}$ direction.
Hence, $F_{21}$ will be in $\hat{j}\times \left(\hat{k}\right)=\hat{i}$ direction.
$\therefore$ it is an repulsive force.
$F_{21}=\frac{2\times {10}^{-7}\times 10\times 2\times 2}{10\times {10}^{-2}}=8\times {10}^{-5}N$