Question

Two long parallel wires are at a distance of 1 metre. Both of them carry one ampere of current. the force of attraction per unit length between the two wires is

• A. $2\times {10}^{-7}N{m}^{-1}$
• B. $2\times {10}^{-8}N{m}^{-1}$
• C. $5\times {10}^{-8}N{m}^{-1}$
• D. ${10}^{-7}N{m}^{-1}$

Answer 1

Answer: (A)

$2\times {10}^{-7}N{m}^{-1}$

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 $F_{perunitlength}=\frac{F_{21}}{l_2}=\frac{{\mu }_0{i}_1{i}_2}{2\pi d}=\frac{2\times {10}^{-7}\times 1\times 1}{1}=2\times {10}^{-7}N$
where $F_{21}$ is the force on wire 2 due to wire 1 and the subscript 2 refers to 2nd wire.
$B_1$ is in $-\hat{k}$ direction.
$l_2$ is in $-\hat{j}$ direction.
Hence, $F_{21}$ will be in $\hat{j}\times (-\hat{k})=-\hat{i}$ direction.
Similarly, $F_{12}$ will be in $\hat{i}$ direction.
Thus, both the wires will attract each other.