Question

Two current-carrying parallel conductors are shown in the figure. The magnitude and nature of force acting between them per unit length will be :

• A. $8\times {10}^{-8}N/m$, attractive
• B. $3.2\times {10}^{-5}N/m$, repulsive
• C. $3.2\times {10}^{-5}N/m$, attractive
• D. $8\times {10}^{-8}N/m$, repulsive

Answer 1

The correct option is C $3.2\times {10}^{-5}N/m$, attractive

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 $F_{perunitlength}=\frac{F_{21}}{l_2}=\frac{{\mu }_0{i}_1{i}_2}{2\pi d}=\frac{2\times {10}^{-7}\times 4\times 4}{10\times {10}^{-2}}=3.2\times {10}^{-5}N$

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.