Question

The magnitude of force per unit length on a wire carrying current $I$ at O due to two semi infinite wires as shown in figure, (if the distance between the long parallel segments of the wire being equal to $L$ and current in them is $I$) is:

• A. $\frac{+{\mu }_0{I}^2}{\pi L}\hat{j}$
• B. $\frac{-{\mu }_0{I}^2}{L}\hat{j}$
• C. $\frac{{\mu }_0{I}^2}{2L}\hat{j}$
• D. $\frac{{\mu }_0{I}^2}{L}\hat{j}$

Answer 1

Answer: (A)

$\frac{+{\mu }_0{I}^2}{\pi L}\hat{j}$

Net Magnetic field at point O due to both the wires shall be
$B=-2{\int }_{{90}^0}^{0^0}\frac{{\mu }_{0}I}{2\pi l}\sin \theta d\theta$ along $-k$
$\therefore B=-\frac{{\mu }_{0}I}{\pi l}$ along $-k$
Now, $F=I(\overrightarrow{dl}\times \overrightarrow{B})$
Where, $\overrightarrow{dl}$ is along $i$
Therefore, $\frac{F}{dl}=\frac{\mu I^2}{\pi l}$ along $j$