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Two long wires carrying current $I_1$ and $I_2$ are arranged as shown in figure. the one carrying current $I_1$ is along the x-axis . the other carrying current $I_2$ is along a line parallel to the y-axis , given by $x=0$ and $z=d$ find the force exacted at $O_2$ because of the wire along the x-axis

$\begin{array}{1 1} 1 \\-1\\ \pm 1 \\ 0 \end{array} $

1 Answer

Solution :
Magnetic field at the point $O_2$ due to the current $I_1$ is
$BO_2 =\large\frac{\mu_0}{2 \pi}. \frac{2I}{d}$
The direction is along y-axis (apply maxwells right hand rule)
The second wire along y -axis
$\therefore $ the angle between $I_2 \overrightarrow{l} $ and $\overrightarrow{BO_2}$ is zero.
This implies $\theta=0^{\circ}$
Force on the wire lying along y-axis
$f= I_2 | \overrightarrow{l} \times \overrightarrow{BO_2}|$
$\quad= I_2 l BO_2 \sin 0^{\circ}=0$
Answer :0
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