Two long parallel wires are at a distance 2d | vedclass

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(A) Let the two wires be located at $x = -d$ and $x = +d$. Both carry current $I$ in the outward direction (positive $z$-axis).Using the right-hand ru

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(A) Let the two wires be located at $x = -d$ and $x = +d$. Both carry current $I$ in the outward direction (positive $z$-axis).Using the right-hand rule,the magnetic field $B$ due to a single wire at distance $r$ is $B = \frac{\mu_0 I}{2\pi r}$.For the wire at $x = -d$,the field is $B_1 = \frac{\mu_0 I}{2\pi (x+d)}$ directed in the negative $y$-direction (downward) for $x > -d$.For the wire at $x = +d$,the field is $B_2 = \frac{\mu_0 I}{2\pi (d-x)}$ directed in the positive $y$-direction (upward) for $x The net magnetic field $B_{net} = B_2 - B_1 = \frac{\mu_0 I}{2\pi} \left( \frac{1}{d-x} - \frac{1}{x+d} ight) = \frac{\mu_0 I}{2\pi} \left( \frac{x+d - (d-x)}{d^2 - x^2} ight) = \frac{\mu_0 I x}{\pi (d^2 - x^2)}$.At $x = 0$,$B_{net} = 0$.As $x 	o d^-$,$B_{net} 	o +\infty$.As $x 	o -d^+$,$B_{net} 	o -\infty$.This behavior corresponds to the graph shown in option $A$.

Two long parallel wires are at a distance $2d$ apart. They carry steady equal currents flowing out of the plane of the paper as shown. The variation of the magnetic field $B$ along the line $XX'$ is given by

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