Two long conductors are arranged as shown abo | vedclass

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Question

(D) For a long cylindrical conductor with uniform current density $J$,the magnetic field at a distance $x$ from the center (where $x < r$) is given by

Vedclass · Accepted Answer

$\mu_0 J d/2$,in the $+y$-direction

Vedclass · Answer

(D) For a long cylindrical conductor with uniform current density $J$,the magnetic field at a distance $x$ from the center (where $x $B(2\pi x) = \mu_0 (J \cdot \pi x^2) \implies B = \frac{\mu_0 J x}{2}$.Let the center of the left cylinder be at $x = -d/2$ and the right cylinder at $x = d/2$. Point $A$ is at the origin $(0,0)$.The field due to the left cylinder (current out of the page) at $A$ is $B_1 = \frac{\mu_0 J (d/2)}{2} = \frac{\mu_0 J d}{4}$ in the $+y$-direction.The field due to the right cylinder (current into the page) at $A$ is $B_2 = \frac{\mu_0 J (d/2)}{2} = \frac{\mu_0 J d}{4}$ in the $+y$-direction.The total magnetic field is $B_{net} = B_1 + B_2 = \frac{\mu_0 J d}{2}$ in the $+y$-direction.

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