A non-planar loop of conducting wire carrying | vedclass

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(D) The magnetic field at point $P(a, 0, a)$ due to the entire loop can be calculated by considering the loop as a combination of two planar loops: $A

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$\frac{1}{\sqrt{2}}(\hat{i} + \hat{k})$

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(D) The magnetic field at point $P(a, 0, a)$ due to the entire loop can be calculated by considering the loop as a combination of two planar loops: $ABCDA$ (in the $xz$-plane) and $AFEBA$ (in the $xy$-plane).$1$. The magnetic field produced by the loop $ABCDA$ at point $P$ is directed along the positive $x$-axis (i.e.,in the $\hat{i}$ direction) due to the right-hand rule.$2$. The magnetic field produced by the loop $AFEBA$ at point $P$ is directed along the positive $z$-axis (i.e.,in the $\hat{k}$ direction) due to the right-hand rule.$3$. Since the geometry of both loops relative to point $P$ is identical,the magnitudes of the magnetic fields produced by both loops are equal.$4$. Let the magnitude of the magnetic field from each loop be $B_0$. The total magnetic field vector at $P$ is $\vec{B} = B_0\hat{i} + B_0\hat{k}$.$5$. The direction of the resultant magnetic field is given by the unit vector $\hat{n} = \frac{B_0\hat{i} + B_0\hat{k}}{\sqrt{B_0^2 + B_0^2}} = \frac{1}{\sqrt{2}}(\hat{i} + \hat{k})$.

$A$ non-planar loop of conducting wire carrying a current $I$ is placed as shown in the figure. Each of the straight sections of the loop is of length $2a$. The magnetic field due to this loop at the point $P(a, 0, a)$ points in the direction:

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