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Question

(A) The current $I$ flows in the negative $z$-direction,i.e.,along $-\hat{k}$.The magnetic field $\vec{B}$ at a point $P(x, y)$ in the $xy$-plane is g

Vedclass · Accepted Answer

$\frac{\mu_0 I (y\hat{i} - x\hat{j})}{2\pi (x^2 + y^2)}$

Vedclass · Answer

(A) The current $I$ flows in the negative $z$-direction,i.e.,along $-\hat{k}$.The magnetic field $\vec{B}$ at a point $P(x, y)$ in the $xy$-plane is given by the right-hand grip rule.The magnitude of the magnetic field at a distance $r = \sqrt{x^2 + y^2}$ from the wire is $B = \frac{\mu_0 I}{2\pi r}$.The direction of the magnetic field is perpendicular to the position vector $\vec{r} = x\hat{i} + y\hat{j}$.Using the cross product $\vec{B} = \frac{\mu_0 I}{2\pi r^2} (\hat{k} 	imes \vec{r})$,where $\hat{k}$ is the direction of current flow $(-I\hat{k})$,we get:$\vec{B} = \frac{\mu_0 (-I)}{2\pi r^2} (\hat{k} 	imes (x\hat{i} + y\hat{j})) = \frac{-\mu_0 I}{2\pi (x^2 + y^2)} (x\hat{j} - y\hat{i}) = \frac{\mu_0 I (y\hat{i} - x\hat{j})}{2\pi (x^2 + y^2)}$.

$A$ long straight wire along the $z$-axis carries a current $I$ in the negative $z$-direction. The magnetic vector field $\vec{B}$ at a point having coordinates $(x, y)$ in the $z = 0$ plane is

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