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Question

(A) The magnetic field $\vec B$ due to a long straight wire carrying current $I$ in the negative $z$-direction (along $-\hat k$) at a point $\vec r =

Vedclass · Accepted Answer

$\frac{{{\mu _0}I\,\left( {y\hat i - x\hat j} \right)}}{{2\pi \,\left( {{x^2} + {y^2}} \right)}}$

Vedclass · Answer

(A) The magnetic field $\vec B$ due to a long straight wire carrying current $I$ in the negative $z$-direction (along $-\hat k$) at a point $\vec r = x\hat i + y\hat j$ is given by the Biot-Savart law or Ampere's law. The direction of the magnetic field is given by the right-hand thumb rule. For a current in the $-\hat k$ direction,the magnetic field lines are clockwise in the $xy$-plane.The magnitude of the magnetic field is $B = \frac{\mu_0 I}{2\pi r}$,where $r = \sqrt{x^2 + y^2}$.The unit vector in the radial direction is $\hat r = \frac{x\hat i + y\hat j}{r}$.The magnetic field vector is $\vec B = B (\hat k 	imes \hat r) = \frac{\mu_0 I}{2\pi r} (\hat k 	imes \frac{x\hat i + y\hat j}{r}) = \frac{\mu_0 I}{2\pi r^2} (x(\hat k 	imes \hat i) + y(\hat k 	imes \hat j))$.Since $\hat k 	imes \hat i = \hat j$ and $\hat k 	imes \hat j = -\hat i$,we get $\vec B = \frac{\mu_0 I}{2\pi (x^2 + y^2)} (x\hat j - y\hat i) = \frac{\mu_0 I (y\hat i - x\hat j)}{2\pi (x^2 + y^2)}$ is incorrect; let's re-evaluate: $\vec B = \frac{\mu_0 I}{2\pi (x^2 + y^2)} (y\hat i - x\hat j)$.Wait,for current in $-\hat k$,the field is $\vec B = \frac{\mu_0 I}{2\pi r^2} (y\hat i - x\hat j)$. Thus,option $A$ is correct.

$A$ 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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