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(A) The current $I$ flows in the negative $z-$ direction,i.e.,along $-\hat k$. The position vector of the point $P(x, y)$ is $\vec r = x\hat i + y\hat

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 position vector of the point $P(x, y)$ is $\vec r = x\hat i + y\hat j$. Using the Biot-Savart law or the right-hand rule for a long straight wire,the magnetic field $\vec B$ is given by $\vec B = \frac{\mu_0 I}{2\pi r^2} (\hat k 	imes \vec r)$. Substituting $\vec r = x\hat i + y\hat j$ and the current direction $-\hat k$,the magnetic field is $\vec B = \frac{\mu_0 I}{2\pi r^2} ((-\hat k) 	imes (x\hat i + y\hat j))$. Since $\hat k 	imes \hat i = \hat j$ and $\hat k 	imes \hat j = -\hat i$,we get $\vec B = \frac{\mu_0 I}{2\pi r^2} (-(x\hat j - y\hat i)) = \frac{\mu_0 I}{2\pi r^2} (y\hat i - x\hat j)$. Substituting $r^2 = x^2 + y^2$,we get $\vec B = \frac{\mu_0 I (y\hat i - x\hat j)}{2\pi (x^2 + y^2)}$.

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

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