A long straight wire along the z-axis carries a current I in the negative z direction. The magnetic vector field mathop Blimits^ to at a

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

a (a) Magnetic field at $P$ is $\overrightarrow B $, perpendicular to $OP$ in the direction shown in figure. So, $\overrightarrow B = B\sin \theta \,\hat i - B\cos \theta \,\hat j$ Here $B = \frac{{{\mu _0}}}{{2\pi }}\frac{I}{r}$ $\sin \theta = \frac{y}{r}$ and $\cos \theta = \frac{x}{r}$ $\overrightarrow B = \frac{{{\mu _0}I}}{{2\pi }} \cdot \frac{1}{{{r^2}}}(y\hat i - x\hat j) = \frac{{{\mu _0}I(y\hat i - x\hat j)}}{{2\pi ({x^2} + {y^2})}}$(as ${r^2} = {x^2} + {y^2}$)

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

A$\frac{{{\mu _o}I\,(y\hat i - x\hat j)}}{{2\pi ({x^2} + {y^2})}}$
B$\frac{{{\mu _o}I\,(x\hat i + y\hat j)}}{{2\pi ({x^2} + {y^2})}}$
C$\frac{{{\mu _o}I\,(x\hat j - y\hat i)}}{{2\pi ({x^2} + {y^2})}}$
D$\frac{{{\mu _o}I\,(x\hat i - y\hat j)}}{{2\pi ({x^2} + {y^2})}}$

IIT 2002, Diffcult

STD 11 Science
NEET
STD 12 - 4. Moving charges and magnetism
Physics

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