Test Index

BITSAT 2015 Paper

Section: Physics

Question : 40 of 150

Marks: +1, -0

\overset{\rightarrow}{B}

Solution:

The wire carries a current I in the negative z-direction. We have to consider the magnetic vector field

\overset{\rightarrow}{B}

at (x, y) in the z = 0 plane.

Magnetic field

\overset{\rightarrow}{B}

is perpendicular to OP.

∴

\overset{\rightarrow}{B}

= b sin θ

\hat{i}

- B cos θ

\hat{j}

sin θ =

\frac{y}{r}

, cos θ =

\frac{x}{r}

, B =

\frac{\mu_0 I}{2\pi r}

∴

\overset{\rightarrow}{B}

\frac{\mu_0 I}{2\pi r^2} (\widehat{yi} - \widehat{xj})

\overset{\rightarrow}{B}

\frac{\mu_0 I (\widehat{yi} - \widehat{xj})}{2\pi (x^2+y^2)}

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