Question

A straight wire is carrying a current of $100A$ as shown in the figure. What is the magnitude of the magnetic field at point $P$ which is at a perpendicular distance $(\sqrt{3}-1)m$ from the wire? (It is also given that ends $A$ and $B$ of the wire subtend angles $30^{{}^{\circ }}$ and $60^{{}^{\circ }}$ respectively at $P$ as shown)

• A. $5\times 10^{-6}T$
• B. $2.5\times 10^{-6}T$
• C. $2.5\times 10^{-5}T$
• D. $8\times 10^{-5}T$

Answer 1

Answer: (A)

Magnetic field due to a straight current carrying conductor is given by
$B=\frac{{\mu }_{0}I}{4\pi d}[sin\alpha +sin\beta ]$
Here $\alpha$ is considered positive when measured clockwise and $\beta$ is considered positive when measured anticlockwise.

P
$\alpha =60^{\circ }$ (clockwise)
$\beta =150^{\circ }$ (anticlockwise)
Also, $\beta =-30^{\circ }$ (clockwise)
$d=\left(\sqrt{3}-1\right)m$ and $i=100A$
Plugging these values in the formula we get
$B=\frac{{\left(10\right)}^{-7}\times 100}{\left(\sqrt{3}-1\right)}\left[\frac{\sqrt{3}}{2}-\frac{1}{2}\right]=5\times {\left(10\right)}^{-6}T$