Question

Find the magnitude and direction of magnetic field at point $P$ due to the current carrying wire as shown in figure below.

• A. -$\frac{{\mu }_{0}I}{4\pi a}[-\frac{\sqrt{3}}{2}+\frac{1}{2}]$
• B. $\frac{{\mu }_{0}I}{2\pi a}[\frac{\sqrt{3}}{2}-\frac{1}{2}]$
• C. $\frac{{\mu }_{0}I}{4\pi a}[\frac{\sqrt{3}}{2}-\frac{1}{2}]$
• D. -$\frac{{\mu }_{0}I}{2\pi a}[-\frac{\sqrt{3}}{2}+\frac{1}{2}]$

Answer 1

The correct option is C $\frac{{\mu }_{0}I}{4\pi a}[\frac{\sqrt{3}}{2}-\frac{1}{2}]$

Magnetic field due to a finite steady current $I$ carrying wire at a point on the perpendicular in the surrounding space is given by $$B=\frac{{\mu }_{0}I}{4\pi d}[\sin \left({\theta }_1\right)+\sin \left({\theta }_2\right)]$$


Magnetic field due to wires $AB$ and $CD$ will be zero.

Thus, Magnetic field due to $BC$ at the given point in space is $\overrightarrow{B}=\frac{{\mu }_{0}I}{4\pi a}[\sin {60}^{\circ }-\sin {30}^{\circ }]$

$\Rightarrow \overrightarrow{B}=\frac{{\mu }_{0}I}{4\pi a}[\frac{\sqrt{3}}{2}-\frac{1}{2}]$

Hence, option (c) is the correct answer.