Question

An equilateral triangle of side $a$ carries a current $i$, then the magnetic field at the vertex $\text{P}$ of triangle is

• A. $\frac{{\mu }_{0}i}{2\sqrt{3}\pi a}\otimes$
• B. $\frac{{\mu }_{0}i}{2\sqrt{3}\pi a}\odot$
• C. $\frac{2\sqrt{3}{\mu }_{0}i}{\pi a}\odot$
• D. zero

Answer 1

The correct option is B $\frac{{\mu }_{0}i}{2\sqrt{3}\pi a}\odot$

The net magnetic field at $\text{P}$ is equal to vector sum of magnetic fields at $\text{P}$ produced by three wires $\text{AP, PB}$ and $\text{AB}$.

For straight wire $\text{AP}$ and $\text{PB}$, the magnetic field produced at $\text{P}$ $=0$

$(\because \overrightarrow{dl}\times \overrightarrow{r}=0,\text{point lies on axis of wire})$

The lnegth $\text{PC}$ is,

${\text{PC = AP cos 30}}^{\circ }=a\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}a}{2}$

The perpendicular distance of $\text{P}$ from straight wire $\text{AB}$ is

$\text{d = PC}=\frac{\sqrt{3}a}{2}$
Using the relation for magnitude of magnetic field at $\text{P}$ due to straight wire $\text{AB}$:
$\text{B}=\frac{{\mu }_{0}i}{4\pi d}[sin{\varphi }_1+sin{\varphi }_2]$

$(\because {\varphi }_1={30}^{\circ }\text{and}{\varphi }_2={30}^{\circ })$

$\Rightarrow B=\frac{{\mu }_{0}i}{4\pi \left(\frac{\sqrt{3}a}{2}\right)}\left[2sin{30}^{\circ }\right]$

or, $B=\frac{2{\mu }_{0}i}{4\pi \sqrt{3}a}=\frac{{\mu }_{0}i}{2\pi \sqrt{3}a}$

From right hand thumb rule the direction of $\overrightarrow{B}$ at $\text{P}$ is perpendicularly outward to the plane $(\odot )$.

$\therefore$ option ($b$) is the correct answer.

$\begin{array}{c}\text{Why this question?}\\ \text{Tip: The direction of}\overrightarrow{B}\text{can be either given by using right hand thumb rule or by obtaining the direction of}\overrightarrow{dl}\times \overrightarrow{r}\text{from biot-savart's law.}\end{array}$