Question

The magnitude of magnetic field (in $SIunits$) at the center of a hexagonal shape coil of side$10cm,50$ turns and carrying current $I\left(Ampere\right)$ in units of$\frac{{\mu }_{0}I}{\pi }$ is

• A. $250\sqrt{3}$
• B. $50\sqrt{3}$
• C. $500\sqrt{3}$
• D. $5\sqrt{3}$

Answer 1

The correct option is C

$500\sqrt{3}$

Step 1. Given data:

The length of each side of a hexagonal shape coil$\left(d\right)=10cm=0.1m$

Number of turns$N=50$.

Current through the coil$=I$

Step 2. Formula used:

Magnetic field, $\left(B\right)=\frac{{\mu }_{0}NI}{4\pi d}\left(\sin {\theta }_1+\sin {\theta }_2\right)$

The magnetic field at the center of a hexagonal shape coil is,

$B=6\times \frac{{\mu }_{0}NI}{4\pi d}\left(\sin {\theta }_1+\sin {\theta }_2\right)$

Step 3. Calculating the magnetic field at the center:

From this figure, we can get,

$\begin{array}{rcl}\tan {30}^{\circ }& =& \frac{1}{\sqrt{3}}\\ \tan {30}^{\circ }& =& \frac{x}{d}\\ \therefore \frac{1}{\sqrt{3}}& =& \frac{x}{d}\\ & \Rightarrow & d=x\sqrt{3}\\ d& =& 0.05\sqrt{3}m\\ & & \end{array}$

The magnetic field at the center of the coil,

$\begin{array}{rcl}B& =& 6\times \frac{{\mu }_{0}NI}{4\pi d}\left(\sin {\theta }_1+\sin {\theta }_2\right)\\ & =& \frac{{\mu }_{0}I}{\pi }\times 6\times \frac{50}{4\times 0.05\sqrt{3}}\times \left(\sin {30}^{\circ }+\sin {30}^{\circ }\right)\\ & =& \frac{{\mu }_{0}I}{\pi }\times 500\sqrt{3}\end{array}$

Thus, the magnetic field at the center of the coil in units of$\frac{{\mu }_{0}I}{\pi }$ is$500\surd 3$

Hence, option C is the correct answer.