Question

Magnitude of magnetic field (in SI units) at the center of a hexagonal shape coil of side $10\text{cm}\&50$ turns and carrying current $I\text{(Ampere)}$ 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 $500\sqrt{3}$


Magnetic field due to one side of the hexagon is,

$B=\frac{{\mu }_{0}I}{4\pi (\frac{\sqrt{3}a}{2})}(\sin {30}^{\circ }+\sin {30}^{\circ })$

$\Rightarrow B=\frac{{\mu }_{0}I}{2\sqrt{3}a\pi }(\frac{1}{2}+\frac{1}{2})=\frac{{\mu }_{0}I}{2\sqrt{3}a\pi }$


Now, magnetic field due to the complete hexagonal coil is,

$B=6\times \frac{{\mu }_{0}I}{2\sqrt{3}a\pi }=\frac{\sqrt{3}{\mu }_{0}I}{a\pi }$

Again, magnetic field at the centre of a hexagonal shape coil of $50$ turns is,

$B=50\times \frac{\sqrt{3}{\mu }_{0}I}{a\pi }[\because a=\frac{10}{100}=0.1\text{m}]$

$B=\frac{500\sqrt{3}{\mu }_{0}I}{\pi }$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.