Question

A current of 1 a is flowing on the sides of an equilateral triangle of side 4.5 10 2 m. The magnetic field at the centre of the triangle will be :

Answer 1

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Answer 2

Answer: The correct answer is $4\times 10^{-5} Wb/m^{2}$.

Explanation:

Each angle of the equilateral triangle is 60 degree. By using angle sum property the angle at the centre of the equilateral triangle is 120 degree. Median will act as the mid point on the side of the triangle.

Calculate the distance of the wire from the centre.

$R=\frac{r}{2\sqrt{3}}$

Here, R is the distance from the centre of the triangle to the point on the side of the equilateral triangle and r is the side of the triangle.

Calculate the magnetic field due to the side BC of the equilateral triangle.

$B_{BC}= \frac{\mu _{0}I}{4\pi R}(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2})$

Here, ${\mu _{0}$ is the absolute permeability of the free space and R is the distance.

$B_{BC}= \sqrt{3}\frac{\mu _{0}I}{4\pi R}$

Calculate the magnetic field at the centre of the triangle.

$B_{net}= 3\sqrt{3}\frac{\mu _{0}I}{4\pi R}$

Put $R=\frac{r}{2\sqrt{3}}$.

$B_{net}= 3\sqrt{3}\frac{\mu _{0}I}{4\pi(\frac{r}{2\sqrt{3}})}$

Put I = 1 A and $BC=r= 4.5\times 10^{-2}$ and $\mu _{0}=4\pi \times 10^{-7} F/m$ .

$B_{net}= 3\sqrt{3}\frac{4\pi \times 10^{-7}(1)}{4\pi(\frac{4.5\times 10^{-2}}{2\sqrt{3}})}$

$B_{net}=4\times 10^{-5} Wb/m^{2}$

Therefore, the magnetic field at the center of the triangle is  $4\times 10^{-5} Wb/m^{2}$.