Animated Solution for Physics - Magnetic Effects of Current: The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is [Take, ${\mu }_0=4\pi \times 10^{-7}{\text{ NA}}^{-2}$]

Around you, a steady current of $10\text{ A}$ is marching along the perimeter. Your mission? To calculate the invisible magnetic field swirling at your feet.

In physics, when faced with a complex geometry, the best strategy is to divide and conquer. Instead of tackling the entire triangle at once, we will isolate just one side and analyze the magnetic field it produces at the center.

Biot-Savart Law To find the magnetic field generated by a finite straight wire, we bring out our heavy artillery: the Biot-Savart Law. For a straight wire, the magnetic field $B$ at a perpendicular distance $r$ is given by:

Here, ${\theta }_1$ and ${\theta }_2$ are the angles subtended by the ends of the wire at our observation point, measured from the perpendicular drop. This equation is the bridge between the geometry of the wire and the physics of the magnetic field.

By pure symmetry, the lines joining the center to the vertices bisect the $60^{\circ }$ angles of the triangle, making them $30^{\circ }$.

If we drop a perpendicular from the center to one of the sides, we form a right-angled triangle. The remaining angle at the center is simply $90^{\circ }-30^{\circ }=60^{\circ }$. Therefore, both ${\theta }_1$ and ${\theta }_2$ are exactly $60^{\circ }$.

What about the perpendicular distance $r$? In an equilateral triangle of side $a$, the distance from the center to any side (the inradius) is given by $r=\frac{a}{2\sqrt{3}}$. Since our side length is $1\text{ m}$, we have $r=\frac{1}{2\sqrt{3}}\text{ m}$.

Let's substitute these raw geometric values into our Biot-Savart equation for one side:

But wait, we have three identical sides! By the Right-Hand Thumb Rule, if you curl your fingers along the direction of the current, your thumb points straight out of the page. This means the magnetic field from all three sides points in the exact same direction. They don't cancel out; they reinforce each other!

So, the net magnetic field is simply three times the field of one side:

Let's crunch the numbers. We know that $\frac{{\mu }_0}{4\pi }=10^{-7}\text{ T}\cdot \text{m/A}$. The term $\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)$ simplifies beautifully to just $\sqrt{3}$.

Bringing the denominator up, our equation transforms into:

Multiplying the constants: $3\times 10\times 2\times 3=180$. So $B_{net}=180\times 10^{-7}\text{ T}=18\times 10^{-6}\text{ T}$.

This gives us our final, elegant answer:

The beauty of this problem lies in how perfectly the geometry of the triangle aligns with the physics of the Biot-Savart law. Once you master this, you can easily generalize it to a square, a hexagon, or any regular polygon!