A current of 1.5 A is flowing through a triangle, of side 9 cm each. The magnetic field at the centr

Q: Why do we multiply the magnetic field of one side by 3?

Because the triangle is equilateral and the centroid is equidistant from all three sides. The current is the same, and the direction of the magnetic field produced by each side at the centroid is identical (into the page). Thus, the total field is the sum of three equal contributions.

Q: How did we find the angles $\theta_1$ and $\theta_2$ to be $60^\circ$?

The lines joining the centroid to the vertices of an equilateral triangle bisect the $60^\circ$ interior angles. In the right-angled triangle formed by the centroid, a vertex, and the midpoint of a side, the angle at the centroid is $180^\circ - 90^\circ - 30^\circ = 60^\circ$.

Q: What is the formula for the perpendicular distance $r$?

In the right-angled triangle $AOD$, $\tan 30^\circ = \frac{r}{l/2}$. Solving for $r$ gives $r = \frac{l}{2\sqrt{3}}$.

Q: How do we determine the direction of the magnetic field?

Using the Right-Hand Grip Rule: point your right thumb in the direction of the current along any side, and your fingers will curl into the page at the location of the centroid. Since the current is clockwise, the net field is directed into the plane.

Q: Can this method be used for other shapes like a square?

Absolutely! For any regular polygon, you can find the field due to one side using the finite wire formula and multiply it by the number of sides. The perpendicular distance and angles will just depend on the specific geometry of the polygon.

Question

A current of $1.5 	ext{ A}$ is flowing through a triangle, of side $9 	ext{ cm}$ each. The magnetic field at the centroid of the triangle is (Assume that, the current is flowing in the clockwise direction.)

Options:

(A) $3 	imes 10^{-7} 	ext{ T}$, outside the plane of triangle

(B) $2\sqrt{3} 	imes 10^{-7} 	ext{ T}$, outside the plane of triangle

(C) $2\sqrt{3} 	imes 10^{-5} 	ext{ T}$, inside the plane of triangle

(D) $3 	imes 10^{-5} 	ext{ T}$, inside the plane of triangle

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine an equilateral triangle with a current flowing clockwise through its sides. The problem asks us to find the magnetic field at the centroid of this triangle.

The centroid is a special point—it is equidistant from all three sides. Because of this perfect symmetry, the magnetic field produced by each side at the centroid will be identical in both magnitude and direction.

Therefore, we don't need to calculate the field for all three sides from scratch. We can simply find the magnetic field due to one side and multiply it by three!

### The Master Equation

To find the magnetic field from a straight finite wire, we rely on the formula derived from the **Biot-Savart Law**:
$$B = \frac{\mu_0 I}{4\pi r} (\sin	heta_1 + \sin	heta_2)$$

Here, $r$ is the perpendicular distance from the wire to the point of interest, and $	heta_1$ and $	heta_2$ are the angles subtended by the ends of the wire at that point.

If we draw a perpendicular from the centroid to one of the sides, it bisects the angle subtended by the entire side. For an equilateral triangle, the total angle subtended by a side at the centroid is $120^\circ$. Thus, the bisected angles are $	heta_1 = 60^\circ$ and $	heta_2 = 60^\circ$.

### Calculating the Perpendicular Distance

Before we can use our master equation, we need to find $r$. Let's look at the small right-angled triangle formed by the centroid, a vertex, and the midpoint of a side.

The angle at the vertex is bisected, so it is $30^\circ$. Using basic trigonometry:
$$	an 30^\circ = \frac{r}{l/2}$$

Since the total side length is $l$, the adjacent side is $l/2$. Solving for $r$, we get:
$$r = \frac{l}{2\sqrt{3}}$$

### The Atomic Compute

Now, let's substitute our values for $r$ and the angles into the magnetic field formula for a single side:
$$B_1 = \frac{\mu_0 I}{4\pi \left(\frac{l}{2\sqrt{3}}ight)} (\sin 60^\circ + \sin 60^\circ)$$

We know that $\sin 60^\circ = \frac{\sqrt{3}}{2}$. Adding them together gives $\sqrt{3}$. Simplifying the expression, we find the magnetic field due to one side:
$$B_1 = \frac{3\mu_0 I}{2\pi l}$$

Since the total magnetic field is three times this value, we have:
$$B_{net} = 3 	imes B_1 = \frac{9\mu_0 I}{2\pi l}$$

### Final Calculation

It's time to plug in the given numerical values. We have $I = 1.5 	ext{ A}$ and $l = 9 	imes 10^{-2} 	ext{ m}$. Remember that $\mu_0 = 4\pi 	imes 10^{-7} 	ext{ T}\cdot	ext{m/A}$.
$$B_{net} = \frac{9 	imes (4\pi 	imes 10^{-7}) 	imes 1.5}{2\pi 	imes 9 	imes 10^{-2}}$$

The $\pi$ cancels out, and the $9$s cancel out beautifully. We are left with:
$$B_{net} = 3 	imes 10^{-5} 	ext{ T}$$

### Determining the Direction

Finally, what about the direction? We use the **Right-Hand Grip Rule**.

If you curl the fingers of your right hand in the clockwise direction of the current, your thumb points directly into the screen. Therefore, the magnetic field is directed **inside the plane** of the triangle.

This matches option (d) perfectly!

Animated Solution for Physics - Magnetic Effects of Current: A current of $1.5 A$ is flowing through a triangle, of side $9 cm$ each. The magnetic field at the centroid of the triangle is (Assume that, the current is flowing in the clockwise direction.)

Select Answer:

Visualized Solution

Analyzing the Geometry

Biot-Savart Law for Finite Wire

Calculating Perpendicular Distance $r$

Field Due to One Side

Total Magnetic Field

Final Calculation & Direction

Generalizing for Regular Polygons

The Sigma Insight: Biot-Savart Law

Analyzing the Setup

The Master Equation

Calculating the Perpendicular Distance

The Atomic Compute

Final Calculation

Determining the Direction

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