A coil in the shape of an equilateral triangle of side 10 cm lies in a vertical plane between the po

Q: Why is the angle $\theta$ taken as $90^\circ$ when the problem says the plane is parallel to the magnetic field?

This is a classic trap! The angle $\theta$ in the torque formula $\tau = MB \sin\theta$ is the angle between the **magnetic field** and the **area vector** of the coil, not the plane of the coil. The area vector is always perpendicular to the plane of the coil. So, if the plane is parallel to the field, the area vector is perpendicular to it, making $\theta = 90^\circ$.

Q: What exactly is the magnetic dipole moment?

Think of the magnetic dipole moment as a measure of how strongly the current loop behaves like a tiny bar magnet. It depends on how much current is flowing ($i$) and how large the loop is ($A$). A larger area or a higher current makes the "magnet" stronger!

Q: Does the shape of the coil matter for the torque formula?

The beautiful thing about the formula $\vec{\tau} = \vec{M} \times \vec{B}$ is that it is universally true for any planar loop, regardless of its shape! Whether it's a circle, a square, or an equilateral triangle, as long as you can calculate its area, the formula works perfectly.

Q: Why did we convert $10 \text{ cm}$ to $0.1 \text{ m}$ and $20 \text{ mT}$ to $20 \times 10^{-3} \text{ T}$?

Physics is unforgiving when it comes to units! To get the final torque in standard SI units (Newton-meters), every single input must be in its base SI unit. Centimeters must become meters, and milli-Teslas must become Teslas. Skipping this step is the most common way to lose marks.

Q: What would happen if the current was reversed?

Reversing the current would flip the direction of the area vector (and thus the magnetic moment) by $180^\circ$. The magnitude of the torque would remain exactly the same, but its direction would reverse, meaning the coil would tend to rotate in the opposite direction.

Question

A coil in the shape of an equilateral triangle of side $10 	ext{ cm}$ lies in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field $20 	ext{ mT}$. The torque acting on the coil when a current of $0.2 	ext{ A}$ is passed through it and its plane becomes parallel to the magnetic field will be $\sqrt{x} 	imes 10^{-5} 	ext{ Nm}$. The value of $x$ is.......... .

Sigma Percentile · Accepted Answer

### The Magic of Magnetic Torque

Imagine you are holding a loop of wire, and you suddenly turn on a strong magnetic field. If there is a current flowing through that wire, it won't just sit there—it will experience a twisting force, a torque, that tries to align it with the magnetic field. This is the exact same principle that makes electric motors spin!

In our problem, we have a coil shaped like an equilateral triangle. It is placed in a vertical plane, and a horizontal magnetic field is passing through it. Our goal is to find the exact torque acting on this triangular coil.

### Visualizing the Geometry: The Crucial Angle

The master equation for the torque $	au$ on a current-carrying loop is given by:

$$ 	au = MB \sin	heta $$

Here, $M$ is the magnetic dipole moment, $B$ is the magnetic field strength, and $	heta$ is the angle between the magnetic field and the **area vector** of the coil.

This is where many students make a critical mistake! The problem states that the plane of the coil is parallel to the magnetic field. It is tempting to plug in $	heta = 0^\circ$. However, the area vector is always perpendicular to the surface of the coil. If the surface is parallel to the field, the area vector must be perpendicular to the field. Therefore, the correct angle to use is $	heta = 90^\circ$.

### The Anatomy of the Coil: Magnetic Moment and Area

The magnetic dipole moment $M$ is a measure of the strength of our current loop acting as a magnet. It is simply the product of the current $i$ and the area $A$ of the loop:

$$ M = iA $$

Since our coil is an equilateral triangle with side length $a$, we can use the standard geometric formula for its area:

$$ A = \frac{\sqrt{3}}{4} a^2 $$

### Bringing It All Together: The Master Equation

Now, let's substitute our expressions for $M$ and $A$ back into the torque equation. We know that $\sin 90^\circ = 1$, so the equation simplifies beautifully:

$$ 	au = i \left( \frac{\sqrt{3}}{4} a^2 ight) B $$

This is our raw setup. It shows exactly how the physical dimensions of the coil and the external field interact to produce the twisting force.

### The Final Execution: Crunching the Numbers

Physics requires strict adherence to units. Before we plug in the numbers, we must ensure everything is in standard SI units:
- Current, $i = 0.2 	ext{ A}$
- Side length, $a = 10 	ext{ cm} = 0.1 	ext{ m}$
- Magnetic field, $B = 20 	ext{ mT} = 20 	imes 10^{-3} 	ext{ T}$

Let's substitute these values into our master equation:

$$ 	au = 0.2 	imes \frac{\sqrt{3}}{4} 	imes (0.1)^2 	imes 20 	imes 10^{-3} $$

Now, we perform the atomic computations step-by-step:

$$ 	au = 0.2 	imes \frac{\sqrt{3}}{4} 	imes 10^{-2} 	imes 20 	imes 10^{-3} $$

$$ 	au = \frac{0.2 	imes 20}{4} 	imes \sqrt{3} 	imes 10^{-5} $$

$$ 	au = \frac{4}{4} 	imes \sqrt{3} 	imes 10^{-5} $$

$$ 	au = \sqrt{3} 	imes 10^{-5} 	ext{ Nm} $$

### The Grand Reveal

The problem states that the torque is equal to $\sqrt{x} 	imes 10^{-5} 	ext{ Nm}$. By comparing our calculated result with this expression, the answer becomes crystal clear:

$$ \sqrt{3} 	imes 10^{-5} = \sqrt{x} 	imes 10^{-5} $$

Therefore, the value of $x$ is exactly **3**.

I know this problem might have looked intimidating at first glance with its geometry and units, but by breaking it down into logical, atomic steps, we conquered it effortlessly!

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Torque on a Current Loop$

$Determining the Angle θ$

$Magnetic Dipole Moment$

$Area of Equilateral Triangle$

$Master Equation Setup$

$Substituting Values$

$Calculation$

$Final Answer$

The Sigma Insight: Magnetic Moment of Current Loop

The Magic of Magnetic Torque

Visualizing the Geometry

The Anatomy of the Coil

Bringing It All Together

The Final Execution

The Grand Reveal

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Torque on a Current Loop

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Determining the Angle θ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Dipole Moment

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Area of Equilateral Triangle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Master Equation Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer