A region in the form of an equilateral triangle (in x-y plane) of height L has a uniform magnetic fi

Q: Why is the overlap area a rhombus initially?

Because you are intersecting an upright triangle with an inverted triangle. The boundaries of the overlap are formed by the two slanted sides of the field triangle at the top, and the two slanted sides of the loop triangle at the bottom, creating a four-sided diamond or rhombus shape.

Q: Why does the graph cross zero at $4L/3$ and not $3L/2$?

The rate of change of area depends on the difference between the width of the loop entering the field and the width leaving it. Because of the specific geometry of the triangles, this rate of change hits zero exactly at $x = 4L/3$. It's a purely geometric result of the area integral.

Q: Why is the positive peak at $2L$ twice as large as the negative peak at $L$?

At $x=L$, the loop is half-way through, and the rate of change of area is governed by the slanted sides. At $x=2L$, the loop is just about to leave the field completely. The flat base of the loop is exiting the widest part of the magnetic field, causing a very rapid decrease in area, which leads to a larger induced EMF.

Q: Can I solve this using motional EMF $E = Bvl$ instead of Faraday's Law?

Yes, you can! The effective length $l$ would be the difference between the horizontal width of the loop at the bottom boundary of the field and the width at the top boundary. However, using the area integral is often less prone to sign errors and geometric mistakes.

Q: Why does the EMF drop to zero immediately after $x=2L$?

At exactly $x=2L$, the top base of the loop aligns with the bottom base of the magnetic field. For any $x > 2L$, the loop is completely outside the magnetic field region. Since the flux becomes zero and stays zero, the induced EMF must also be zero.

Question

A region in the form of an equilateral triangle (in $x-y$ plane) of height $L$ has a uniform magnetic field $\vec{B}$ pointing in the $+z$-direction. A conducting loop PQR, in the form of an equilateral triangle of the same height $L$, is placed in the $x-y$ plane with its vertex P at $x=0$ in the orientation shown in the figure. At $t=0$, the loop starts entering the region of the magnetic field with a uniform velocity $\vec{v}$ along the $+x$-direction. The plane of the loop and its orientation remain unchanged throughout its motion.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When tackling electromagnetic induction problems involving moving geometric shapes, the first and most crucial step is to perfectly visualize the geometry. We have a uniform magnetic field confined to a region shaped like an equilateral triangle. Let's call this the 'field triangle'. Its vertex is at the origin ($x=0$), and its base is at $x=L$. Because the $+x$-direction is downwards in the figure, this triangle is upright ($\Delta$).

Now, look at the conducting loop PQR. It is also an equilateral triangle of height $L$. However, its vertex P is at $x=0$, and its base QR is at $x=-L$. This means the loop is an inverted triangle ($
abla$). As the loop moves downwards with velocity $v$, it begins to penetrate the field triangle. The induced EMF is governed by Faraday's Law: $E = -\frac{d\Phi}{dt} = -Bv \frac{dA}{dx}$, where $A(x)$ is the overlapping area.

### Phase 1: The Rhombus Overlap ($0 \le x \le L$)

As the inverted loop enters the upright field region, the overlapping area is bounded by the slanted sides of both triangles. If you draw this out, the intersection of a $
abla$ and a $\Delta$ forms a perfect rhombus.

To find the area rigorously, we integrate the width of the overlap. At any depth $y$, the width of the field is $w_f = \frac{2y}{\sqrt{3}}$, and the width of the loop is $w_l = \frac{2(x-y)}{\sqrt{3}}$. The overlap width is simply the minimum of these two. They are equal exactly at the midpoint $y = x/2$.

Integrating this gives the area:
$$A(x) = \int_0^{x/2} \frac{2y}{\sqrt{3}} dy + \int_{x/2}^x \frac{2(x-y)}{\sqrt{3}} dy = \frac{x^2}{2\sqrt{3}}$$

Differentiating this area with respect to $x$ gives $\frac{dA}{dx} = \frac{x}{\sqrt{3}}$. Therefore, the induced EMF is:
$$E = -Bv \frac{x}{\sqrt{3}}$$
This is a straight line with a negative slope, reaching a negative peak of $-\frac{BvL}{\sqrt{3}}$ at $x=L$.

### Phase 2: The Hexagon Overlap ($L \le x \le 2L$)

Once $x$ exceeds $L$, the flat base of the loop enters the magnetic field. The overlapping region is no longer a simple rhombus; it becomes a hexagon bounded by the flat top of the loop, the flat bottom of the field, and four slanted sides.

We adjust our integration limits to span from the loop's base ($y=x-L$) to the field's base ($y=L$). The area function becomes:
$$A(x) = \int_{x-L}^{x/2} \frac{2y}{\sqrt{3}} dy + \int_{x/2}^L \frac{2(x-y)}{\sqrt{3}} dy = \frac{1}{\sqrt{3}} \left( \frac{x^2}{2} - 2(x-L)^2 ight)$$

Differentiating this new quadratic area function yields a linear rate of change:
$$\frac{dA}{dx} = \frac{4L - 3x}{\sqrt{3}}$$
Thus, the EMF for this phase is:
$$E = -Bv \frac{4L - 3x}{\sqrt{3}}$$

### The Ultimate Discriminator

This linear equation is the key to choosing the correct graph. Let's find where the EMF becomes zero:
$$4L - 3x = 0 \implies x = \frac{4L}{3} \approx 1.33L$$
This zero-crossing is strictly to the left of $1.5L$ (or $3L/2$). If we look at the given options, Option B shows the graph crossing exactly at the $3L/2$ tick mark. **Option A**, however, shows the zero-crossing correctly shifted to the left of $3L/2$. Furthermore, at $x=2L$, the EMF reaches a positive peak of $+\frac{2BvL}{\sqrt{3}}$, which is exactly twice the magnitude of the negative peak, a feature perfectly depicted in Option A.

Select Answer:

Visualized Solution

Visualizing the Geometry

Faraday's Law Setup

Phase 1: $0 \leq x \leq L$

EMF for Phase 1

Phase 2: $L \leq x \leq 2 L$

EMF for Phase 2

Analyzing the Graph

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

Analyzing the Setup

Phase 1

Phase 2

The Ultimate Discriminator

Similar Questions

A bar magnet is passing through a conducting loop of radius $R$ with velocity $v$ . The radius of the bar magnet is such that it just passes through the loop. The induced emf in the loop can be represented by the approximate curve

List-I

List-II

At time $t = 0$ magnetic field of 1000 gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500 gauss, in the next 5 s, then induced emf in the loop is

A planar loop of wire rotates in a uniform magnetic field. Initially at $t = 0$ , the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of $10 s$ about an axis in its plane, then the magnitude of induced emf will be maximum and minimum respectively at