A conducting wire of parabolic shape, initially y = x^2, is moving with velocity V = V_0 i in a non-

Q: Why do we use $d\vec{l} = dx\hat{i} + dy\hat{j}$ instead of just $dl$?

Because the magnetic force acts in a specific direction (here, the negative y-direction). Using the vector form $d\vec{l}$ allows the dot product to automatically extract the component of the wire's length that actually cuts the magnetic field lines.

Q: Why does the x-component of the wire's length not contribute to the EMF?

The wire is moving in the x-direction. The magnetic force on the charges ($\vec{V} \times \vec{B}$) points in the y-direction. Since the force is perpendicular to the x-axis, no work is done moving charges along the horizontal segments of the wire.

Q: What is the physical meaning of the projected length in this context?

The projected length on the y-axis represents the 'effective' length of the wire that is actively sweeping through the magnetic field. Any horizontal curves or bends don't contribute to the EMF.

Q: Why does replacing the parabolic wire with a straight wire not change the EMF?

Because the EMF only depends on the vertical limits of the wire (from $y=0$ to $y=L$). As long as the starting and ending y-coordinates are the same, the path taken between them doesn't matter for this specific magnetic field.

Q: How do we handle the non-uniform magnetic field during integration?

Since the magnetic field only varies with $y$, and our differential element is $dy$, we can simply substitute the expression for $B$ as a function of $y$ directly into our integral and evaluate it using standard polynomial rules.

Question

A conducting wire of parabolic shape, initially $y = x^2$, is moving with velocity $\vec{V} = V_0\hat{i}$ in a non-uniform magnetic field $\vec{B} = B_0\left(1 + \left(\frac{y}{L}\right)^\beta\right)\hat{k}$, as shown in figure. If $V_0, B_0, L$ and $\beta$ are positive constants and $\Delta\phi$ is the potential difference developed between the ends of the wire, then the correct statement(s) is/are:

Options:

(A) $|\Delta\phi|$ remains the same if the parabolic wire is replaced by a straight wire, $y = x$ initially, of length $\sqrt{2}L$

(B) $|\Delta\phi|$ is proportional to the length of the wire projected on the y-axis.

(C) $|\Delta\phi| = \frac{1}{2}B_0V_0L$ for $\beta = 0$

(D) $|\Delta\phi| = \frac{4}{3}B_0V_0L$ for $\beta = 2$

Sigma Percentile · Accepted Answer

The problem of a moving wire in a non-uniform magnetic field is a classic test of your understanding of motional EMF. It beautifully combines vector calculus with electromagnetic induction. Let's embark on this thrilling journey to decode the potential difference developed across the ends of this parabolic wire!

### Analyzing the Setup
Imagine you are observing a parabolic wire, defined by the equation $y = x^2$, gliding smoothly towards the right with a constant velocity $\vec{V} = V_0\hat{i}$.

The space it moves through is permeated by a magnetic field pointing directly at you, out of the screen. But this isn't just any magnetic field; it's non-uniform, varying with the y-coordinate as $\vec{B} = B_0\left(1 + \left(\frac{y}{L}ight)^\betaight)\hat{k}$.

### The Master Equation for Motional EMF
To find the potential difference, we must rely on the fundamental law of motional EMF for a small element of the wire. The EMF $d\phi$ across a tiny length element $d\vec{l}$ is given by the dot product of the magnetic force per unit charge and the length element:
$$d\phi = (\vec{V} 	imes \vec{B}) \cdot d\vec{l}$$

First, let's evaluate the cross product of the velocity and the magnetic field. Since the velocity is purely in the x-direction and the magnetic field is in the z-direction, their cross product will point in the negative y-direction:
$$\vec{V} 	imes \vec{B} = (V_0\hat{i}) 	imes (B\hat{k}) = -V_0 B \hat{j}$$

### The Magic of the Dot Product
Here is where the magic happens. The length element of the wire can be written in its general vector form as $d\vec{l} = dx\hat{i} + dy\hat{j}$.

When we take the dot product of our cross product result with this length element, the x-component completely vanishes! 
$$d\phi = (-V_0 B \hat{j}) \cdot (dx\hat{i} + dy\hat{j}) = -V_0 B dy$$

This beautifully implies that the motional EMF depends **only** on the vertical projection $dy$ of the wire, and is completely independent of its horizontal shape!

### Performing the Integration
Now, we integrate this expression to find the total magnitude of the potential difference $|\Delta\phi|$. Our limits of integration will naturally be from $y = 0$ to $y = L$.
$$|\Delta\phi| = \int_0^L V_0 B_0 \left(1 + \left(\frac{y}{L}ight)^\betaight) dy$$

I know this integral might look slightly intimidating with the $\beta$ exponent, but let's take a breath. It's just a simple polynomial integration. Integrating term by term, we get:
$$|\Delta\phi| = V_0 B_0 \left[ y + \frac{y^{\beta+1}}{(\beta+1)L^\beta} ight]_0^L$$

Substituting the upper limit $L$, we arrive at our master equation:
$$|\Delta\phi| = V_0 B_0 \left( L + \frac{L^{\beta+1}}{(\beta+1)L^\beta} ight) = B_0 V_0 L \left( 1 + \frac{1}{\beta+1} ight)$$

### Decoding the Options
With our master equation in hand, the options fall like dominoes.

Look closely at the equation: the EMF is directly proportional to $L$. What is $L$? It is exactly the projected length of the wire on the y-axis. Thus, **Option (B) is absolutely correct**.

Let's test the specific cases for $\beta$. If we substitute $\beta = 0$, the equation yields $|\Delta\phi| = 2 B_0 V_0 L$, which means Option (C) is incorrect. However, substituting $\beta = 2$ gives us $|\Delta\phi| = \frac{4}{3} B_0 V_0 L$, perfectly matching **Option (D)**.

Finally, what if we replace the parabola with a straight wire $y = x$ of length $\sqrt{2}L$? The endpoints of this straight wire would still be $(0,0)$ and $(L,L)$. Since we proved that the EMF only depends on the y-limits, the integral remains completely unchanged! Therefore, **Option (A) is also correct**.

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Visualized Solution

$A na l y z in g t h e S e t u p$

$M o t i o na l E M F F or m u l a$

$C r oss P r o d u c t$

$D o tP r o d u c tw i t h d l$

$S e tt in g u pt h e I n t e g r a l$

$P er f or min g t h e I n t e g r a t i o n$

$M a s t er E q u a t i o n$

$C h ec k in g O pt i o n B$

$C h ec k in g O pt i o n s C an d D$

$C h ec k in g O pt i o n A$

The Sigma Insight: Motional EMF

Analyzing the Setup

The Master Equation for Motional EMF To find the potential difference, we must rely on the fundamental law of motional EMF for a small element of the wire

The Magic of the Dot Product Here is where the magic happens

Performing the Integration Now, we integrate this expression to find the total magnitude of the potential difference $∣Δ ϕ ∣$

Decoding the Options

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Visualized Solution

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

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