A horizontal straight wire 20 m long extending from East to West is falling with a speed of 5.0 m/s,

Q: Why do we only consider the horizontal component of the Earth's magnetic field?

The vertical component of the Earth's magnetic field ($B_V$) points downwards, which is parallel to the velocity of the falling wire. The cross product of parallel vectors is zero ($\vec{v} \times \vec{B}_V = 0$), so the vertical component does not contribute to the motional EMF.

Q: What if the wire was aligned North-South instead of East-West?

If the wire was aligned North-South, its length would be parallel to the horizontal magnetic field ($B_H$). In that case, falling downwards would not cut the $B_H$ lines, and no EMF would be induced by the horizontal component.

Q: How do we know which end of the wire becomes positive?

You can use Fleming's Right-Hand Rule or the Lorentz force equation $\vec{F} = q(\vec{v} \times \vec{B})$. With velocity down and magnetic field North, the force on positive charges is towards the East. Thus, the East end becomes positive.

Q: Does the mass or material of the wire affect the induced EMF?

No, the induced motional EMF depends purely on the kinematics (velocity and length) and the external magnetic field. The mass or resistivity of the wire does not change the generated EMF, though it would affect the current if a closed circuit were formed.

Q: Why is it called 'motional' EMF?

It is called 'motional' because the electromotive force is generated purely due to the physical motion of the conductor through a stationary magnetic field. This is in contrast to 'transformer' EMF, which is generated by a time-varying magnetic field passing through a stationary loop.

Question

A horizontal straight wire 20 m long extending from East to West is falling with a speed of 5.0 m/s, at right angles to the horizontal component of the earth's magnetic field $0.30 	imes 10^{-4} 	ext{ Wb/m}^2$. The instantaneous value of the emf induced in the wire will be

Options:

(A) 6.0 mV

(B) 3 mV

(C) 4.5 mV

(D) 1.5 mV

Sigma Percentile · Accepted Answer

Have you ever wondered what happens when a simple piece of wire falls through the air? It seems like a purely mechanical event—gravity pulls it down, and air resistance pushes back. But because we live on Earth, there is an invisible, omnipresent force field surrounding us: the Earth's magnetic field.

When a conductor moves through this magnetic field, the laws of electromagnetism awaken. This seemingly mundane act of falling transforms the wire into a tiny, transient battery. This phenomenon is known as **Motional Electromotive Force (EMF)**.

### The 3D Geometry of the Fall

To truly understand this problem, we must first build a clear mental model of the three-dimensional space. Imagine you are standing outside, facing North.

The wire is described as extending from East to West. Let's align this with our X-axis. So, the length vector $\vec{l}$ points along the East-West line.

Next, the wire is falling. Gravity pulls it straight down towards the center of the Earth. This downward velocity vector $\vec{v}$ aligns with our negative Z-axis.

Finally, we have the Earth's magnetic field. The problem specifically mentions the **horizontal component** of the Earth's magnetic field, denoted as $\vec{B}_H$. Magnetic field lines on Earth generally point from the geographic South Pole towards the geographic North Pole. Therefore, $\vec{B}_H$ points straight ahead of you, along the Y-axis.

Take a moment to visualize these three vectors:
1. Length $\vec{l}$ is East-West.
2. Velocity $\vec{v}$ is Down.
3. Magnetic Field $\vec{B}_H$ is North.

Notice the beautiful symmetry here? All three vectors are mutually perpendicular to each other! This orthogonal relationship is the absolute sweet spot for generating maximum motional EMF.

### The Master Equation

Why does an EMF get generated in the first place? It all comes down to the fundamental constituents of the wire: electrons.

As the wire falls, every single free electron inside it is also falling with velocity $\vec{v}$. These electrons are moving charges traversing a magnetic field $\vec{B}_H$. According to the laws of physics, they experience a magnetic force known as the **Lorentz Force**, given by the equation:

$$ \vec{F}_m = q(\vec{v} 	imes \vec{B}) $$

Because $\vec{v}$ (down) and $\vec{B}_H$ (North) are perpendicular, the cross product yields a force that pushes the electrons along the length of the wire (East-West). As electrons accumulate at one end, they leave a net positive charge at the other. This separation of charge creates an electric field $\vec{E}$ inside the wire, which in turn exerts an electric force $\vec{F}_e = q\vec{E}$ that opposes the magnetic force.

Equilibrium is reached almost instantly when the electric force perfectly balances the magnetic force ($qE = qvB$). The potential difference created by this electric field across the length $l$ of the wire is our induced EMF:

$$ e = E \cdot l = (vB) \cdot l = B l v $$

This elegant equation, $e = B l v$, is our master key. It holds true precisely because $\vec{l}$, $\vec{v}$, and $\vec{B}$ are all mutually perpendicular.

### Executing the Calculation

Now that we have our theoretical foundation, let's bring in the numbers. The problem provides us with the following values:

The horizontal magnetic field is $B_H = 0.30 	imes 10^{-4} 	ext{ Wb/m}^2$.
The length of the wire is $l = 20 	ext{ m}$.
The falling speed is $v = 5.0 	ext{ m/s}$.

We substitute these directly into our master equation:

$$ e = B_H l v $$

$$ e = (0.30 	imes 10^{-4}) 	imes 20 	imes 5.0 $$

When faced with a string of numbers, it is always best to look for easy pairings. Notice the $20$ and the $5.0$. Multiplying them together gives a clean, round number:

$$ 20 	imes 5.0 = 100 $$

Now, substitute this back into the expression:

$$ e = 0.30 	imes 10^{-4} 	imes 100 $$

Multiplying by $100$ is equivalent to shifting the decimal point two places to the right. So, $0.30$ becomes $30$:

$$ e = 30 	imes 10^{-4} 	ext{ V} $$

### The Final Polish

We have our answer, but it is not quite in the format of the given options. The options are presented in millivolts ($	ext{mV}$). We need to convert our result.

Recall that $1 	ext{ mV} = 10^{-3} 	ext{ V}$. Let's manipulate our scientific notation to reveal this factor:

$$ 30 	imes 10^{-4} 	ext{ V} = 3.0 	imes 10 	imes 10^{-4} 	ext{ V} $$

$$ e = 3.0 	imes 10^{-3} 	ext{ V} $$

And there it is! The $10^{-3} 	ext{ V}$ perfectly translates to millivolts:

$$ e = 3 	ext{ mV} $$

The instantaneous value of the induced EMF is exactly $3 	ext{ mV}$.

### The Way Forward: Finding the Polarity

We have successfully solved the problem, but true mastery comes from asking the next logical question: **Which end of the wire is positive?**

We can determine this using **Fleming's Right-Hand Rule**, which is designed specifically for generator effects like motional EMF.

1. Point your **First finger** in the direction of the Magnetic Field (North).
2. Point your **Thumb** in the direction of the Motion (Down).
3. Your **Middle finger** will now point in the direction of the induced conventional current.

If you contort your hand into this position, your middle finger will point towards the **East**.

Conventional current flows from the negative terminal to the positive terminal *inside* the source of EMF (just like inside a battery). Therefore, the positive charges are being pushed towards the East end, making the **East end at a higher potential** than the West end.

This simple falling wire is not just a piece of metal; it is a dynamic interplay of gravity, magnetism, and electricity!

Select Answer:

Visualized Solution

$Visualizing the 3D Setup$

$The Magnetic Field$

$Motional EMF Formula$

$Substituting Values$

$Simplifying the Math$

$Calculating the Magnitude$

$Final Answer$

$Food for Thought$

The Sigma Insight: Motional EMF

The 3D Geometry of the Fall

The Master Equation

Executing the Calculation

The Final Polish

The Way Forward

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