A rod of mass $m$ and resistance $R$ slides smoothly over two parallel perfectly conducting wires kept sloping at an angle $\theta$ with respect to the horizontal. The circuit is closed through a perfect conductor at the top. There is a constant magnetic field $\mathbf{B}$ along the vertical direction. If the rod is initially at rest, find the velocity of the rod as a function of time.

> [!TLDR] > The rod accelerates under gravity but experiences a magnetic drag force proportional to its velocity. Solving the equation of motion gives $v(t) = \frac{mgR \sin\theta}{B^2 L^2 \cos^2\theta} \left(1 - e^{-t/\tau}\right)$, where $\tau = \frac{mR}{B^2 L^2 \cos^2\theta}$. ## Why this is a motional EMF problem The rod slides down the sloping wires, cutting the vertical magnetic field. As it moves, the area of the loop changes, inducing an EMF. This EMF drives a current through the rod, and the current-carrying rod experiences a magnetic force. That force opposes the motion (Lenz's law), so the rod doesn't fall freely — it reaches a terminal velocity. The key insight: the induced EMF depends on the component of velocity perpendicular to the field, and the magnetic force depends on the component of current perpendicular to the field. Both involve the geometry of the slope. > [!FORMULA] > Motional EMF for a rod of length $L$ moving with velocity $v$ in a field $B$: $\mathcal{E} = B L v_\perp$, where $v_\perp$ is the component of velocity perpendicular to both the rod and the field. ## Step-by-step solution **1. Set up the geometry** The wires are at angle $\theta$ to the horizontal. The rod slides along them, so its velocity $v$ is directed down the slope. The magnetic field $\mathbf{B}$ is vertical (downward, say). The rod's velocity has two components relative to the vertical field: - A component parallel to $\mathbf{B}$: $v \sin\theta$ — this does NOT contribute to motional EMF. - A component perpendicular to $\mathbf{B}$: $v \cos\theta$ — this is what matters. The rod itself is horizontal (perpendicular to the plane of the page, if we draw the slope). Its length $L$ is the separation between the two parallel wires. > [!TIP] > Always resolve velocity into components parallel and perpendicular to the magnetic field. Only the perpendicular component induces EMF. **2. Find the induced EMF** The motional EMF is: $$\mathcal{E} = B L (v \cos\theta)$$

A rod of mass m and resistance R slides smoothly over two parallel perfectly conducting wires kept sloping at

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Concept understanding — Motional Emf

Motional Emf

When a conductor moves through a magnetic field, the free charges inside it experience a magnetic force. This force pushes the charges along the conductor, setting up a potential difference across its ends — a motional emf. It is electromagnetic induction viewed from a moving conductor rather than from a changing field.

Origin: the Lorentz force

Consider a straight rod of length $l$ moving with velocity $\vec{v}$ perpendicular to a uniform field $\vec{B}$ . Each free charge $q$ feels a force

\vec{F} = q\,(\vec{v}\times\vec{B})

of magnitude $qvB$ directed along the rod. Charges pile up at the ends until the electric field they create balances the magnetic force. The resulting potential difference — the motional emf — is

\varepsilon = Bvl

Consistency with Faraday's law

Let the rod of length $l$ slide along parallel conducting rails, sweeping out a distance $x$ . The enclosed area is $A = lx$ , so the flux is $\Phi = Blx$ . Then

\varepsilon = -\frac{d\Phi}{dt} = -Bl\frac{dx}{dt} = -Blv

The magnitude $Bvl$ matches the Lorentz-force result, showing that the flux rule and the force picture agree.

Worked example

A rod of length $0.4\,\text{m}$ moves at $5\,\text{m/s}$ perpendicular to a field of $0.5\,\text{T}$ :

\varepsilon = Bvl = 0.5\times5\times0.4 = 1\,\text{V}

If the circuit resistance is $2\,\Omega$ , the induced current is $I = \varepsilon/R = 0.5\,\text{A}$ .

Force and energy …

Why this formula?

Motional EMF: Why the Formula Holds

Let's build this from first principles — understanding the why before the formula.

The Core Idea

Motional EMF arises when a conductor moves through a magnetic field. The key insight: moving charges in a magnetic field experience a magnetic force, which acts like a battery pushing charges around the conductor.

Step 1: The Force on a Moving Charge

A charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ feels the Lorentz magnetic force:

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

This force is perpendicular to both velocity and magnetic field.

Step 2: What Happens Inside a Moving Conductor

Consider a straight metal rod of length $L$ moving with constant velocity $\vec{v}$ perpendicular to a uniform magnetic field $\vec{B}$ (pointing into the page).

Free electrons in the rod are moving with the rod at velocity $\vec{v}$ .
Each electron experiences a magnetic force:

\vec{F}_m = -e (\vec{v} \times \vec{B})

(negative sign because electron charge is $-e$ )

This force pushes electrons along the rod — say, toward one end.

Step 3: Charge Separation Creates an Electric Field

As electrons accumulate at one end, that end becomes negatively charged, leaving the other end positively charged.

This charge separation creates an internal electric field $\vec{E}$ inside the rod, pointing from positive to negative end.
The electric field exerts an opposing force on the electrons:

\vec{F}_e = -e \vec{E}

Step 4: Equilibrium — The "Battery" is Formed

Charge keeps moving until the electric force balances the magnetic force:

\vec{F}_e + \vec{F}_m = 0

-e \vec{E} - e (\vec{v} \times \vec{B}) = 0

\vec{E} = - (\vec{v} \times \vec{B})

Magnitude-wise (for perpendicular $\vec{v}$ and $\vec{B}$ ):

E = vB

Step 5: From Electric Field to EMF

The motional EMF $\mathcal{E}$ is the work done per unit charge to move a test charge from one end to the other:

\mathcal{E} = \int_{\text{negative}}^{\text{positive}} \vec{E} \cdot d\vec{l}

For a uniform field along the rod of length $L$ :

\mathcal{E} = E \cdot L = vBL

The Key Formula

Motional EMF for a straight conductor moving perpendicular to $\vec{B}$ :

$\boxed{\mathcal{E} = B L v}$

Why This Makes Physical Sense

Quantity	Role
$B$	Stronger magnetic field → larger force on charges
$L$	Longer conductor → more charge separation possible

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The rod accelerates under gravity but experiences a magnetic drag force proportional to its velocity. Solving the equation of motion gives $v(t) = \frac{mgR \sin\theta}{B^2 L^2 \cos^2\theta} \left(1 - e^{-t/\tau}\right)$ , where $\tau = \frac{mR}{B^2 L^2 \cos^2\theta}$ .

Why this is a motional EMF problem

The rod slides down the sloping wires, cutting the vertical magnetic field. As it moves, the area of the loop changes, inducing an EMF. This EMF drives a current through the rod, and the current-carrying rod experiences a magnetic force. That force opposes the motion (Lenz's law), so the rod doesn't fall freely — it reaches a terminal velocity.

The key insight: the induced EMF depends on the component of velocity perpendicular to the field, and the magnetic force depends on the component of current perpendicular to the field. Both involve the geometry of the slope.

L

Step-by-step solution

1. Set up the geometry

The wires are at angle $\theta$ to the horizontal. The rod slides along them, so its velocity $v$ is directed down the slope. The magnetic field $\mathbf{B}$ is vertical (downward, say).

The rod's velocity has two components relative to the vertical field:

A component parallel to $\mathbf{B}$ : $v \sin\theta$ — this does NOT contribute to motional EMF.
A component perpendicular to $\mathbf{B}$ : $v \cos\theta$ — this is what matters.

The rod itself is horizontal (perpendicular to the plane of the page, if we draw the slope). Its length $L$ is the separation between the two parallel wires.

Tip

Always resolve velocity into components parallel and perpendicular to the magnetic field. Only the perpendicular component induces EMF.

2. Find the induced EMF

The motional EMF is:

\mathcal{E} = B L (v \cos\theta)

This EMF drives current through the circuit. The total resistance is $R$ (rod's resistance; wires are perfect conductors).

3. Find the induced current

By Ohm's law:

I = \frac{\mathcal{E}}{R} = \frac{B L v \cos\theta}{R}

The direction of current is such that it opposes the motion (Lenz's law). For a rod sliding down, the current flows in a direction that produces an upward magnetic force along the slope.

4. Find the magnetic force on the rod

The rod carries current $I$ in a magnetic field $B$ . The magnetic force on a current-carrying conductor is:

\mathbf{F}_m = I (\mathbf{L} \times \mathbf{B})

Here $\mathbf{L}$ is along the rod (horizontal), and $\mathbf{B}$ is vertical. The cross product gives a force perpendicular to both — which lies in the plane of the slope. But we need the component along the slope (the direction of motion).

The magnitude of the magnetic force is:

F_m = I L B

But this force is horizontal (perpendicular to the rod and to $\mathbf{B}$ ). To find its component along the slope, we project it. The horizontal force makes an angle $\theta$ with the slope direction (since the slope is at angle $\theta$ to horizontal). So the component along the slope (upward, opposing motion) is:

F_{m,\parallel} = I L B \cos\theta

Substitute $I$ :

$F_{m,\parallel} = \frac{B L v \cos\theta}{R} \cdot L B \cos\theta = \frac{B^2 L^2 v \cos^2\theta}{R}$ …

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Concept understanding — Motional Emf

Motional Emf

Origin: the Lorentz force

Consistency with Faraday's law

Worked example

Force and energy …

Why this formula?

Motional EMF: Why the Formula Holds

The Core Idea

Step 1: The Force on a Moving Charge

Step 2: What Happens Inside a Moving Conductor

Step 3: Charge Separation Creates an Electric Field

Step 4: Equilibrium — The "Battery" is Formed

Step 5: From Electric Field to EMF

The Key Formula

Why This Makes Physical Sense

Why this is a motional EMF problem

Step-by-step solution

Method: Motional EMF with Angled Geometry (Rod on an Incline)

Steps

Why this is a motional EMF problem

Step-by-step solution

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