(a) State Lenz's law. A metallic rod of length L is rotated about an axis passing through its end M perpendicular to its length, with a constant angular velocity $\omega$ in a uniform magnetic field $B$ parallel to the axis. Obtain an expression for the emf induced between its ends. OR (b) Define self-inductance of a coil. Derive an expression for the self-inductance of a long solenoid of cross-sectional area A and length $l$, having $n$ turns per unit length.

(a) State Lenz's law. A metallic rod of length L is rotated about an axis passing through its end M perpendicu

Question

\omega

(OR)

(b) Define self-inductance of a coil. Derive an expression for the self-inductance of a long solenoid of cross-sectional area A and length

l

, having

n

turns per unit length.

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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

Part (a): Lenz's law says the induced emf opposes the flux change; a rod of length $L$ rotating at $\omega$ in field $B$ (parallel to the axis) develops $\mathcal{E}=\tfrac12 B\omega L^2$ . Part (b): self-inductance is flux linkage per unit current, and for a long solenoid $L=\mu_0 n^2 lA$ .

Lenz's law and the rotating rod

Lenz's law. The direction of any induced emf/current is such that it opposes the change in flux causing it — the negative sign in Faraday's law $\mathcal{E}=-\,d\Phi_B/dt$ . It is a statement of energy conservation.

Rod rotating about one end. The rod (length $L$ ) spins with angular velocity $\omega$ about an axis through $M$ , in a uniform field $B$ parallel to that axis.

A point at distance $x$ from the axis has speed $v=\omega x$ , perpendicular to both the rod and $B$ .
The motional emf of a small element $dx$ is

d\mathcal{E}=Bv\,dx=B(\omega x)\,dx.

Integrate along the rod:

\mathcal{E}=\int_0^L B\omega x\,dx=B\omega\left[\frac{x^2}{2}\right]_0^L=\frac12 B\omega L^2.

Tip

The $L^2$ dependence follows because the speed grows linearly to $\omega L$ at the tip; the average speed $\tfrac{\omega L}{2}$ times $BL$ reproduces $\tfrac12 B\omega L^2$ .

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Concept understanding — Self-Inductance of a Solenoid

Self-Inductance of a Solenoid: From Intuition to Formula

Imagine you push a heavy door. It doesn't resist your push once it's moving — but it does resist you trying to change its speed suddenly. That resistance to change is inertia. A solenoid carrying current behaves the same way: it "wants" to keep its current steady, and fights any attempt to change it.

This property is called self-inductance. The solenoid generates a back emf that opposes the change in its own current — not the current itself, but the change in current. That's the core idea.

Why does a solenoid oppose current changes?

A solenoid is a long coil of wire. When current flows through it, it produces a magnetic field inside. If you try to increase the current, the magnetic field strengthens. But a changing magnetic field induces an emf in the coil itself (Faraday's law). By Lenz's law, this induced emf opposes the change that caused it — so it pushes back against the rising current.

If you try to decrease the current, the field weakens, and the induced emf tries to keep the current flowing. The solenoid acts like an electrical "flywheel."

The precise statement

Self-inductance $L$ is defined by the relation:

\mathcal{E} = -L \frac{dI}{dt}

where $\mathcal{E}$ is the induced back emf, and $\frac{dI}{dt}$ is the rate of change of current. The negative sign tells you the emf opposes the change.

For a solenoid, $L$ depends only on its geometry and the core material — not on the current. The formula is:

L = \mu_0 n^2 A l

L = \mu_0 n^2 A l

Let's unpack each symbol:

$\mu_0$ — permeability of free space ( $4\pi \times 10^{-7}$ H/m). It's a universal constant that tells you how strongly a vacuum responds to magnetic fields.
$n$ — number of turns per unit length (turns/m). More turns per metre means a stronger field per ampere, so more inductance.
$A$ — cross-sectional area of the solenoid (m²). A wider coil encloses more magnetic flux.
$l$ — length of the solenoid (m). Longer solenoid means more total turns, hence more inductance.

Where does $L = \mu_0 n^2 A l$ come from?

Start with the magnetic field inside a long solenoid:

B = \mu_0 n I

The magnetic flux through one turn is $BA = \mu_0 n I A$ . For all $N = nl$ turns, the total flux linkage is:

\Phi_{\text{total}} = N \cdot BA = (nl)(\mu_0 n I A) = \mu_0 n^2 A l I

By definition, self-inductance is the constant of proportionality between flux linkage and current:

\Phi_{\text{total}} = L I

Comparing, you get:

L = \mu_0 n^2 A l

Note

This formula assumes an ideal solenoid — infinitely long, with a uniform field inside and zero field outside. Real solenoids are close approximations if $l \gg \sqrt{A}$ .

What does a larger $L$ mean?

A solenoid with high $L$ strongly resists changes in current. If you try to switch the current on quickly, the back emf is large, so the current rises slowly. If you short-circuit the solenoid, the current doesn't drop instantly — it decays gradually.

This is why inductors are used in filters, chokes, and timing circuits. They smooth out current variations. …

Part (b) — Self-inductance of a long solenoid …