A fan of metal blades of length 0.4 m rotates normal to a magnetic field of $4 \times 10^{-3}$ T. If the induced emf between the centre and edge of the blade is 0.02 V, determine the rate of rotation of the blade.

A fan of metal blades of length 0.4 m rotates normal to a magnetic field of 4 × 10⁻³ T. If the induced emf bet

IV. Numerical Problems · Q7

4 \times 10^{-3}

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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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Step 1. For a rod (blade) of length l = 0.4 m rotating about one end, the emf induced between centre and edge is $\varepsilon=\tfrac12B\omega l^2$ .

Step 2. Given $B=4\times10^{-3}$ T, $\varepsilon=0.02$ V: $0.02 = \tfrac12(4\times10^{-3})\omega(0.4)^2$ .

Step 3. $(0.4)^2=0.16$ ; $\tfrac12(4\times10^{-3})(0.16)=3.2\times10^{-4}$ ; so $\omega = 0.02/(3.2\times10^{-4}) \approx 62.5$ rad/s. …

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Q.A fan of metal blades of length 0.4 m rotates normal to a magnetic field of 4×10−34 \times 10^{-3}4×10−3 T. If the induced emf between the centre and edge of the blade is 0.02 V, determine the rate of rotation of the blade.

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Q.A fan of metal blades of length 0.4 m rotates normal to a magnetic field of $4 \times 10^{-3}$ T. If the induced emf between the centre and edge of the blade is 0.02 V, determine the rate of rotation of the blade.

Unlock everything free for 14 days