A thin semi-circular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magn

I. Multiple Choice Questions · Q2

v

(a) Zero

(b)

\dfrac{\pi B r v}{2}

and P is at higher potential

(c)

\pi B r v

and R is at higher potential

(d)

2Brv

and R is at higher potential

A thin semi-circular conducting ring PQR of radius r, its flat diameter horizontal at the top with ends P (left) and R (right) and the — Class 12 Physics question — Figure

Puducherry TnboardTextbookSubjectiveImportance★★★★★

5% · 6/132 Questions

✓ Free question

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

Once current $I$ flows, the field exerts a retarding force $F = BIl$ on the rod, opposing its motion (Lenz's law). To keep the rod moving at constant speed, an external agent must supply power

P = Fv = BIl\,v = \varepsilon I

exactly equal to the electrical power dissipated in the circuit — energy is conserved.

Note

Motional emf arises only from the component of velocity perpendicular to $\vec{B}$ . Motion parallel to the field produces no emf.

Motional emf, derived from the Lorentz force on charges in a moving conductor, is a key numerical topic within the NCERT Class 12 Physics chapter on electromagnetic induction, tested in CBSE boards and JEE Main. Searches for "motional emf formula and derivation class 12 physics" will find this rod-on-rails explanation, consistent with Faraday's flux rule, matches the NCERT-prescribed derivation.

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
$v$	Faster motion → larger magnetic force → larger EMF

Alternative Derivation: Faraday's Law

The same result comes from Faraday's law of induction:

\mathcal{E} = -\frac{d\Phi_B}{dt}

For a rod of length $L$ moving with speed $v$ through a field $B$ , the area swept per second is $Lv$ , so:

\frac{d\Phi_B}{dt} = B \cdot \frac{dA}{dt} = B L v

Thus:

\mathcal{E} = B L v

Both approaches give the same answer — confirming consistency.

Important Exam Points

Direction: Use Fleming's right-hand rule (generator rule) to find polarity.
General formula (when $\vec{v}$ and $\vec{B}$ are not perpendicular):

\mathcal{E} = B L v \sin\theta

where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$ .

EMF is induced only while the conductor moves — stop the motion, stop the EMF.

Bottom line: Motional EMF is simply the magnetic force acting on moving charges inside a conductor, creating a charge separation that acts like a battery. The formula $\mathcal{E} = BLv$ is a direct consequence of balancing magnetic and electric forces.

Step 1. For a conductor moving with velocity v in a field B, the motional emf between any two of its points depends only on the STRAIGHT-LINE distance between those points along the direction perpendicular to both v and B -- not on the actual (possibly curved) path length of the conductor between them.

Step 2. For the semicircular ring PQR of radius r, falling with speed v, the two ends P and R are separated by the straight-line (diameter) distance $PR=2r$ , regardless of the semicircular arc's own length.

Step 3. The potential difference across the effective 'straight rod of length 2r' is therefore $\varepsilon = B(2r)v = 2Brv$ .

Step 4. The direction (which end is at higher potential) follows from the same $\vec v\times\vec B$ reasoning used for a straight rod, giving R at the higher potential for this geometry.

Step 5. Eliminating the others: (a) zero would be true only if there were no relative motion or no field component perpendicular to v; (b) and (c) both incorrectly involve $\pi r$ (as if the arc LENGTH, not the chord, mattered).

✓Final answer

2Brv

←Previous question · evaluationQ1 — An electron moves on a straight line path XY as shown in the figure. T…→Next question · evaluationQ3 — The flux linked with a coil at any instant t is given by PhiB = (10t² …

Q.A thin semi-circular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in the figure. The potential difference developed across the ring when its speed is vvv, is (NEET 2014)

🎓Unlock everything free for 14 days

Q.A thin semi-circular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in the figure. The potential difference developed across the ring when its speed is $v$ , is (NEET 2014)

Unlock everything free for 14 days