ODBAC is a fixed rectangular conducting frame of negligible resistance in which the segment CO is left open (n

NCERT Exemplar · Q20

Q. ODBAC is a fixed rectangular conducting frame of negligible resistance in which the segment CO is left open (not connected). O is the mid-point of the frame's base. Taking O as the origin with the base along the x-axis, the frame occupies the rectangle bounded by the base from C(-l, 0) through O(0, 0) to D(+l, 0), the right side D(+l, 0) to B(+l, l), the top side B(+l, l) to A(-l, l), and the left side A(-l, l) to C(-l, 0) — a rectangle of width 2l and height l with O at the middle of its base. A straight conductor OP is pivoted at O and rotates in the plane of the frame with constant angular speed ω, its outer end P always maintaining sliding electrical contact with the frame ABDC. A uniform magnetic field B is directed perpendicular to the plane of the frame. The rotating conductor OP has resistance λ per unit length. Find the current in the rotating conductor as OP turns through 180° from the base direction OD to the base direction OC.

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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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A rod rotating about one end in a field B generates emf $\tfrac12 B\omega r^2$ , where r is its instantaneous length in contact. Here r changes with angle because the tip slides along the rectangle's sides. Dividing the emf by the rod's resistance $\lambda r$ gives $i=\dfrac{B\omega r}{2\lambda}$ , a piecewise function of the swept angle.

Concept

For a conductor of length r rotating with angular speed ω about one end in a uniform field B (perpendicular to the plane), the motional emf between its ends is

\varepsilon=\int_0^r B\,\omega\,x\,dx=\tfrac12 B\omega r^2.

The rotating rod's own resistance is $\lambda r$ (resistance per unit length × length), and the fixed frame ABDC has negligible resistance, so the current is

i=\frac{\varepsilon}{\lambda r}=\frac{\tfrac12 B\omega r^2}{\lambda r}=\frac{B\omega r}{2\lambda}.

How r depends on the angle

Let $\theta=\omega t$ be the angle of OP measured from OD (the +x direction), sweeping through the upper half of the frame. The half-width and the height are both $l$ . The tip P meets:

the right side ( $x=l$ ) while $0\le\theta\le 45^\circ$ : $r\cos\theta=l\Rightarrow r=\dfrac{l}{\cos\theta}$ ;
the top side ( $y=l$ ) while $45^\circ\le\theta\le135^\circ$ : $r\sin\theta=l\Rightarrow r=\dfrac{l}{\sin\theta}$ ;
the left side ( $x=-l$ ) while $135^\circ\le\theta\le180^\circ$ : $r|\cos\theta|=l\Rightarrow r=\dfrac{l}{|\cos\theta|}$ .

Current in each stage ( $\theta=\omega t$ )

$i=\frac{B\omega l}{2\lambda\cos\omega t}\quad\left(0\le\omega t\le\tfrac{\pi}{4}\right),$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. A bicycle wheel with 10 spokes is rotating at a rate of 2 Cycle Per Second perpendicular to the horizontal component of the earth's magnetic field. This produces an induced emf 'E' between the axle and rim of the wheel. If the number of spokes is doubled, then the value of induced emf will be$
(a) 4E
(b) 2E
(c) E
(d) E/2
›Reveal solutionSolution

Each spoke is an independent conducting rod rotating about the same axle in the same field, so each develops the SAME emf; connecting more of them in parallel between axle and rim does not add up their emfs, so E stays unchanged.

For a single conducting rod of length R rotating with angular speed omega in a field B (perpendicular to the plane of rotation), the motional emf between the centre and the rim is E = (1/2) B omega R^2. Every spoke, being identical in length and rotating at the same rate in the same field, develops this same emf E between the axle and the rim. All spokes are connected between the same two po …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. A conducting rod of length l, rotates about one of its ends in a uniform magnetic field B, with a constant angular velocity ω. If the plane of rotation is perpendicular to B, the e.m.f. induced between the ends of rod is ______.$
(a) (1/2)Bωl²
(b) Bωl²
(c) 2Bωl²
(d) Bωl
›Reveal solutionSolution

A rod rotating about one end sweeps out a circle; summing the motional emf $Bv\,dr$ over its length gives $\varepsilon = \tfrac12 B\omega l^2$ .

Consider a small element of the rod at distance $r$ from the pivoted end, of length $dr$ . Its linear speed is $v = \omega r$ (perpendicular to the rod, in the plane of rotation, hence also perpendicular to $B$ ). The motional emf induced across this element is:

$d\varepsilon = Bv\,dr = B\omega r\,dr$

…

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CBSE 2026Set ANNUAL1 mark
$Q. Match the Column A with Column B and write the correct pair. Column A: Motional emf. Column B:$
(i) μ₀nI,
(ii) Volt × second,
(iii) μ₀nI/2,
(iv) Volt × meter,
(v) Volt × meter⁻¹,
(vi) Volt.
›Reveal solutionSolution

Motional emf ε = Bvl is a voltage; its unit is volt, option (vi).

When a conductor of length l moves with velocity v perpendicular to a magnetic field B, an emf is induced across it: ε = Bvl. This is an electromotive force, so its …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. A conductor of length 'l' is moving with velocity 'v' parallel to a magnetic field of intensity 'B'. The induced e.m.f. in the conductor will be$
(a) lvB
(b) (1/2) lvB
(c) zero
(d) (1/2) l^2 vB
›Reveal solutionSolution

Motional EMF in a moving conductor comes from the magnetic force on its free charges, which depends on v x B; if v is parallel to B this cross product vanishes.

The motional EMF induced in a straight conductor of length l moving with velocity v in field B is:

emf = (v x B) . l = B v l sin(phi)

where phi is the angle between v and B. Here the conductor moves parallel to the magnetic field, so phi = 0 degrees, and sin(0) = 0.

emf = B v l sin(0) = 0

…

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CBSE 2023Set 55/1/11 markMCQ
$r$
(a) magnetic field $(B)$
(b) velocity $(v)$
(c) resistance $(r)$
(d) length of PQ
›Reveal solutionSolution
The induced emf in a moving conductor in a uniform magnetic field is purely a motional emf given by $\mathcal{E} = B l v$ , which depends only on the magnetic field $B$ , the length $l$ of the moving arm, and its velocity $v$ . The resistance $r$ of the arm does not appear in this expression — it only determines the current that flows. Hence the correct answer is (c).

The key idea here is the distinction between induced emf and induced current. Many students mix them up, especially when a problem mentions resistance. Let's clear that up first.

When a conductor moves in a magnetic field, the free electrons inside it experience a magnetic Lorentz force $q(\vec{v} \times \vec{B})$ . This force pushes charges along the conductor, creating a potential difference — that potential difference is the motional emf. It is a direct consequence of the motion and the field, nothing else.

The resistance of the conductor only comes into the picture when you ask: "How much current flows as a result of this emf?" That's Ohm's law: $I = \mathcal{E}/R$ . But the emf itself is independent of the resistance.

Now let's walk through the problem step by step.

Identify the source of emf. The arm PQ is the only part of the loop that is moving. The rest of the loop (PSRQ) is stationary and has negligible resistance. So the entire induced emf in the loop is generated across PQ.

Write the expression for motional emf. For a straight conductor of length $l$ moving with velocity $v$ perpendicular to a uniform magnetic field $B$ , the motional emf is:

$\mathcal{E} = B l v$
This is derived from the work done per unit charge by the magnetic force: $F_m = q v B$ , so the electric field set up inside the conductor is $E = vB$ , and over length $l$ , the potential difference is $E l = B l v$ .

Check each option against this formula.

(a) magnetic field $B$ — appears in $\mathcal{E} = B l v$ . So emf does depend on it.

(b) velocity $v$ — appears directly. So emf does depend on it.

(d) length of PQ — that's $l$ in the formula. So emf does depend on it.

(c) resistance $r$ — does not appear in $\mathcal{E} = B l v$ . So emf does not depend on it. …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. Direction of current induced in a wire moving in a magnetic field is found using$
(1) Fleming's left hand rule
(2) Fleming's right hand rule
(3) Ampere's rule
(4) none of these
›Reveal solutionSolution

Fleming's right-hand rule gives the direction of INDUCED current (a motional-EMF/generator situation); the left-hand rule instead gives the direction of FORCE on a current-carrying conductor (a motor situation).

…

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CBSE 2022Set ANNUAL1 mark
$Q. When a metal rod of length l is placed normal to a uniform magnetic field B and moved with a velocity v perpendicular to the field, the induced emf (called motional emf) across its end is ............ .$
›Reveal solutionSolution

The induced (motional) emf across the ends of the rod is $\varepsilon = Bvl$ .

When a conducting rod of length $l$ moves with velocity $v$ perpendicular to a uniform magnetic field $B$ (with $B$ , $v$ and the rod's length mutually perpendicular), each free charge experiences a magnetic force $qvB$ . This separates charge until an electric field balances it, producin …

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CBSE 2020Set 55/2/11 mark
$l$
›Reveal solutionSolution
When a rod moves parallel to a magnetic field, the velocity and field are aligned, so the magnetic flux through any loop remains constant and no emf is induced. The answer is zero.

Why motional emf depends on perpendicular motion

Motional emf arises when a conductor cuts through magnetic field lines. The physical picture: as the rod moves, the magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ pushes charge carriers along the rod, creating a potential difference. This force—and hence the emf—depends critically on the component of velocity perpendicular to the magnetic field.

The motional emf in a straight rod is given by
$\mathcal{E} = \int (\vec{v} \times \vec{B}) \cdot d\vec{l}$
For a uniform field and velocity, this simplifies to
$\mathcal{E} = (\vec{v} \times \vec{B}) \cdot \vec{l}$
where $\vec{l}$ is the length vector along the rod. The cross product $\vec{v} \times \vec{B}$ measures how much the velocity is perpendicular to the field. When $\vec{v}$ and $\vec{B}$ are parallel (or antiparallel), this cross product vanishes.

Step-by-step analysis

Identify the geometry

The rod of length $l$ is parallel to $\vec{B}$ , and it moves with velocity $\vec{v}$ along the direction of $\vec{B}$ . So $\vec{v} \parallel \vec{B}$ .

Compute the cross product

Since $\vec{v}$ and $\vec{B}$ point in the same (or exactly opposite) direction, the angle $\theta$ between them is either $0°$ or $180°$ . The magnitude of the cross product is

$|\vec{v} \times \vec{B}| = vB \sin\theta = vB \cdot 0 = 0$

Evaluate the motional emf Substituting into the emf formula, …
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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

Concept

How r depends on the angle

Current in each stage ( $\theta=\omega t$ )

Method: A Rotating Conductor Whose Contact Length Changes as It Sweeps a Boundary

Steps

Why motional emf depends on perpendicular motion

Step-by-step analysis

Concept

How r depends on the angle

Current in each stage (θ=ωt\theta=\omega tθ=ωt)

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Current in each stage ( $\theta=\omega t$ )

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