Motional EMF

Q: Does a rod need a closed circuit to have a motional emf?

No. An isolated rod moving in a field still has Bvl volts between its ends, because the charge separation happens inside the rod itself. A closed path is needed only for a current to flow, and therefore for any force or heating to appear.

Q: Why does a rod on rails slow down exponentially?

The retarding force is B²l²v/(R + r), which is proportional to the speed itself. A force proportional to velocity always gives exponential decay, with time constant τ = (m(R + r))/(B²l²), exactly as for motion through a viscous fluid.

PhysicsElectromagnetic InductionFor JEE aspirants

Motional emf is the voltage that appears across a conductor simply because it is moving through a magnetic field. For a straight rod of length $l$ sliding at speed $v$ perpendicular to a field $B$ , the motional emf is $ε = B v l$ , and the rod behaves exactly like a small battery. Everything in this chapter that involves a sliding rod, a rotating rod or a wire near a current-carrying wire is built on that one result. It is a heavily tested area in JEE Main, JEE Advanced and NEET.

On this page1

ε = B v l

2Where the emf comes from3Angles and bent wires4Rod on rails5How the rod moves6Rotating rod7Near a long wire

Key Formulas - Quick Reference

★ Must learn Straight rod, $v$ , $B$ and the rod all mutually perpendicular: $ε = B v l$
Rod making an angle $θ$ with $v$ : $ε = B v l sin θ$ ; in general $ε = \int (v \times B) \cdot d l$
★ Must learn Bent or irregular wire from $A$ to $B$ : only the straight separation across $v$ counts, $ε = B v l_{⊥}$
★ Must learn Rod on rails: $i = \frac{B v l}{R + r}$ , retarding force $F = B i l = \frac{B ^{2} l ^{2} v}{R + r}$
Power: $P = F v = \frac{B ^{2} l ^{2} v ^{2}}{R + r} = i^{2} (R + r)$ , all of it appearing as heat
Rod projected with speed $u$ and then left alone: $v = u e^{- t / τ}$ with $τ = \frac{m ( R + r )}{B ^{2} l ^{2}}$
★ Must learn Constant force $F$ from rest: $v = v_{T} (1 - e^{- t / τ})$ with terminal speed $v_{T} = \frac{F ( R + r )}{B ^{2} l ^{2}}$
Rails on an incline $θ$ with a vertical field: $v_{T} = \frac{m g R sin θ}{B ^{2} l ^{2} cos ^{2} θ}$
★ Must learn Rod rotating about one end: $ε = \frac{1}{2} B ω l^{2} = B π l^{2} f$ ; about its centre the two ends are at the same potential
Rod of length $l$ parallel to a long wire, at distance $x$ , moving away at speed $v$ : $ε = \frac{μ _{0} i v l}{2 π x}$

1. A Moving Rod Behaves Like a Battery

Faraday's law needs a changing flux. When a rod slides along two rails, nothing about the field changes; what changes is the area of the circuit. The emf produced this way is called motional emf, and it is the easiest kind to calculate because the geometry does all the work.

Figure 1: In a time

d t

the rod sweeps an area

d A = l v d t

, so the flux through the circuit changes by

d ϕ = B l v d t

and

∣ ε ∣ = \frac{d ϕ}{d t} = B v l

In a short time $d t$ the rod of length $l$ moving at speed $v$ sweeps out an area $d A = l v d t$ .
The field is perpendicular to that area, so the flux through the circuit changes by $d ϕ = B d A = B l v d t$ .
Faraday's law then gives
$∣ ε ∣ = \frac{d ϕ}{d t} = B v l$

For a straight conductor of length $l$ moving with speed $v$ perpendicular to itself, in a field $B$ perpendicular to both,

ε = B v l

with $B$ in tesla, $v$ in $m s^{- 1}$ and $l$ in metre giving the emf in volt. An equivalent way to say it: the rod cuts $B v l$ field lines per second.

Exam Trick

Check the units before you check the algebra. A tesla is $V s m^{- 2}$ , so $B v l$ has units $V s m^{- 2} \times m s^{- 1} \times m = V$ . Any answer of yours that is not in volt has a slip in it somewhere.

2. Where the emf Actually Comes From

The swept-area argument gives the right number but hides the physics. Look at what happens to one free electron inside the rod.

Figure 2:

q (v \times B)

pushes positive carriers towards

P

until the electric pull balances it:

q v B = q E

, so

E = v B

and

V_{P} - V_{Q} = E l = B v l

. With

v

to the right and

B

into the page, the top end is positive.

Every free carrier inside the rod is dragged along at speed $v$ , so it feels a magnetic force $q (v \times B)$ directed along the rod.
Charge therefore piles up: positive at one end ( $P$ ), negative at the other ( $Q$ ).
The separated charge sets up an electric field $E$ inside the rod, pointing from $P$ to $Q$ , which pushes the carriers back.
Within nanoseconds the two forces balance: $q v B = q E$ , so $E = v B$ .
The potential difference across the rod is then
$V_{P} - V_{Q} = E l = B v l$

This is why a rod that is not part of a circuit still has a motional emf across its ends. The emf is a property of the rod and its motion; a closed circuit is needed only if you want a current to flow. An isolated rod simply sits there with $B v l$ volts between its ends and no current at all.

Flux view (Faraday)

Look at the circuit: the rod changes the enclosed area, $ε = B \frac{d A}{d t} = B v l$ . Needs a closed path to define the area.

Force view (Lorentz)

Look at the rod: $q (v \times B)$ pushes carriers along it, $ε = \int (v \times B) \cdot d l$ . Works for an isolated rod too.

Key idea

A conductor moving across a field is a battery:

ε = B v l

between its ends, positive end given by

q v \times B

, even with no circuit attached.

3. When the Geometry is Not Ideal

3.1 The rod at an angle

The formula $ε = B v l$ assumes the rod, its velocity and the field are all mutually perpendicular. If the rod is tilted, only the part of it lying across the velocity sweeps new area.

Figure 3: Only the part of the rod lying across the velocity sweeps new area. With

θ

between the rod and

v

(and

B

perpendicular to both),

ε = B v l sin θ

With $θ$ measured between the rod and $v$ (and $B$ still perpendicular to both of them),

ε = B v l sin θ

so a rod dragged along its own length produces nothing at all.

3.2 The general formula

For any element $d l$ of a conductor moving with velocity $v$ in a field $B$ , the contribution to the emf is $(v \times B) \cdot d l$ , so

ε = \int (v \times B) \cdot d l

Every special case on this page is this integral with a particular geometry put in.

3.3 Bent wires and effective length

What if the conductor is not straight? Consider a rigid closed loop moving in a uniform field: the flux through it never changes, so the emf round the whole loop is zero. That single fact settles everything.

Figure 4: The emf round a rigid closed loop in a uniform field is zero, so every path from

A

B

gives the same emf. Only the separation across

v

counts:

ε = B v l_{⊥}

If the emf round the closed loop is zero, then going from $A$ to $B$ along one path must give the same emf as going along any other path.
So a bent, kinked or wiggly wire from $A$ to $B$ gives exactly the same emf as the straight line $A B$ .
Only the component of $A B$ perpendicular to $v$ counts: $ε = B v l_{⊥}$ .

Exam Trick

When a question shows an ugly shape moving in a uniform field, do not integrate. Join the two end points with a straight line, drop the component along $v$ , and multiply by $B v$ . The messy figure is there to waste your time.

Quick Recall: tap to check

A rod is dragged along its own length across a field. What emf?

Zero: it sweeps no area.

A rigid closed loop moves at constant velocity in a uniform field. What is the total emf round it?

Zero: its flux never changes.

A semicircular wire of radius

r

moves perpendicular to its diameter in a field

B

. What emf?

ε = B v (2 r)

: only the straight separation across

v

counts.

With

v

to the right and

B

into the page, which end of a vertical rod is positive?

The top end:

v \times B

points up.

4. Rod on Rails: the Complete Circuit

Close the rails with a resistance and the emf can drive a current. The single most useful habit in this chapter is to redraw the picture as a circuit: the rod is a cell of emf $B v l$ whose internal resistance is the rod's own resistance $r$ .

Figure 5: Replace the moving rod by a cell of emf

ε = B v l

(

P

positive) with the rod's own resistance

r

inside it. The current

i = \frac{B v l}{R + r}

flows anticlockwise, and

F = B i l

on the rod points against

v

Quantity	Expression	Notes
emf	$ε = B v l$	independent of the resistance
Current	$i = \frac{B v l}{R + r}$	$r$ is the rod's own resistance
Force on the rod	$F = B i l = \frac{B ^{2} l ^{2} v}{R + r}$	always opposes the motion (Lenz's law)
Power supplied	$P = F v = \frac{B ^{2} l ^{2} v ^{2}}{R + r}$	by whoever pulls the rod
Heat produced	$P = i^{2} (R + r)$	equal to the power supplied, always
Charge flowed	$q = \frac{B l Δ x}{R + r}$	$Δ x$ is the distance moved

Notice the last two rows. The power you put in and the heat that comes out are equal, with nothing left over. That is not a coincidence, it is Lenz's law: the magnetic force on the rod points backwards, so work has to be done against it, and that work is precisely what appears in the resistance.

If the rails themselves have resistance, or there are several resistors, none of the physics changes. Work out the emf from $B v l$ , work out the total circuit resistance as you would in any d.c. problem, and divide.

Figure 6: Rod-on-rails problems in one chart. A resistor gives a force proportional to

v

(exponentials, terminal speed); a capacitor gives constant acceleration and no terminal speed.

Key idea

Redraw every rod-on-rails problem as a circuit: a cell of emf

B v l

(internal resistance

r

) driving

i = \frac{B v l}{R + r}

; the force

\frac{B ^{2} l ^{2} v}{R + r}

always opposes

v

5. How the Rod Moves

Once you know that the retarding force is $F = \frac{B ^{2} l ^{2} v}{R + r}$ , which is proportional to $v$ , the mechanics is the same as motion through a viscous fluid. Write $K = \frac{B ^{2} l ^{2}}{R + r}$ , so that the retarding force is simply $K v$ .

5.1 Rod projected and then left alone

Newton's second law: $m \frac{d v}{d t} = - K v$ .
Separate and integrate from $u$ to $v$ :
$\int_{u}^{v} \frac{d v}{v} = - \frac{K}{m} \int_{0}^{t} d t$
Therefore
$v = u e^{- K t / m} = u e^{- t / τ}, τ = \frac{m}{K} = \frac{m ( R + r )}{B ^{2} l ^{2}}$

The rod never quite stops, but after one time constant it has only $37%$ of its speed left.

5.2 Rod pulled by a constant force

Newton's second law: $m \frac{d v}{d t} = F - K v$ .
Integrating from rest:
$v = \frac{F}{K} (1 - e^{- K t / m}) = v_{T} (1 - e^{- t / τ})$
The terminal speed is reached when the retarding force balances the applied one:
$v_{T} = \frac{F}{K} = \frac{F ( R + r )}{B ^{2} l ^{2}}$

Figure 7: Both curves have time constant

τ = \frac{m ( R + r )}{B ^{2} l ^{2}}

. Projected and left alone,

v = u e^{- t / τ}

; pulled by a steady force,

v = v_{T} (1 - e^{- t / τ})

with

v_{T} = \frac{F ( R + r )}{B ^{2} l ^{2}}

5.3 Rails on an incline

The favourite JEE version puts the rails on a slope of angle $θ$ with the field vertical. Two cosines then appear, and missing either one is the classic error.

Figure 8: Two cosines. The flux uses

B cos θ

(so

i = \frac{B l v cos θ}{R}

), and the horizontal force

F = B i l

has only

F cos θ

along the slope. Balancing

m g sin θ

gives

v_{T} = \frac{m g R sin θ}{B ^{2} l ^{2} cos ^{2} θ}

Only the component of $B$ perpendicular to the plane of the rails, $B cos θ$ , produces an emf: $ε = (B cos θ) v l$ .
The current is $i = \frac{B v l cos θ}{R}$ , and the force on the rod, $F = B i l$ , is horizontal.
Only the component of that force along the slope resists the sliding, which brings in a second $cos θ$ :
$F_{slope} = \frac{B ^{2} l ^{2} v cos ^{2} θ}{R}$
At the steady speed this balances $m g sin θ$ :
$v_{T} = \frac{m g R sin θ}{B ^{2} l ^{2} cos ^{2} θ}$

JEE Advanced

A capacitor instead of a resistor. Close the rails with a capacitance $C$ and no resistance. The charge on it is $q = C\varepsilon = CBvl$, so the current is

i = \frac{d q}{d t} = C B l \frac{d v}{d t} = C B l a

The magnetic force is $B i l = C B^{2} l^{2} a$ , so Newton's second law gives

F - C B^{2} l^{2} a = ma \Rightarrow a = \frac{F}{m + C B ^{2} l ^{2}}

The acceleration is constant, not exponential, and the capacitor makes the rod behave as if it had extra mass $C B^{2} l^{2}$ . A resistor dissipates energy and gives a terminal speed; a capacitor only stores it and gives none.

Figure 9: Same rod, same force. A resistor dissipates energy, so

v \to v_{T}

; a capacitor only stores it, so

a = \frac{F}{m + C B ^{2} l ^{2}}

stays constant and there is no terminal speed.

6. The Rotating Rod

A rod turning about one end is not a rigid translation, so $B v l$ cannot be used directly: every point of the rod moves at a different speed. Integrate instead.

Figure 10: Each element

d r

contributes

B (ω r) d r

, so

ε = \int_{0}^{l} B ω r d r = \frac{1}{2} B ω l^{2}

. The emf grows as

x^{2}

: the inner half gives

\frac{1}{4}

of it, the outer half

\frac{3}{4}

(Solved Example 11).

Take an element of length $d r$ at a distance $r$ from the pivot. It moves with speed $v = ω r$ perpendicular to itself.
Its contribution to the emf is $d ε = B (ω r) d r$ .
Add up along the rod:
$ε = \int_{0}^{l} B ω r d r = \frac{1}{2} B ω l^{2}$

Written with the frequency of rotation $f$ , since $ω = 2 π f$ , this is $ε = B π l^{2} f$ , which is just $B$ times the area swept per second. Two consequences worth memorising:

The emf between the pivot and a point at distance $x$ is $\frac{1}{2} B ω x^{2}$ , so it grows as the square of the distance, not linearly. The outer half of a rod carries three quarters of the emf.
For a rod of length $2 l$ rotating about its centre, each half develops $\frac{1}{2} B ω l^{2}$ with the same sign relative to the centre, so the two ends sit at the same potential and the emf between the ends is zero.

Exam Trick

A rotating rod is a cell of emf $\frac{1}{2} B ω l^{2}$ . If its far end slides on a conducting ring and a resistance $R$ joins the centre to the ring, the current is simply $i = \frac{B ω l ^{2}}{2 ( R + r )}$ , where $r$ is the resistance of the rod. Draw the equivalent cell and the rest is ordinary circuit work.

Figure 11: A rotating rod is a cell of emf

\frac{1}{2} B ω l^{2}

. With its outer end sliding on a ring and

R

joined from the centre to the ring,

i = \frac{B ω l ^{2}}{2 ( R + r )}

Quick Recall: tap to check

What is the emf of a rod of length

l

turning about one end at

ω

in a field

B

\frac{1}{2} B ω l^{2}

, which is

B

times the area swept per second.

The same rod turns about its centre. What is the emf between its two ends?

Zero: both halves push charge the same way relative to the centre.

A rod on rails is pulled by a constant force through a resistor. Does it keep accelerating?

No, it reaches

v_{T} = \frac{F ( R + r )}{B ^{2} l ^{2}}

. With a capacitor instead, it does keep accelerating.

Rails on an incline, field vertical: how many factors of

cos θ

appear in the terminal speed?

Two: one from the flux (

B cos θ

), one from resolving the horizontal force.

Key idea

A rotating rod is a cell of emf

\frac{1}{2} B ω l^{2}

; the emf grows as the square of the distance from the pivot.

7. A Rod Moving Near a Long Straight Wire

Here the field is not uniform: it falls off as $\frac{μ _{0} i}{2 π x}$ . Whether you can use $B v l$ directly depends on how the rod is placed.

Figure 12: Parallel rod: one value of

B

, so

ε = μ_{0} i v l /2 π x

. Rod pointing away from the wire:

B

falls along it, so integrate

(μ_{0} i v /2 π x) d x

from

a

a + l

to get a logarithm.

7.1 Rod parallel to the wire

Every point of the rod is at the same distance $x$ , so the field is the same all along it and no integration is needed:

ε = B v l = \frac{μ _{0} i v l}{2 π x}

The emf falls off as the rod is carried further away.

7.2 Rod perpendicular to the wire

Now the field varies along the rod, so take an element $d x$ at distance $x$ from the wire and add up:

ε = \int_{a}^{a + l} \frac{μ _{0} i v}{2 π x} d x = \frac{μ _{0} i v}{2 π} ln (\frac{a + l}{a})

A logarithm in the answer is the signature of a rod pointing away from the wire; a plain $\frac{1}{x}$ is the signature of a rod lying parallel to it. If your answer has the wrong one of the two, you have set the integral up along the wrong direction.

Figure 13: Mind map of motional emf. Every branch starts from

ε = \int (v \times B) \cdot d l

8. Solved Examples

Solved Example 1

A straight conductor

0.40 m

long moves at

7.0 m s^{- 1}

in a magnetic field of

0.90 Wb m^{- 2}

. Find the largest emf that can be induced in it, and say how it must be oriented to get that value.

Solution:

Given: $l = 0.40 m$ , $v = 7.0 m s^{- 1}$ , $B = 0.90 T$ .

In general $ε = B v l sin θ$ , which is largest when $sin θ = 1$ , that is when the rod moves perpendicular to its own length.

ε_{m a x} = B v l = (0.90) (7.0) (0.40) = 2.52 V

Answer: $2.52 V$ , with the rod, its velocity and the field all mutually perpendicular.

Solved Example 2

State the emf induced in each of these cases: (a) a rod moving parallel to a uniform field

B

; (b) a rod dragged along its own length across a field; (c) a rigid closed triangular loop moving with constant velocity in a uniform field.

Solution:

(a) If $v$ is parallel to $B$ then $v \times B = 0$ , so no force acts along the rod and $ε = 0$ .

(b) The rod sweeps no new area, or equivalently $θ = 0$ in $B v l sin θ$ , so $ε = 0$ .

(c) A rigid loop moving in a uniform field encloses the same flux at every instant, so $ε = 0$ round the loop. Individual sides do have emfs, but they cancel exactly.

Answer: all three are zero. These three pictures appear again and again as distractors; recognising them saves a minute each time.

Solved Example 3

An irregular, bent wire joins two points

A

and

B

and moves with velocity

v

in a uniform field

B

perpendicular to the plane of the motion. The straight line

A B

has length

l

and makes an angle

θ

with

v

. Find the emf between

A

and

B

Solution:

Key idea. Complete the circuit by adding the straight line $A B$ . The closed loop so formed is rigid and moves in a uniform field, so the flux through it is constant and the emf round it is zero.

Therefore the emf from $A$ to $B$ along the bent wire equals the emf from $A$ to $B$ along the straight line, and only the part of $A B$ across $v$ matters:

ε = B v l_{⊥} = B v l sin θ

Answer: $ε = B v l sin θ$ , where $l$ is the straight distance $A B$ . The actual shape of the wire is irrelevant.

Solved Example 4

A square wire loop of side

10 cm

and resistance

1.0 Ω

moves at a steady speed

v_{0}

with only its front arm inside a region of field

B = 2.0 Wb m^{- 2}

perpendicular to its plane (the rest of the loop is outside the field). The loop is connected through leads of negligible resistance to a network of five

3 Ω

resistors forming a balanced Wheatstone bridge. What speed gives a steady current of

1.0 mA

in the loop?

Solution:

Reduce the network. The bridge is balanced, so no current flows in the central arm and it can be removed. Two arms of $3 + 3 = 6 Ω$ then sit in parallel:

R_{net} = \frac{6 \times 6}{6 + 6} = 3 Ω

Total resistance: $R = 3 + 1 = 4 Ω$ .

Use the motional emf: $ε = B v_{0} l$ and $i = \frac{ε}{R}$ , so

v_{0} = \frac{i R}{B l} = \frac{( 1.0 \times 1 0 ^{- 3} ) ( 4 )}{( 2.0 ) ( 0.10 )} = 2.0 \times 1 0^{- 2} m s^{- 1}

Answer: $v_{0} = 2.0 cm s^{- 1}$ , Only the arm inside the field acts as a cell of emf $B v_{0} l$ ; spotting the balanced bridge first is what makes this a one-minute question.

Solved Example 5

Two long parallel rails a distance

l

apart are joined at one end by a resistance

R

. A rod

C D

across them is pulled at a constant speed

v

in a field

B

perpendicular to the plane of the rails. Find the work that must be done per second to keep the rod moving, neglecting the resistance of the wires.

Solution:

Motional emf: $ε = B v l$ .
Current: $i = \frac{B v l}{R}$ .
Magnetic force on the rod: $F_{M} = B i l = \frac{B ^{2} l ^{2} v}{R}$ , opposing the motion.
Since the speed is constant, the applied force equals $F_{M}$ , so the power needed is
$P = F_{M} v = \frac{B ^{2} l ^{2} v ^{2}}{R}$

Answer: $P = \frac{B ^{2} l ^{2} v ^{2}}{R}$ . A check: $i^{2} R = (\frac{B v l}{R})^{2} R = \frac{B ^{2} l ^{2} v ^{2}}{R}$ , the same thing. The magnetic field does no work itself; it only passes mechanical energy on as heat.

Solved Example 6

A rod of mass

m

and resistance

r

rests on frictionless, resistanceless rails a distance

l

apart, closed by a resistance

R

, in a field

B

perpendicular to the plane of the rails. It is given an initial speed

u

and released. Find its speed as a function of time.

Solution:

At speed $v$ the current is $i = \frac{B v l}{R + r}$ and the retarding force is $B i l = \frac{B ^{2} l ^{2} v}{R + r}$ .
Newton's second law, with no applied force:
$m \frac{d v}{d t} = - \frac{B ^{2} l ^{2}}{R + r} v = - K v$
Separate the variables and integrate:
$\int_{u}^{v} \frac{d v}{v} = - \frac{K}{m} \int_{0}^{t} d t \Rightarrow ln \frac{v}{u} = - \frac{K t}{m}$

v = u e^{- K t / m}, K = \frac{B ^{2} l ^{2}}{R + r}

Answer: $v = u exp (- \frac{B ^{2} l ^{2} t}{m ( R + r )})$ . The rod slows down for ever but never quite stops, and the total heat produced over all time is $\frac{1}{2} m u^{2}$ , the whole of its kinetic energy.

Solved Example 7

The same rod starts from rest and a constant force

F

is now applied along the rails. Find its speed as a function of time and its final speed.

Solution:

Newton's second law:
$m \frac{d v}{d t} = F - K v, K = \frac{B ^{2} l ^{2}}{R + r}$
Separate and integrate from rest:
$\int_{0}^{v} \frac{d v}{F - K v} = \frac{1}{m} \int_{0}^{t} d t \Rightarrow - \frac{1}{K} ln \frac{F - K v}{F} = \frac{t}{m}$
Rearranging,
$v = \frac{F}{K} (1 - e^{- K t / m})$

Answer: $v = \dfrac{F(R + r)}{B^{2}l^{2}}\left(1 - e^{-B^{2}l^{2}t/m(R + r)}\right)$, rising towards a terminal speed $v_{T} = \frac{F ( R + r )}{B ^{2} l ^{2}}$ . Setting $\frac{d v}{d t} = 0$ in the first line gives $v_{T}$ immediately, without any integration.

Solved Example 8

A square wire of side

l

, mass

m

and resistance

R

slides without friction down two parallel rails lying on a plane inclined at

θ

to the horizontal. A uniform vertical field

B

fills the region. Show that the rod reaches a steady speed

v = \frac{m g R sin θ}{B ^{2} l ^{2} cos ^{2} θ}

Solution:

Step 1: which part of $B$ matters? Only the component perpendicular to the plane of the rails threads the circuit, and that is $B cos θ$ . So the emf at speed $v$ is

ε = (B cos θ) v l

Step 2: current.

i = \frac{B v l cos θ}{R}

Step 3: force. The force on the rod, $F = B i l$ , is horizontal. Its component along the slope is $F cos θ$ , so the retarding force up the slope is

F_{slope} = \frac{B ^{2} l ^{2} v cos ^{2} θ}{R}

Step 4: steady speed. Setting this equal to the driving force $m g sin θ$ ,

\frac{B ^{2} l ^{2} v cos ^{2} θ}{R} = m g sin θ \Rightarrow v = \frac{m g R sin θ}{B ^{2} l ^{2} cos ^{2} θ}

Answer: as required. The two factors of $cos θ$ come from two different places, one from the flux and one from resolving the force. Losing either gives the standard wrong answer.

Solved Example 9

A rod of length

l

is held parallel to a long straight wire carrying a constant current

i

, and is moved directly away from the wire with speed

v

. Find the emf induced when the rod is at a distance

x

from the wire. What changes if the rod is instead perpendicular to the wire, with its near end at distance

a

Solution:

Parallel case. Every point of the rod is at the same distance $x$ , so the field is uniform along it:

B = \frac{μ _{0} i}{2 π x} \Rightarrow ε = B v l = \frac{μ _{0} i v l}{2 π x}

Perpendicular case. Now $B$ changes along the rod, so take an element $d x$ at distance $x$ from the wire, moving with the same speed $v$ :

d ε = \frac{μ _{0} i v}{2 π x} d x \Rightarrow ε = \frac{μ _{0} i v}{2 π} \int_{a}^{a + l} \frac{d x}{x} = \frac{μ _{0} i v}{2 π} ln (\frac{a + l}{a})

Answer: $\frac{μ _{0} i v l}{2 π x}$ for the parallel rod and $\frac{μ _{0} i v}{2 π} ln \frac{a + l}{a}$ for the perpendicular one. Same wire, same speed, completely different form, purely because of the orientation.

Solved Example 10

A metal rod

1.5 m

long rotates about one end in a vertical plane at right angles to the magnetic meridian, at

20

revolutions per second. Taking the horizontal component of the earth's field as

B_{H} = 0.32 \times 1 0^{- 4} T

, find the emf induced between the ends of the rod.

Solution:

Given: $l = 1.5 m$ , $f = 20 Hz$ , $B_{H} = 0.32 \times 1 0^{- 4} T$ . The rod turns in a plane perpendicular to $B_{H}$ , so the full horizontal component is effective.

Angular speed: $ω = 2 π f = 2 π (20) = 125.7 rad s^{- 1}$ .

Rotating-rod formula:

ε = \frac{1}{2} B_{H} ω l^{2} = \frac{1}{2} (0.32 \times 1 0^{- 4}) (125.7) (1.5)^{2}

ε = 4.5 \times 1 0^{- 3} V

Answer: $ε \approx 4.5 mV$ . The same answer follows from $ε = B \times (area swept per second) = B_{H} π l^{2} f$ , which is often quicker.

Solved Example 11

A rod

P Q

of length

2 l

rotates in a uniform field

B

about the end

P

, with

B

perpendicular to the plane of rotation.

M

is the midpoint of the rod. If the emf between

P

and

Q

100 V

, find the emf between

M

and

Q

Solution:

Emf from the pivot out to a distance $x$ : $ε_{P x} = \frac{1}{2} B ω x^{2}$ , so it goes as $x^{2}$ .

Whole rod: $\varepsilon_{PQ} = \dfrac{1}{2}B\omega(2l)^{2} = 2B\omega l^{2} = 100\,\text{V}$.

Inner half: $ε_{P M} = \frac{1}{2} B ω l^{2}$ , which is one quarter of $ε_{P Q}$ , that is $25 V$ .

Outer half: the emfs of the two halves add along the rod, so

ε_{M Q} = ε_{P Q} - ε_{P M} = 100 - 25 = 75 V

Answer: $75 V$ . The outer half carries three quarters of the emf even though it is only half the length, because of the $x^{2}$ dependence.

Solved Example 12

A rod of length

l

rotates with angular speed

ω

about one end, which is at a distance

a

from a long straight wire carrying a constant current

i

. The rod lies along the line pointing away from the wire at the instant shown. Find the emf induced in the rod at that instant.

Solution:

Set up the element. Take a piece $d x$ of the rod at a distance $x$ from the pivot. It is therefore at a distance $(x + a)$ from the wire, where the field is

B = \frac{μ _{0} i}{2 π ( x + a )}

and it moves perpendicular to the rod with speed $ω x$ .

Contribution:

d ε = B (ω x) d x = \frac{μ _{0} iω}{2 π} \cdot \frac{x}{x + a} d x

Integrate after splitting the fraction, using $\frac{x}{x + a} = 1 - \frac{a}{x + a}$ :

ε = \frac{μ _{0} iω}{2 π} \int_{0}^{l} (1 - \frac{a}{x + a}) d x = \frac{μ _{0} iω}{2 π} [l - a ln (\frac{a + l}{a})]

Answer: $\varepsilon = \dfrac{\mu_{0}i\omega}{2\pi} \left[l - a\ln\dfrac{a + l}{a}\right]$. Check the limit $a \to \infty$ : the bracket tends to zero, as it must, because a distant wire produces no field at the rod.

Solved Example 13

A vertical rod

P Q

, with

P

at the top, slides to the right on horizontal rails in a uniform field pointing out of the page. Which end is at the higher potential, and which way does the current flow inside the rod?
(A)

P

higher; current from

Q

P

(B)

Q

higher; current from

P

Q

(C)

P

higher; current from

P

Q

(D)

Q

higher; current from

Q

P

Solution:

Force on a positive carrier: $v \times B$ with $v$ to the right and $B$ out of the page points down, so positive charge collects at $Q$ and $Q$ is the positive terminal.

The rod is the source. Inside any source the current flows from its negative terminal to its positive terminal, so from $P$ to $Q$ ; outside, it returns from $Q$ round the circuit to $P$ .

Answer: (B). Reverse the field (into the page) and $P$ becomes positive, as in the charge-separation figure.

Solved Example 14

A conducting rod of length

l

rotates with angular speed

ω

about a perpendicular axis through its centre, in a uniform field

B

parallel to the axis. The emf between the centre and one end, and between the two ends, are
(A)

\frac{1}{2} B ω l^{2}

and

0

(B)

\frac{1}{8} B ω l^{2}

and

0

(C)

\frac{1}{8} B ω l^{2}

and

\frac{1}{4} B ω l^{2}

(D)

\frac{1}{2} B ω l^{2}

and

B ω l^{2}

Solution:

Each half has length $\frac{l}{2}$ and turns about the centre, so centre to end: $\frac{1}{2} B ω (\frac{l}{2})^{2} = \frac{1}{8} B ω l^{2}$ .

Both ends are at the same potential relative to the centre, so the emf between the ends is zero.

Answer: (B).

Solved Example 15

A metal rod

1.0 m

long rotates about one end at

120

revolutions per minute in a plane perpendicular to a uniform field of

0.50 T

. Find the emf between its ends.

Solution:

Angular speed: $ω = 2 π \times \frac{120}{60} = 4 π = 12.6 rad s^{- 1}$ .

emf: $ε = \frac{1}{2} B ω l^{2} = \frac{1}{2} (0.50) (12.6) (1.0)^{2} = 3.14 V$ .

Answer: $ε \approx 3.1 V$ . Check with $B π l^{2} f = (0.50) π (1.0)^{2} (2) = 3.14 V$ .

Solved Example 16

Two smooth rails

0.50 m

apart lie on a plane inclined at

3 0^{\circ}

and are joined at the bottom by a

2.0 Ω

resistor. A rod of mass

0.10 kg

slides down them in a vertical field of

0.60 T

. Find its terminal speed and the current at that speed, and check the energy balance (

g = 9.8 m s^{- 2}

Solution:

Terminal speed:

v_{T} = \frac{m g R sin θ}{B ^{2} l ^{2} cos ^{2} θ} = \frac{( 0.10 ) ( 9.8 ) ( 2.0 ) ( 0.50 )}{( 0.60 ) ^{2} ( 0.50 ) ^{2} ( 0.75 )} = 14.5 m s^{- 1}

Current: $i = \frac{B l v _{T} cos θ}{R} = \frac{( 0.60 ) ( 0.50 ) ( 14.5 ) ( 0.866 )}{2.0} = 1.89 A$ .

Check forces: $F = B i l = 0.566 N$ (horizontal); up the slope $F cos θ = 0.49 N = m g sin θ$ .

Check energy: $m g sin θ v_{T} = (0.49) (14.5) = 7.1 W$ and $i^{2} R = (1.89)^{2} (2.0) = 7.1 W$ .

Answer: $v_{T} \approx 14.5 m s^{- 1}$ , $i \approx 1.9 A$ . Dropping one $cos θ$ gives $12.6 m s^{- 1}$ and dropping both gives $10.9 m s^{- 1}$ , the standard wrong answers.

Practice Questions

An aeroplane with a wingspan of $20 m$ flies horizontally at $250 m s^{- 1}$ where the vertical component of the earth's field is $5.0 \times 1 0^{- 5} T$ . Find the emf between its wing tips.Answer: $ε = B v l = 0.25 V$ . Only the vertical component matters, because the wings are horizontal and the velocity is horizontal.
A rod of length $0.50 m$ slides at $4.0 m s^{- 1}$ on rails closed by $2.0 Ω$ , in a field of $0.80 T$ perpendicular to the rails. Find the emf, the current, the force needed and the power.Answer: $ε = 1.6 V$ , $i = 0.80 A$ , $F = 0.32 N$ , $P = 1.28 W$ (which equals $i^{2} R$ ).
A rod of length $l$ lies perpendicular to a long wire carrying current $i$ , with its near end at distance $a$ , and moves parallel to the wire with speed $v$ . Find the emf induced in it.Answer: $ε = \frac{μ _{0} i v}{2 π} ln \frac{a + l}{a}$ , from integrating $\frac{μ _{0} i v}{2 π x} d x$ from $a$ to $a + l$ .
A rod $P Q$ of length $l$ and resistance $r$ slides at speed $v$ on rails closed at both ends by resistances $R_{1}$ and $R_{2}$ . Find the current in the rod.Answer: $R_{1}$ and $R_{2}$ are in parallel across the rod, so $i = \frac{B v l}{r + \frac{R _{1} R _{2}}{R _{1} + R _{2}}}$ .
A rod of length $2 L$ and resistance $2 r$ rotates at $ω$ about its centre, with both ends touching a conducting ring of negligible resistance. A resistance $R$ joins the centre to the ring. Find the current in $R$ .Answer: Each half is a cell of emf $\frac{1}{2} B ω L^{2}$ with resistance $r$ , and the two are in parallel, so $i = \frac{B ω L ^{2}}{2 ( R + \frac{r}{2} )}$ .
Show that for a rod rotating about one end, the potential difference between the pivot and a point at distance $x$ along the rod is $\frac{1}{2} B ω x^{2}$ .Answer: Integrate $d ε = B ω r d r$ from $0$ to $x$ . The potential varies as the square of the distance, so half the length carries only a quarter of the emf.
A rod of length $0.20 m$ and resistance $0.25 Ω$ moves at $5.0 m s^{- 1}$ on rails closed by $1.0 Ω$ in a field of $0.50 T$ . Find the potential difference across the rod's ends.Answer: $ε = 0.50 V$ , $i = 0.40 A$ , so the terminal voltage is $i R = 0.40 V$ , less than the emf by $i r = 0.10 V$ .

Common Mistakes to Avoid

Watch out

Using $ε = B v l$ for a rotating rod. Every point moves at a different speed, so the answer is $\frac{1}{2} B ω l^{2}$ , not $B ω l \cdot l$ .
Measuring $θ$ in $B v l sin θ$ from the wrong pair. It is the angle between the rod and its velocity, with $B$ perpendicular to both.
Integrating along a bent wire instead of replacing it with the straight line joining its ends and taking the component across $v$ .
Forgetting the rod's own resistance $r$ . The total circuit resistance is $R + r$ , and $r$ behaves exactly like an internal resistance.
Using only one factor of $cos θ$ for rails on an incline in a vertical field. One comes from the flux, the second from resolving the force along the slope.
Thinking the magnetic field does the work. It does none; the applied force supplies all the energy and the field only routes it into the resistance.
Writing $B v l$ for a rod near a long straight wire when the rod points away from it. There the field varies along the rod and an integral (a logarithm) is needed.
Expecting a terminal speed when the rails are closed by a capacitor. A capacitor gives constant acceleration $\frac{F}{m + C B ^{2} l ^{2}}$ , not a limiting speed.

Frequently Asked Questions

What is motional emf?

Motional emf is the voltage produced across a conductor because it is moving through a magnetic field. The magnetic force pushes the free carriers along the conductor until the charge they pile up stops them. For a rod perpendicular to both $v$ and $B$ , the motional emf is $B v l$ .

Is motional emf different from the emf given by Faraday's law?

No, they are two views of the same thing. Faraday's law looks at the changing area of the circuit; the motional picture looks at the magnetic force on the carriers inside the rod. For a rod on rails both give $B v l$ , and any disagreement means an error somewhere.

Does a rod need a closed circuit to have a motional emf?

No. An isolated rod moving in a field still has $B v l$ volts between its ends, because the charge separation happens inside the rod itself. A closed path is needed only for a current to flow, and therefore for any force or heating to appear.

Why is the emf of a rotating rod half B omega l squared?

Because the speed is not the same everywhere. An element at distance $r$ moves at $ω r$ and contributes $B ω r d r$ . Integrating from $0$ to $l$ gives $\frac{1}{2} B ω l^{2}$ . The factor of one half is the average of a quantity that grows linearly along the rod.

Does the magnetic field do work on the rod?

Never. A magnetic force is always perpendicular to the velocity of the charge it acts on, so it can do no net work. The person or motor pulling the rod supplies every joule; the field simply converts that mechanical work into electrical energy and then heat.

Why does a rod on rails slow down exponentially?

The retarding force is $\frac{B ^{2} l ^{2} v}{R + r}$ , which is proportional to the speed itself. A force proportional to velocity always gives exponential decay, with time constant $τ = \frac{m ( R + r )}{B ^{2} l ^{2}}$ , exactly as for motion through a viscous fluid.

What motional emf questions appear in NEET?

NEET keeps to direct substitution: the emf across an aircraft's wings, a rod on rails, a rotating rod using $\frac{1}{2} B ω l^{2}$ , and the force or power for a rod pulled at constant speed. Recognising the zero-emf pictures is worth an easy mark.

What is asked in JEE Main and JEE Advanced?

JEE Main favours rod-on-rails circuits and rods near a long straight wire. JEE Advanced adds calculus: exponential decay and growth of speed, rails on an incline with a vertical field, a capacitor in place of the resistor, and rotating rods in non-uniform fields.

Previous year questions on Motional EMF

14 questions from past papers, each with a step-by-step solution.

Show all 14 questions

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