Chapter 6 · Electromagnetic Induction

6.2 Motional EMF and Energy Considerations

A conductor moving through a magnetic field has an emf $\varepsilon=Blv$ across it, because the field pushes its free charges along it. In a closed circuit the induced current brakes the motion, and the work done against that braking force becomes heat.

When an electric train or car brakes, its motors can be switched to work as generators: the wheels keep the motor's coils turning in a magnetic field, the induced current is fed back into the supply or the battery, and the magnetic force on that current slows the vehicle down. Indian Railways' modern electric locomotives return a sizeable fraction of their braking energy to the overhead line in this way. The same physics, run the other way, is how every generator works — mechanical work pushes conductors through a magnetic field, and electrical energy comes out.

In Section 6.1 the flux through a circuit changed because a magnet moved or a current changed. Here the field is steady and the conductor moves. We find the emf of a moving rod directly from the magnetic force on its free charges — and see that it agrees with Faraday's law — then the emf of a rotating rod, the force that opposes the motion, and the full energy account: where the work goes, how much charge flows, and why a rod pushed by a steady force reaches a terminal velocity.

Motional EMF from the Magnetic Force

Take a straight metal rod PQ of length $l$ moving with velocity $\mathbf v$ through a uniform magnetic field $\mathbf B$, with the rod, $\mathbf v$ and $\mathbf B$ mutually perpendicular. Every free electron in the rod moves with the rod, so it feels the magnetic force $q\mathbf v\times\mathbf B$ (Chapter 4), of size $evB$, directed along the rod. The electrons drift towards one end, leaving the other end positive. The charges build up until the electric field they create inside the rod balances the magnetic push: $eE=evB$.

The emf is the work done per unit charge by the non-electrostatic force in carrying a charge from one end to the other. Here that force is the magnetic force $qvB$, acting over the length $l$, so the work per unit charge is $vBl$. The rod behaves like a cell of emf $Blv$ — with no chemicals in it.

The same answer from Faraday's law. Let the rod slide along a pair of parallel rails a distance $l$ apart, joined at one end, so that it closes a rectangular circuit. If the rod is a distance $x$ from the joined end, the area of the circuit is $lx$ and the flux through it is $\Phi_B=Blx$. As the rod moves, $x$ grows at the rate $dx/dt=v$: $$|\varepsilon|=\frac{d\Phi_B}{dt}=Bl\frac{dx}{dt}=Blv .$$ The two pictures — force on the charges, and rate of change of flux — always agree.

📘 Definition 6.2 — Motional emf
The emf induced across a conductor because it moves through a magnetic field is called a motional emf. Its source is the magnetic force $q\mathbf v\times\mathbf B$ on the free charges in the conductor. A conductor moving parallel to $\mathbf B$, or along its own length, has no motional emf.
📐 Result 6.4 — EMF of a moving rod

A straight rod of length $l$ moving with speed $v$ in a uniform field $B$, with the rod, the velocity and the field mutually perpendicular, has an emf across its ends $$\varepsilon=Blv .$$ If the velocity (still perpendicular to the rod) makes an angle $\theta$ with $\mathbf B$, only the component $v\sin\theta$ perpendicular to the field counts: $\varepsilon=Blv\sin\theta$. The end towards which the magnetic force pushes positive charges is the positive terminal.

🧭 Problem-solving strategy — the direction of a motional emf

Fleming's right-hand rule. Hold the thumb, first finger and second finger of the right hand mutually at right angles. Point the thumb along the motion of the conductor and the first finger along the field; the second finger then points along the induced current in the conductor (from its negative to its positive end inside the rod, as in a cell). Equivalently, the force $q\mathbf v\times\mathbf B$ on a positive charge points along the current. Check the answer with Lenz's law: the current must oppose the change of flux through the circuit.

(Fleming's left-hand rule, from Chapter 4, gives the force on a current; the right-hand rule gives the current produced by motion. Motors: left; generators: right.)

✏️ Example 1 — The emf across a train axle
A train runs at $90$ km/h on level track at a place where the vertical component of the Earth's magnetic field is $4.0\times10^{-5}$ T. The steel axle joining a pair of wheels is $1.5$ m long. (a) Find the emf between the ends of the axle. (b) Could a voltmeter riding on the train, connected to the two ends of the axle, show this emf?
Which component? The axle is horizontal and moves horizontally, perpendicular to its length. Only the field component perpendicular to both — the vertical component — contributes.
(a) emf $v=90\ \text{km/h}=25$ m/s. $\varepsilon=Blv$ $=(4.0\times10^{-5})(1.5)(25)$ $=1.5\times10^{-3}$ V $=1.5$ mV.
(b) Voltmeter No. The voltmeter's leads move with the train through the same field and have exactly the same emf induced in them, so the emfs round the closed loop cancel. The emf could only be detected by a circuit part of which is at rest — for instance, the rails (which is how some train-detection circuits work, the rails being insulated in sections).
✏️ Example 2 — A rod sliding on rails
Two long horizontal metal rails, $0.40$ m apart, are joined at their left ends by a $2.0\ \Omega$ resistor. A uniform field of $0.50$ T points vertically downwards through the circuit (into the page in a top view). A metal rod lying across the rails is pulled to the right at a steady $5.0$ m/s. The rod and rails have negligible resistance and there is no friction. Find (a) the emf, (b) the current and its direction, (c) the force needed to keep the rod moving, and (d) the power supplied by the pulling agent and the power dissipated in the resistor.
(a) emf $\varepsilon=Blv=(0.50)(0.40)(5.0)=1.0$ V.
(b) Current $I=\dfrac{\varepsilon}{R}=\dfrac{1.0}{2.0}=0.50$ A. Direction: $\mathbf v$ to the right, $\mathbf B$ into the page, so $q\mathbf v\times\mathbf B$ on a positive charge points up the page along the rod. The current flows up through the rod, left along the top rail, down through the resistor and back along the bottom rail — anticlockwise in the top view. Check (Lenz): the area, and so the into-the-page flux, is increasing; an anticlockwise current makes a field out of the page, opposing the increase ✓.
(c) Force The current $I$ in the rod, in the field, feels a force $BIl$ $=(0.50)(0.50)(0.40)$ $=0.10$ N. By Fleming's left-hand rule it points to the left — opposing the motion. To keep $v$ constant, the agent must pull with $0.10$ N to the right.
(d) Power Agent: $P=Fv=(0.10)(5.0)=0.50$ W. Resistor: $I^2R=(0.50)^2(2.0)=0.50$ W. They are equal: all the mechanical work becomes heat.

Rotating Rods and Discs

Now let a metal rod of length $l$ rotate with angular velocity $\omega$ about one end O, in a plane perpendicular to a uniform field $B$ — like the spoke of a wheel. Different parts of the rod now move at different speeds: a point at distance $r$ from O has speed $v=\omega r$, which grows steadily from $0$ at the pivot to $\omega l$ at the tip. Two short arguments give the emf between O and the tip.

(i) Average speed. The emf of each short piece is $B\,v\,\delta r$, proportional to its speed. Since $v$ grows linearly along the rod, the total is the same as if the whole rod moved at the average speed $\tfrac12\omega l$: $\varepsilon=Bl\cdot\tfrac12\omega l$. (Adding the pieces, $\int_0^lB\omega r\,dr$, gives the same result.)

(ii) Flux swept. In time $dt$ the rod turns through $\omega\,dt$ and sweeps out a thin sector of area $\tfrac12l^2\omega\,dt$. If the rod is part of a circuit (a sliding contact on a circular rim, for instance), the flux through the circuit changes by $B\cdot\tfrac12l^2\omega\,dt$, so $\varepsilon=\tfrac12B\omega l^2$.

📐 Result 6.5 — EMF of a rod rotating about one end

A rod of length $l$ rotating with angular speed $\omega$ about one end, in a plane perpendicular to a uniform field $B$, has an emf between its ends $$\varepsilon=\tfrac12B\omega l^2=B\,l\,\frac{\omega l}{2}.$$ With $\omega=2\pi\nu$ for $\nu$ revolutions per second, $\varepsilon=\pi\nu Bl^2$. A metal disc of radius $l$ spinning about its axis parallel to $\mathbf B$ (a Faraday disc or homopolar generator) has the same emf between its centre and its rim — it is like many spokes in parallel.

✏️ Example 3 — A spinning rod and a spinning disc
(a) A copper rod $1.0$ m long rotates about one end at $5.0$ revolutions per second in a plane perpendicular to a uniform field of $0.20$ T. Find the emf between its ends. (b) A copper disc of radius $10$ cm spins at $3000$ revolutions per minute about its axis, which is parallel to a field of $0.50$ T. Find the emf between its centre and its rim. (c) If the rod of part (a) rotated about its midpoint, what would the emf be between its two ends?
(a) Rod $\omega=2\pi(5.0)\approx31.4$ rad/s. $\varepsilon=\tfrac12B\omega l^2$ $=\tfrac12(0.20)(31.4)(1.0)^2$ $\approx3.1$ V (exactly $\pi$ volts).
(b) Disc $\omega=\dfrac{2\pi\times3000}{60}\approx314$ rad/s. $\varepsilon=\tfrac12(0.50)(314)(0.10)^2$ $\approx0.79$ V. A Faraday disc gives a steady (DC) emf, but a small one — it needs a huge current to be useful.
(c) Midpoint Each half, of length $l/2$, has an emf $\tfrac12B\omega(l/2)^2$ from the centre outwards, so both ends are at the same potential relative to the centre: the emf between the ends is zero. (Between the centre and either end it is $\tfrac18B\omega l^2\approx0.79$ V.)

Energy Considerations

Return to the rod on rails, now with a resistance $R$ in the circuit (the resistor plus the rod and rails). The emf $Blv$ drives a current $I=\dfrac{Blv}{R}$. This current flows through the rod, which is in the field, so the rod feels a magnetic force $BIl$ — and by Lenz's law that force opposes the motion. It acts like a brake whose strength grows with speed: $$F_\text{mag}=BIl=\frac{B^2l^2v}{R}.$$ To keep the rod moving at constant speed, an external agent must push with an equal and opposite force, doing work at the rate $$P=F_\text{mag}v=\frac{B^2l^2v^2}{R}=\frac{\varepsilon^2}{R}=I^2R .$$ That is exactly the rate at which heat is produced in the resistance. Mechanical work is converted first into electrical energy and then into heat — energy is conserved, as Lenz's law guarantees. (The magnetic force on a moving charge never does work itself; it is the agent's push, transmitted through the magnetic force, that drives the current.)

📐 Principle 6.6 — Energy in motional induction

For a rod of length $l$ moving with speed $v$ perpendicular to a uniform field $B$ in a closed circuit of total resistance $R$:

  • induced current $I=\dfrac{Blv}{R}$;
  • retarding magnetic force on the rod $F=\dfrac{B^2l^2v}{R}$, opposite to $\mathbf v$;
  • power that must be supplied to keep $v$ constant, equal to the Joule heating: $P=\dfrac{B^2l^2v^2}{R}=I^2R$;
  • charge that flows while the rod moves a distance $\Delta x$: $Q=\dfrac{\Delta\Phi_B}{R}=\dfrac{Bl\,\Delta x}{R}$.

If the circuit is open there is an emf but no current, no retarding force, and no energy is used.

Terminal velocity. Suppose a constant force $F$ (a steady pull, or gravity on a rod sliding down vertical rails) acts on a rod of mass $m$ starting from rest. At first the rod is slow, the braking force is small and the rod accelerates. As it speeds up the braking force grows, until it equals $F$. Then the net force is zero and the rod moves on at a constant terminal velocity, just as a raindrop falling through air does.

📐 Result 6.7 — Terminal velocity of a rod on rails

A rod pushed by a constant force $F$ along frictionless rails (the rest as in Principle 6.6) reaches the terminal velocity at which $\dfrac{B^2l^2v}{R}=F$: $$v_t=\frac{FR}{B^2l^2}.$$ For a rod of mass $m$ falling down vertical rails with the field perpendicular to the plane of the rails, $F=mg$ and $v_t=\dfrac{mgR}{B^2l^2}$. At terminal velocity all the work done by $F$ becomes heat: $Fv_t=I^2R$.

✏️ Example 4 — A rod that coasts to rest
A rod of mass $0.10$ kg lies across smooth horizontal rails $1.0$ m apart in a vertical field of $0.50$ T; the circuit resistance is $0.25\ \Omega$. The rod is given a push so that it starts moving along the rails at $4.0$ m/s, and is then left alone. Find (a) its initial deceleration, (b) the total charge that flows before it stops, (c) the distance it travels, and (d) the heat produced.
(a) Deceleration $F=\dfrac{B^2l^2v_0}{R}$ $=\dfrac{(0.25)(1.0)(4.0)}{0.25}$ $=4.0$ N, so $a=\dfrac Fm=40\ \text{m/s}^2$. The force falls as the rod slows, so the deceleration does not stay at this value.
(b) Charge In each short interval the magnetic force gives an impulse $BIl\,\Delta t=Bl\,\Delta Q$. The total impulse must remove all the momentum: $Bl\,Q=mv_0$, so $Q=\dfrac{mv_0}{Bl}$ $=\dfrac{(0.10)(4.0)}{(0.50)(1.0)}$ $=0.80$ C.
(c) Distance From Principle 6.6, $Q=\dfrac{Bl\,\Delta x}{R}$, so $\Delta x=\dfrac{QR}{Bl}$ $=\dfrac{(0.80)(0.25)}{0.50}$ $=0.40$ m.
(d) Heat All the kinetic energy becomes heat: $\tfrac12mv_0^2=\tfrac12(0.10)(4.0)^2=0.80$ J.
Check In Interactive 1 choose coasting, $B=0.50$ T and $R=0.25\ \Omega$: the rod stops after $0.40$ m and the heat bar reaches $0.80$ J.
✏️ Example 5 — Terminal velocity on vertical rails
A metal rod of mass $20$ g slides without friction down a pair of long vertical rails $0.50$ m apart, which are joined at the top through a $1.0\ \Omega$ resistor. A horizontal field of $0.80$ T is perpendicular to the plane of the rails. Find (a) the terminal velocity, (b) the current at terminal velocity, and (c) the rate at which heat is produced then. Take $g=9.8\ \text{m/s}^2$.
(a) Balance $mg=\dfrac{B^2l^2v_t}{R}$, so $v_t=\dfrac{mgR}{B^2l^2}$ $=\dfrac{(0.020)(9.8)(1.0)}{(0.80)^2(0.50)^2}$ $=\dfrac{0.196}{0.16}$ $\approx1.2$ m/s.
(b) Current At terminal velocity $BIl=mg$: $I=\dfrac{0.196}{(0.80)(0.50)}=0.49$ A. (Check: $\dfrac{Blv_t}{R}=\dfrac{(0.40)(1.225)}{1.0}=0.49$ A ✓.)
(c) Power $I^2R=(0.49)^2(1.0)\approx0.24$ W, equal to the rate of loss of gravitational potential energy $mgv_t=(0.196)(1.225)\approx0.24$ W ✓. The kinetic energy no longer changes.
Interactive 1 — Sliding Rod on Rails: Braking Force, Terminal Velocity and Energy

A rod of mass $0.10$ kg slides on smooth rails $1.0$ m apart, joined through a resistance $R$, in a field into the screen. In pulled mode a constant force $F$ starts it from rest; in coasting mode it is launched at $4.0$ m/s and left alone. The motion is found by stepping Newton's second law, $m\,dv/dt=F-B^2l^2v/R$, forward in small time steps. Watch the green pull and the red magnetic brake, the speed graph levelling off at $v_t$, and the energy bars: the work done always equals kinetic energy + heat.

0.501.000.50
time; speed $v$—
emf $Blv$; current—
magnetic brake $B^2l^2v/R$—
work in = KE + heat—

Loops Moving through a Field

A closed loop moving through a uniform field has no net emf: its two sides perpendicular to the motion have equal emfs $Blv$ that cancel round the loop (equivalently, the flux through it does not change). An emf appears only while the loop is entering or leaving a field region, when one side is in the field and the other is not. Then $\varepsilon=Blv$, with $l$ the length of the side crossing the edge of the field, and the braking force and the heat follow from Principle 6.6.

✏️ Example 6 — A loop pulled through a field region
A rectangular loop has sides $l=5.0$ cm (perpendicular to its motion) and $b=10$ cm (along its motion), and resistance $0.50\ \Omega$. It is pulled at a steady $1.0$ m/s through a region $20$ cm wide containing a uniform field of $0.60$ T perpendicular to the loop; outside the region there is no field. (a) Find the emf and current while the loop is entering the region, while it is wholly inside, and while it is leaving. (b) Find the force needed in each stage. (c) Find the total heat produced and check it against the work done.
(a) emf Entering: only the leading side is in the field: $\varepsilon=Blv$ $=(0.60)(0.050)(1.0)$ $=0.030$ V, $I=\dfrac{0.030}{0.50}=0.060$ A, for $\dfrac{b}{v}=0.10$ s. Wholly inside: the flux is constant, $\varepsilon=0$, for $\dfrac{0.20-0.10}{1.0}=0.10$ s. Leaving: $0.030$ V again, with the current in the opposite sense (the flux is now decreasing), for $0.10$ s.
(b) Force Entering or leaving: $F=BIl$ $=(0.60)(0.060)(0.050)$ $=1.8\times10^{-3}$ N, opposing the motion in both cases (Lenz). Wholly inside: no current, no force.
(c) Energy Heat $=I^2R\times(\text{time with current})$ $=(0.060)^2(0.50)(0.20)$ $=3.6\times10^{-4}$ J. Work $=F\times(\text{distance with current})$ $=(1.8\times10^{-3})(0.20)$ $=3.6\times10^{-4}$ J ✓.
Check Interactive 2 starts with this loop and field region.
Interactive 2 — A Loop Crossing a Field Region

A loop $10$ cm long and $5.0$ cm high (resistance $0.50\ \Omega$) is pulled at constant speed through a region of field $0.60$ T into the screen. The graphs, plotted against the position of the loop's leading edge, show the flux through the loop, the emf and the force the agent must apply. Make the region narrower than the loop and see what changes; double the speed and compare the heat.

201.0
leading edge at—
flux $\Phi_B$—
emf; current—
force needed; heat so far—
⚠️ Common mistakes with motional emf

1. Using the wrong field component. Only the component of $\mathbf B$ perpendicular to both the rod and its velocity contributes. A horizontal rod moving horizontally "sees" only the vertical component of the Earth's field.

2. Using the tip speed for a rotating rod. The emf is $Bl\times(\text{average speed})=\tfrac12B\omega l^2$, not $B\omega l^2$.

3. Mixing up the hand rules. Right hand for the current produced by motion (generator); left hand for the force on a current (motor).

4. Forgetting that the force depends on $v$. The brake $B^2l^2v/R$ falls as the rod slows, so the motion is not uniformly decelerated — use momentum (charge) or energy, not $v^2=u^2+2as$.

5. A loop wholly inside a uniform field. No net emf and no current, however fast it moves.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — eddy currents, and how the rod approaches $v_t$

Not in the CBSE syllabus. Eddy currents are listed by JEE Main and NEET (not JEE Advanced); the rod's approach to terminal velocity applies Faraday's and Lenz's laws, which all three list.

Eddy currents. Induced currents are not confined to wires. When a solid piece of metal moves through a magnetic field, or sits in a changing field, currents swirl round inside it in closed loops — eddy currents. By Lenz's law they oppose the motion or change that causes them. They are used for magnetic damping (the coil of a moving-coil meter is wound on a metal frame so the needle settles quickly; a metal plate swinging between magnet poles stops in a few swings), for electromagnetic brakes on some trains and roller coasters, and for heating (induction furnaces and induction cooktops). They are a nuisance in the iron cores of transformers and motors, where they waste energy as heat, so the cores are built from thin insulated sheets (laminations) that cut the eddy-current paths. Cutting slots in a swinging metal plate makes it swing much longer for the same reason. Syllabus note: eddy currents are in the NEET and JEE Main syllabi, but not in the CBSE syllabus or the JEE Advanced syllabus.

The approach to terminal velocity (uses integration). Newton's law $m\dfrac{dv}{dt}=F-\dfrac{B^2l^2}{R}v$ has the solution $$v=v_t\left(1-e^{-t/\tau}\right),\qquad \tau=\frac{mR}{B^2l^2},$$ so the rod reaches $63\%$ of $v_t$ after one time constant $\tau$. With no force, a launched rod slows as $v=v_0e^{-t/\tau}$: in theory it never quite stops, yet it travels only the finite distance $v_0\tau=\dfrac{mv_0R}{B^2l^2}$ (Example 4: $0.40$ m). Interactive 1 integrates the same equation step by step.

Exercises

I
Motional emf of a straight rod.
  • 1 Derive the expression $\varepsilon=Blv$ for the emf across a rod moving perpendicular to a uniform field, using the magnetic force on its free charges.
  • 2 A rod $25$ cm long moves at $8.0$ m/s perpendicular to its length and to a field of $0.40$ T. Find the emf across it.
  • 3 What is the emf across the rod of part 2 if it moves (a) along its own length, (b) parallel to the field?
  • 4 A rod $0.50$ m long moves at $6.0$ m/s in a field of $0.30$ T. The velocity is perpendicular to the rod but makes $30^\circ$ with the field. Find the emf.
  • 5 A car's axle is $1.6$ m long and the car travels at $72$ km/h where the vertical component of the Earth's field is $4.5\times10^{-5}$ T. Find the emf between the ends of the axle.
  • 6 An aircraft with a wingspan of $40$ m flies horizontally due north at $900$ km/h where the vertical component of the Earth's field is $5.0\times10^{-5}$ T, pointing downwards. Find the emf between the wing tips. Which wing tip is positive?
  • 7 A horizontal rod lying north–south moves due east through a field pointing vertically upwards. Use Fleming's right-hand rule to find which end of the rod is positive.
1 Each free charge $q$ moves with the rod at $v$, so it feels a magnetic force $qvB$ along the rod. The work done by this force in carrying the charge along the length $l$ is $qvBl$; the work per unit charge, the emf, is $\varepsilon=Blv$. (In equilibrium the separated charges make an electric field $E=vB$ inside the rod, and $V=El=Blv$ across the open ends.)
2 $\varepsilon=Blv=(0.40)(0.25)(8.0)=0.80$ V.
3 (a) Zero: the force $q\mathbf v\times\mathbf B$ is then perpendicular to the rod and cannot move charges along it. (b) Zero: $\mathbf v\times\mathbf B=0$.
4 $\varepsilon=Blv\sin30^\circ$ $=(0.30)(0.50)(6.0)(0.50)$ $=0.45$ V.
5 $v=20$ m/s; $\varepsilon=(4.5\times10^{-5})(1.6)(20)$ $\approx1.4\times10^{-3}$ V.
6 $v=250$ m/s; $\varepsilon=(5.0\times10^{-5})(40)(250)=0.50$ V. Velocity north, field down: $\mathbf v\times\mathbf B$ points west, so positive charges collect at the west (left) wing tip.
7 Thumb east (motion), first finger up (field): the second finger points south. The current inside the rod flows towards the south end, so the south end is positive. (Check: $\hat{\text{east}}\times\hat{\text{up}}=\hat{\text{south}}$.)
II
Rotating rods and discs.
  • 1 Show that the emf between the ends of a rod of length $l$ rotating about one end with angular speed $\omega$, in a plane perpendicular to a field $B$, is $\tfrac12B\omega l^2$.
  • 2 A rod $0.50$ m long rotates about one end at $10$ rad/s in a plane perpendicular to a field of $0.40$ T. Find the emf.
  • 3 A metal disc of radius $10$ cm spins at $3000$ rpm with its axis parallel to a field of $0.50$ T. Find the emf between its axle and its rim.
  • 4 A rod of length $l$ rotates about its midpoint in a plane perpendicular to $\mathbf B$. Find the emf between its ends, and between its midpoint and one end.
  • 5 Modelling. The metal blades of a ceiling fan are $0.60$ m long and turn at $180$ rpm in a horizontal plane. The vertical component of the Earth's field is $4.0\times10^{-5}$ T. Find the emf between the centre and the tip of a blade.
  • 6 At what angular speed must a rod $1.0$ m long rotate about one end, in a field of $0.50$ T perpendicular to its plane of rotation, to give an emf of $1.0$ V?
1 In time $dt$ the rod sweeps a sector of area $\tfrac12l^2\,\omega\,dt$, so the flux through a circuit containing it changes at the rate $B\cdot\tfrac12l^2\omega$: $\varepsilon=\tfrac12B\omega l^2$. (Or: speed grows linearly from $0$ to $\omega l$; average $\tfrac12\omega l$; $\varepsilon=Bl\cdot\tfrac12\omega l$.)
2 $\varepsilon=\tfrac12(0.40)(10)(0.50)^2=0.50$ V.
3 $\omega=\dfrac{2\pi\times3000}{60}\approx314$ rad/s; $\varepsilon=\tfrac12(0.50)(314)(0.10)^2$ $\approx0.79$ V.
4 Between the ends: zero (the two halves give equal emfs from the centre outwards). Between the midpoint and an end: $\tfrac12B\omega\left(\tfrac l2\right)^2=\tfrac18B\omega l^2$.
5 $\omega=\dfrac{2\pi\times180}{60}\approx18.8$ rad/s; $\varepsilon=\tfrac12(4.0\times10^{-5})(18.8)(0.60)^2$ $\approx1.4\times10^{-4}$ V — about $0.14$ mV, harmless.
6 $\omega=\dfrac{2\varepsilon}{Bl^2}=\dfrac{2(1.0)}{(0.50)(1.0)^2}=4.0$ rad/s.
III
Forces, power and energy (frictionless rails, field perpendicular to the plane of the rails).
  • 1 A rod slides on rails $0.20$ m apart in a field of $1.2$ T; the circuit resistance is $3.0\ \Omega$. It is pulled at a steady $2.5$ m/s. Find the emf, the current, the force needed and the power supplied. Check that the power equals $I^2R$.
  • 2 Show in general that the power supplied by the agent pulling a rod at constant speed equals the rate of Joule heating in the circuit.
  • 3 In part 1 the speed is doubled. By what factors do the current, the force and the power change?
  • 4 A rod of mass $50$ g falls between vertical rails $0.40$ m apart, joined through $0.80\ \Omega$, with a horizontal field of $0.50$ T perpendicular to the rails. Find its terminal velocity.
  • 5 A constant force of $0.30$ N pulls a rod along rails $0.50$ m apart in a field of $0.60$ T; the circuit resistance is $1.5\ \Omega$. Find the terminal velocity and the current then.
  • 6 A rod moves $0.40$ m along rails $0.25$ m apart in a field of $0.80$ T; the circuit resistance is $2.0\ \Omega$. How much charge flows? Does it depend on the speed?
  • 7 A rod given a push along frictionless rails joined through a resistor slows down and stops. Where does its kinetic energy go?
  • 8 The resistor is removed so that the rails are not joined at all. The rod is given the same push. What happens now, and why?
Check: rod on rails
emf $Blv$—
current—
force needed—
power $Fv$ = $I^2R$—
1 $\varepsilon=(1.2)(0.20)(2.5)=0.60$ V; $I=\dfrac{0.60}{3.0}=0.20$ A; $F=BIl$ $=(1.2)(0.20)(0.20)$ $=0.048$ N; $P=Fv=0.12$ W, and $I^2R=(0.20)^2(3.0)=0.12$ W ✓.
2 $I=\dfrac{Blv}{R}$; the agent's force balances $BIl$, so $P=BIlv$ $=I\cdot(Blv)=I\varepsilon=I\cdot IR=I^2R$.
3 $I\propto v$: $\times2$. $F=\dfrac{B^2l^2v}{R}\propto v$: $\times2$. $P\propto v^2$: $\times4$.
4 $v_t=\dfrac{mgR}{B^2l^2}$ $=\dfrac{(0.050)(9.8)(0.80)}{(0.50)^2(0.40)^2}$ $=\dfrac{0.392}{0.040}$ $=9.8$ m/s.
5 $v_t=\dfrac{FR}{B^2l^2}$ $=\dfrac{(0.30)(1.5)}{(0.60)^2(0.50)^2}$ $=5.0$ m/s; $I=\dfrac{F}{Bl}=\dfrac{0.30}{0.30}=1.0$ A.
6 $Q=\dfrac{Bl\,\Delta x}{R}$ $=\dfrac{(0.80)(0.25)(0.40)}{2.0}$ $=0.040$ C, whatever the speed (it depends only on the change of flux).
7 The induced current feels a magnetic force opposing the motion; the rod's kinetic energy is converted into electrical energy and then into heat in the resistance (and the rails and rod).
8 There is still an emf $Blv$ across the rod, but the circuit is open, so no current flows and there is no magnetic braking force: the rod slides on at constant speed (no friction) and no energy is converted.
IV
Loops in field regions, and harder problems.
  • 1 A square loop of side $10$ cm and resistance $0.20\ \Omega$ is pulled out of a region of uniform field $0.40$ T at a steady $2.0$ m/s, one side parallel to the edge of the region. Find the emf, the current, the force needed, the time taken and the heat produced; check the heat against the work done.
  • 2 Sketch graphs of the flux through the loop, the emf and the applied force against time as a loop crosses a field region wider than itself at constant speed.
  • 3 The loop of part 1 is pulled out at $4.0$ m/s instead. How do the heat produced and the charge that flows compare with part 1?
  • 4 A rod of mass $0.20$ kg on smooth horizontal rails $0.50$ m apart, in a vertical field of $0.40$ T, with circuit resistance $0.10\ \Omega$, is launched at $3.0$ m/s. Find the charge that flows, the distance travelled and the heat produced before it stops.
  • 5 Modelling. A simple model of a magnetic brake on a roller-coaster car treats a copper fin under the car as a rod $0.50$ m long moving through a field of $1.0$ T, in a circuit of resistance $0.010\ \Omega$. The car (mass $500$ kg) enters the brake at $20$ m/s. Find the initial braking force, the initial deceleration and the initial rate of heating. Why does such a brake never lock the wheels?
1 Only the side still in the field contributes: $\varepsilon=(0.40)(0.10)(2.0)=0.080$ V; $I=\dfrac{0.080}{0.20}=0.40$ A; $F=BIl$ $=(0.40)(0.40)(0.10)$ $=0.016$ N; time $=\dfrac{0.10}{2.0}=0.050$ s; heat $=I^2Rt$ $=(0.16)(0.20)(0.050)$ $=1.6\times10^{-3}$ J; work $=F\times0.10\ \text{m}=1.6\times10^{-3}$ J ✓.
2 Flux: rises linearly (entering), flat (wholly inside), falls linearly (leaving), zero outside. emf: a constant value while entering, zero inside, the same size with opposite sign while leaving. Force: a constant value (opposing the motion) while entering and while leaving, zero otherwise — it does not change sign, because the brake always opposes the motion.
3 Heat $=\dfrac{B^2l^2v^2}{R}\times\dfrac{b}{v}=\dfrac{B^2l^2vb}{R}\propto v$: doubled, $3.2\times10^{-3}$ J. Charge $=\dfrac{\Delta\Phi_B}{R}$ $=\dfrac{(0.40)(0.010)}{0.20}$ $=0.020$ C: the same.
4 $Q=\dfrac{mv_0}{Bl}=\dfrac{0.60}{0.20}=3.0$ C; $\Delta x=\dfrac{QR}{Bl}=\dfrac{0.30}{0.20}=1.5$ m; heat $=\tfrac12mv_0^2=0.90$ J.
5 $F=\dfrac{B^2l^2v}{R}$ $=\dfrac{(1.0)^2(0.50)^2(20)}{0.010}$ $=500$ N; $a=\dfrac{500}{500}=1.0\ \text{m/s}^2$; $P=Fv=1.0\times10^4$ W $=10$ kW. The force is proportional to speed, so it fades smoothly as the car slows and vanishes at rest — the brake can never lock or jerk (a mechanical brake is needed to hold the car still).

Original problems in the style of the CBSE board paper, JEE Main, JEE Advanced and NEET. Try each one against the clock (about 2 minutes for an MCQ, 5 minutes for a long answer) before opening the solution.

🎯 Exam corner — motional emf and energy BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks A conducting rod of length $l$ moves with velocity $v$ perpendicular to a uniform magnetic field $B$ and to its own length. Derive an expression for the emf across it, and evaluate it for $l=0.30$ m, $v=5.0$ m/s, $B=0.60$ T.

Derivation Imagine the rod closing a rectangular circuit on rails; at distance $x$ the flux is $\Phi_B=Blx$, so $|\varepsilon|=\dfrac{d\Phi_B}{dt}=Bl\dfrac{dx}{dt}=Blv$. (Or use the force $qvB$ on the free charges along length $l$.)
Answer $\varepsilon=(0.60)(0.30)(5.0)=0.90$ V. Exam tip: draw the rails and mark $x$ — the diagram earns a mark.

E2. JEE Main (Numerical value.) A metal rod $2.0$ m long rotates about one of its ends with angular speed $5.0$ rad/s in a plane perpendicular to a uniform magnetic field of $0.30$ T. Find the emf, in volts, between the ends of the rod.

Formula $\varepsilon=\tfrac12B\omega l^2$.
Answer $\varepsilon=\tfrac12(0.30)(5.0)(2.0)^2=\mathbf{3.0}$ V. Exam tip: forgetting the $\tfrac12$ gives $6.0$ — usually one of the options.

E3. NEET A conducting rod of mass $m$ and length $l$ slides down a pair of smooth vertical conducting rails joined at the top through a resistance $R$. A uniform horizontal field $B$ is perpendicular to the plane of the rails. The terminal speed of the rod is (A) $\dfrac{mgR}{Bl}$ (B) $\dfrac{mgR}{B^2l^2}$ (C) $\dfrac{B^2l^2}{mgR}$ (D) $\dfrac{mg}{B^2l^2R}$

Balance $mg=BIl=\dfrac{B^2l^2v_t}{R}$.
Answer (B). Exam tip: check by dimensions — only (B) is a speed: $\dfrac{\text{N}\,\Omega}{\text{T}^2\,\text{m}^2}$ $=\dfrac{\text{N}\,\Omega}{(\text{N/(A m)})^2\,\text{m}^2}$ $=\dfrac{\Omega\,\text{A}^2}{\text{N}}$ $=\dfrac{\text{W}}{\text{N}}$ $=\text{m/s}$.

E4. JEE Advanced (One or more correct.) A rod of mass $m$ lies across smooth horizontal rails a distance $l$ apart, joined through a resistance $R$, in a uniform vertical field $B$. The rod is given a speed $v_0$ along the rails and released. (A) The total charge that flows through $R$ is $\dfrac{mv_0}{Bl}$. (B) The total heat produced is $\tfrac12mv_0^2$. (C) The distance travelled by the rod is $\dfrac{mv_0R}{B^2l^2}$. (D) The rod comes to rest at time $t=\dfrac{mR}{B^2l^2}$.

Impulse $\sum BIl\,\Delta t=BlQ=mv_0$, so $Q=\dfrac{mv_0}{Bl}$ (A) ✓.
Distance $Q=\dfrac{Bl\,\Delta x}{R}\Rightarrow\Delta x=\dfrac{QR}{Bl}=\dfrac{mv_0R}{B^2l^2}$ (C) ✓. Energy: all the kinetic energy becomes heat (B) ✓.
Time The braking force is proportional to $v$, so $v$ decays exponentially, $v=v_0e^{-t/\tau}$ with $\tau=\dfrac{mR}{B^2l^2}$; at $t=\tau$ the speed is $v_0/e$, not zero. (D) ✗.
Answer (A), (B), (C). Exam tip: with a speed-dependent force, use momentum for charge and energy for heat; avoid the equations of uniform acceleration.

E5. NEET JEE Main A solid copper plate swinging as a pendulum between the poles of a strong magnet comes to rest after a few swings, but the same plate with several deep slots cut in it swings for much longer. This is because (A) the slots reduce the mass of the plate (B) the slots break up the paths of the induced eddy currents, reducing the damping (C) copper is magnetised by the field and the slots weaken it (D) the slots increase the air resistance.

Concept Beyond the CBSE syllabus, but in NEET and JEE Main (eddy currents — see the 🚀 box). The moving plate carries induced currents that, by Lenz's law, oppose its motion; cutting slots forces the currents into narrow, high-resistance paths, so they are much weaker.
Answer (B). Exam tip: the same idea explains laminated transformer cores.

E6. Board · Case study A rectangular loop of wire, $20$ cm long and $10$ cm wide, with resistance $0.40\ \Omega$, lies in a uniform field of $0.80$ T perpendicular to its plane. Its $10$ cm side is along the straight edge of the field region, and the loop is pulled out of the field, perpendicular to that edge, at a steady $0.50$ m/s. (a) Find the induced emf and current. (2 marks) (b) Find the force needed to pull it at constant speed. (1 mark) (c) Find the heat produced while the loop leaves the field. (1 mark)

(a) The $10$ cm side inside the field is the one that "cuts" the field: $\varepsilon=Blv$ $=(0.80)(0.10)(0.50)$ $=0.040$ V; $I=\dfrac{0.040}{0.40}=0.10$ A.
(b) $F=BIl$ $=(0.80)(0.10)(0.10)$ $=8.0\times10^{-3}$ N, in the direction of the motion (to balance the magnetic brake).
(c) The loop takes $\dfrac{0.20}{0.50}=0.40$ s to leave: heat $=I^2Rt$ $=(0.10)^2(0.40)(0.40)$ $=1.6\times10^{-3}$ J $=F\times0.20$ m ✓. Exam tip: the length that matters for $Blv$ is the side parallel to the edge; the other side sets the time.
🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes

Research question. A wave-energy buoy drives a coil up and down past a row of magnets — a linear generator. What load resistance should it be connected to, to deliver the most power?

Why it matters. Ocean waves carry enormous energy along coasts such as those of Scotland, Portugal, Australia and India's west coast, and test sites (for example the European Marine Energy Centre in Orkney) trial devices that convert the bobbing of a buoy directly into electricity with a moving coil — motional emf on a large scale. India's ocean-technology institutes have run wave-powered navigation buoys. Matching the generator to its load decides how much of the captured energy reaches the grid.

What to do.

  1. Model the generator as a "rod" of effective length $L=Nl$ (the total length of coil wire crossing the field) moving at speed $v$ through a field $B$, so that $\varepsilon=BLv$. Illustrative design values: $N=200$ turns, $l=0.30$ m, $B=0.80$ T, coil resistance $R_c=5.0\ \Omega$; the buoy's speed varies as $v=v_0\sin\omega t$ with $v_0=1.0$ m/s. Find the peak emf.
  2. With a load $R_L$ the current is $I=\dfrac{\varepsilon}{R_c+R_L}$ (treat the coil like a cell with internal resistance, Chapter 3). Show that the average power delivered to the load is $\bar P=\dfrac{\varepsilon_0^2}{2}\cdot\dfrac{R_L}{(R_c+R_L)^2}$ (the average of $\sin^2\omega t$ is $\tfrac12$).
  3. Tabulate $\bar P$ for $R_L=2,\ 5,\ 10$ and $20\ \Omega$. Which load gives the most power? Make a conjecture and test it with two more values.
  4. Find the peak magnetic force on the coil, $BLI$, for the best load. Who supplies the work done against this force, and what does the force do to the buoy's motion?

Tools: Spreadsheet or calculator; optionally a graph of $\bar P$ against $R_L$.

Hand in: The peak emf; your table and graph of $\bar P$ against $R_L$; the best load with a one-line justification; the peak force; and a paragraph on the energy chain from wave to load.

Going further. Real waves are irregular, and a stronger magnetic brake (a smaller $R_L$) changes how the buoy moves. Why might the best load for a real buoy differ from your answer? Look up "power take-off" in wave-energy engineering.

Instructor note: $\varepsilon_0=BLv_0=0.80\times60\times1.0=48$ V. $\bar P$: $R_L=2\ \Omega$: $47$ W; $5\ \Omega$: $57.6$ W; $10\ \Omega$: $51.2$ W; $20\ \Omega$: $36.9$ W. The maximum is at $R_L=R_c$ (maximum-power transfer), $\bar P_\text{max}=\varepsilon_0^2/(8R_c)$. Peak current at $R_L=5\ \Omega$ is $4.8$ A, so the peak force is $BLI=0.80\times60\times4.8\approx230$ N — supplied by the waves; the force damps the buoy's motion, which is why the best electrical load and the best mechanical behaviour interact.

🧪 Self-check quiz

0/5 correct

Q1. A rod of length $l$ moves with speed $v$; the rod, its velocity and a uniform field $B$ are mutually perpendicular. The emf across the rod is

Q2. A rod of length $l$ rotates with angular speed $\omega$ about one end in a plane perpendicular to a field $B$. The emf between its ends is

Q3. A rod is pulled at constant speed along rails joined through a resistor, in a uniform field. If the speed is doubled, the power needed becomes

Q4. The direction of the current induced in a conductor moving through a magnetic field is given by

Q5. A straight rod moves along the direction of a uniform magnetic field. The emf induced across it is

← 6.1 Magnetic Flux, Faraday's Law and Lenz's Law6.3 Self and Mutual Inductance and the AC Generator →