CLASS 12 · CHAPTER 6 · ELECTROMAGNETISM

Electromagnetic Induction

Slide this rod twice as fast and the current obediently doubles. But the force shoving back against your hand only doubles too, while the power you must supply to keep it moving quadruples, Lenz's law collects its energy toll exactly, and it collects more of it the faster you go.

01 · See it

Watch it happen

Rod velocity v: 8.0 m/s

At v = 8.0 m/s, keeping the rod moving takes a push of F = 0.16 N, supplying P = 1.28 W. Double that speed to 16.0 m/s and the force only doubles, to 0.32 N, but the power you must supply quadruples, to 5.12 W: Lenz’s law collects its energy toll exactly, and it collects more of it the faster you go.

1.60 V

EMF ε

0.80 A

Current I

0.16 N

Retarding force F

1.28 W

Power P

Drag the slider to speed the rod up. Watch all four readouts together: EMF and current climb in step with v, the retarding force you’d have to push against climbs in step with v too, but power climbs much faster than either, exactly tracking v². The explanatory text below the canvas spells out the doubled-speed comparison with the actual numbers at your current setting.

02 · Derive it

Where the formula comes from

Faraday’s law says any change in the magnetic flux $\Phi$ linked with a circuit induces an EMF proportional to how fast that flux changes:

Faraday's law of induction

\varepsilon = -\dfrac{d\Phi}{dt}

A sliding rod is a special case of this. As the rod moves, the loop’s enclosed area grows, so the flux through it grows too, and carrying out $d\Phi/dt$ for this geometry collapses to a strikingly simple result:

Motional EMF

\textcolor{#e08a1e}{\varepsilon} = BLv

With the circuit closed through external resistance $R$ , Ohm’s law gives the induced current:

Induced current

I = \dfrac{\varepsilon}{R} = \dfrac{BLv}{R}

That current, flowing through the rod inside the field, feels its own force, $F = BIL$ . By Lenz’s law this force opposes the rod’s motion, so it’s exactly what you must push against to keep v constant:

Retarding force

F = BIL = \dfrac{B^2L^2v}{R}

The power you must supply to overcome it is $Fv$ , and it must equal the power dissipated in $R$ , $I^2R$ , by energy conservation. Check that they agree:

Power, two ways, must match

Fv = \dfrac{B^2L^2v^2}{R} \;=\; I^2R = \left(\dfrac{BLv}{R}\right)^2 R \;=\; \dfrac{B^2L^2v^2}{R}

Look at the exponents of $v$ in the two boxed results: force carries $v^1$ , power carries $v^2$ . That single difference is the whole chapter.

03 · Break it

Where the shortcut stops working

It’s tempting to treat induced current as something you get for free: the rod moves, a current appears, nothing seems to be spent. Lenz’s law exists specifically to kill that intuition. Every induced effect opposes the change producing it, which means the induced current always creates a mechanical force pushing back against whatever is moving the rod. You cannot induce a current without something, somewhere, doing work against that pushback.

The subtlety most students miss under exam pressure is how fast that pushback grows. Double the rod’s speed and EMF doubles ( $\varepsilon=BLv$ ), current doubles ( $I=\varepsilon/R$ ), and the retarding force also merely doubles ( $F=BIL$ , linear in $v$ throughout). It’s natural to assume the effort to maintain that speed also just doubles. It doesn’t. Power is force times velocity, $P=Fv$ , and since $F$ itself is proportional to $v$ , power carries $v\times v=v^2$ : doubling speed quadruples the power you must supply.

This is not a quirk of the formula, it’s energy conservation enforcing itself. Every joule you supply by pushing harder is dissipated as heat in $R$ , not less, not more, confirmed exactly by $Fv=I^2R$ above. A bicycle dynamo barely resists at walking pace but visibly drags at a sprint; regenerative brakes grab gently at low speed and bite hard at highway speed. Same mechanism, same $v$ versus $v^2$ split, every time.

04 · Master it

Apply it under exam conditions

Q1. A rod of length 0.5 m slides at 6 m/s along rails in a field B = 1.0 T, closed through R = 2.5 Ω. Find the EMF, current, retarding force, and power supplied.

\varepsilon = BLv = (1.0)(0.5)(6) = \textcolor{#e08a1e}{3.0\text{ V}}

I = \dfrac{\varepsilon}{R} = \dfrac{3.0}{2.5} = \textcolor{#e08a1e}{1.2\text{ A}}

F = BIL = (1.0)(1.2)(0.5) = \textcolor{#e08a1e}{0.6\text{ N}}

P = I^2R = (1.2)^2(2.5) = \textcolor{#e08a1e}{3.6\text{ W}}

Check: $Fv = (0.6)(6) = 3.6\text{ W}$ , matching P exactly, as energy conservation demands.

Q2. A different rod, length 0.25 m, slides at 10 m/s in a field B = 0.4 T, closed through R = 0.5 Ω. Find F and P, then find both again if the speed is doubled to 20 m/s, and compare how each one scaled.

At v = 10 m/s:

\varepsilon = (0.4)(0.25)(10) = 1.0\text{ V}, \quad I = \dfrac{1.0}{0.5} = 2.0\text{ A}

F = (0.4)(2.0)(0.25) = \textcolor{#e08a1e}{0.2\text{ N}}, \quad P = (2.0)^2(0.5) = \textcolor{#e08a1e}{2.0\text{ W}}

At v = 20 m/s (speed doubled):

\varepsilon = (0.4)(0.25)(20) = 2.0\text{ V}, \quad I = \dfrac{2.0}{0.5} = 4.0\text{ A}

F = (0.4)(4.0)(0.25) = \textcolor{#e08a1e}{0.4\text{ N}}, \quad P = (4.0)^2(0.5) = \textcolor{#e08a1e}{8.0\text{ W}}

F went from 0.2 N to 0.4 N, exactly doubled. P went from 2.0 W to 8.0 W, exactly quadrupled, precisely the v versus v² split the derivation predicts.

05 · FAQs

Quick answers

Why does a bicycle dynamo, or regenerative braking, feel so much harder to pedal or brake the faster you go, not just a little harder?+

Because the force resisting you grows linearly with speed, but the power you must supply grows with speed squared. Force F = B²L²v/R only doubles when v doubles, but power P = F·v doubles again on top of that, so it quadruples. At low speed the extra effort is barely noticeable; at high speed it becomes dramatic, exactly the v² term you're feeling in your legs or brake pads.

Which direction does the induced current actually flow?+

Whichever direction makes its own magnetic effects oppose the change that created it, that's Lenz's law, and it's really just energy conservation in disguise. For the sliding rod, as it moves the loop's area (and hence flux) keeps changing, so the induced current flows in the direction that creates a force on the rod opposing its motion. If the current instead flowed the other way and helped the rod along, you'd get motion that feeds its own energy with no input, which is the real reason Lenz's law can't be violated.

If I stop pushing the rod, does it just keep sliding forever at constant velocity?+

No. The moment you stop supplying the force F = B²L²v/R, that same force (now unopposed) decelerates the rod. As it slows, the induced EMF, current, and retarding force all shrink too, so the slowing itself tapers off rather than stopping abruptly, but the rod never sustains its own motion. A moving rod in a field is not a free energy source; it's a converter, and it only converts while something keeps feeding it mechanical work.

What if the rails aren't closed by a resistor, is there still an EMF?+

Yes, the EMF ε = BLv exists the instant the rod moves through the field, regardless of whether the circuit is closed. It's set entirely by B, L, and v, not by R. What changes with an open circuit is the current: with no closed path, I = ε/R becomes I = 0, so there's no retarding force and no power dissipated, exactly like a battery sitting on a shelf has an EMF across its terminals but drives no current until a circuit is connected.

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