PYQ Vault

JEE Mains Physics · Formula sheet

Electromagnetic Induction formulas

10 formulas, 1 reference table and 34 common traps for JEE Mains Physics Electromagnetic Induction, grouped by subtopic.

Full notes with worked examples

Magnetic Flux, Faraday's Law and Lenz's Law

Learn this subtopic in the notes

Induced emf from a changing magnetic flux

Flux and Faraday's law

Φ=NBAcos⁡θε=−dΦdt\Phi = NBA\cos\theta \qquad \varepsilon = -\frac{d\Phi}{dt}

Charge, heat and power from an induced current

Induced charge and power

Q=N ΔΦRP=ε2RQ = \frac{N\,\Delta\Phi}{R} \qquad P = \frac{\varepsilon^{2}}{R}

Lenz's law and the direction of the induced current

SituationWhat the flux doesInduced current or effect
Field into the page, increasingFlux into the page growsAnticlockwise, so its own field points out of the page
Field into the page, decreasingFlux into the page fallsClockwise, so its own field points into the page
North pole pushed towards a loopFlux from the magnet growsNear face of the loop becomes a north pole; magnet repelled
North pole pulled away from a loopFlux from the magnet fallsNear face becomes a south pole; magnet attracted back
Bar magnet passing right through a loopRises as it enters, falls as it leavesTwo emf pulses of opposite sign, with a gap while it is inside
Coil moved through a uniform fieldUnchangedNo emf and no current
Coil rotated in a uniform fieldChanges with the angleAlternating emf
Field reversed in directionChanges by twice BAEmf while it reverses
Magnet dropped down a long copper tubeChanges in every ring of the tubeEddy currents brake it; it falls at a nearly constant speed
A non-magnetic bar of the same size falls freely and arrives first.
The induced current opposes the change in flux, never the flux itself.

Common traps

Angle measured from the plane

The cosine in NBA cos θ takes the angle between the field and the NORMAL. If a question gives the angle between the field and the plane of the loop, the cosine of that angle is the wrong factor: use its sine.

Using the solenoid's area for a loop inside it

A small loop inside a long solenoid links only the field over its own area. The flux is μ₀nI times the loop's area; the solenoid's larger cross-section does not enter.

Putting the time into the flux instead of its derivative

The emf at time t is dΦ/dt evaluated at t, not Φ(t) divided by t. Differentiate first, then substitute.

Reversing the field is not 'no change'

When a field of size B turns to the opposite direction, the flux goes from +NBA to −NBA. The change is 2NBA, so the charge is twice that of simply removing the field.

Charge does not depend on the time taken

Q = NΔΦ/R has no time in it. A time given in such a question is there only for the average emf or current, not for the charge.

Power of a sinusoidal emf uses half the square of the peak

The average of sin² over a cycle is one half, so the mean power is ε₀²/2R. Using ε₀²/R doubles the answer.

Opposing the flux instead of its change

A growing flux into the page gives an induced field out of the page, but a shrinking flux into the page gives an induced field INTO the page. The current opposes the change, so it can point the same way as the external field.

Motion alone does not induce an emf

A coil translated through a uniform, steady field, even with changing speed, keeps the same flux. Only a change of field, area, angle or direction induces an emf.

Eddy currents need a conductor

An insulator carries no eddy currents, so it feels no magnetic braking. The drag on a falling magnet comes from currents in the metal tube around it.

Motional EMF: Rods, Rails and Moving Loops

Learn this subtopic in the notes

Motional emf of a moving rod

Motional emf

ε=BlvBV=Bsin⁡δ, BH=Bcos⁡δ\varepsilon = Blv \qquad B_V = B\sin\delta,\ B_H = B\cos\delta

Force, power and terminal speed for a rod on rails

Rod on rails

I=BlvRF=B2l2vRvt=mgRB2l2I = \frac{Blv}{R} \qquad F = \frac{B^{2}l^{2}v}{R} \qquad v_t = \frac{mgR}{B^{2}l^{2}}

Loop crossing a field boundary or a non-uniform field

Loop in a non-uniform field

ε=(Bfront−Bback) lv\varepsilon = (B_{\text{front}} - B_{\text{back}})\,lv

Common traps

Using the total field instead of the cut component

Aircraft wings and a horizontal rod moving horizontally cut only the vertical component, B sin δ. A falling horizontal wire cuts only the horizontal component, B cos δ.

Sine and cosine of the dip swapped

The dip is the angle the field makes with the HORIZONTAL. So the horizontal component is B cos δ and the vertical one is B sin δ.

Leaving the speed in km/h or the field in gauss

Blv gives volts only with tesla, metres and metres per second. 180 km/h is 50 m/s, and 0.5 gauss is 5 × 10⁻⁵ T.

Using the rod's length instead of the rail gap

Only the part of the rod between the rails carries current. A rod longer than the gap still has l equal to the gap in Blv and in BIl.

Forgetting that the force goes as v

The magnetic drag B²l²v/R grows with speed. That is why a falling rod reaches a terminal speed instead of accelerating at g for ever.

Work done is not the stored energy

Pulling a loop out at constant speed stores nothing. All the work appears as heat, I²R t, in the loop's resistance.

An emf while the loop is fully inside

Inside a uniform field the flux through a moving loop is constant, so the emf is zero, however fast it moves. The emf appears only while an edge is crossing the boundary.

Adding the two sides in a non-uniform field

The front and back sides drive current in opposite senses round the loop. Their emfs subtract; the net is the field difference times lv.

Using the arc for a ring at an edge

For a ring crossing a straight boundary, the effective length is the straight chord along the boundary, not the arc of the ring inside the field.

Rotating Coils, Rods and Discs

Learn this subtopic in the notes

Emf of a coil rotating in a magnetic field

AC generator

ε=NBAωsin⁡ωtε0=NBAω\varepsilon = NBA\omega\sin\omega t \qquad \varepsilon_0 = NBA\omega

Emf of a rod or disc rotating about one end

Rotating rod or disc

ε=12Bωl2\varepsilon = \frac12 B\omega l^{2}

Common traps

Leaving rpm unconverted

NBAω needs ω in rad/s. Multiply rpm by 2π/60, and revolutions per second by 2π. Using the raw rpm gives an answer about ten times too big.

Zero flux does not mean zero emf

When the plane lies along the field, no lines pass through the coil, but the flux is changing fastest. That is the instant of PEAK emf.

Angle with the plane versus angle with the normal

With the normal at angle θ to B, the emf is ε₀ sin θ. With the plane at angle α to B, it is ε₀ cos α. Check which angle the question gives.

Forgetting the half

The pieces of a rotating rod move at speeds from zero to ωl, so the average speed is ωl/2. The emf is ½Bωl², not Bωl².

Adding the emfs of fan blades

Every blade runs from the hub to a tip, so the blades are connected in parallel. The emf between hub and tips is that of one blade, whatever the number of blades.

Total field for a horizontal fan

A ceiling fan turns in a horizontal plane, so only the vertical component B sin δ is perpendicular to that plane.

Self and Mutual Inductance, Energy and LR Circuits

Learn this subtopic in the notes

Self-inductance and the back emf

Self-inductance

ε=−LdIdtL=μ0n2Al\varepsilon = -L\frac{dI}{dt} \qquad L = \mu_0 n^{2}Al

Mutual inductance of two coils

Mutual inductance

ε2=−MdI1dtM=BcentreI asmall\varepsilon_2 = -M\frac{dI_1}{dt} \qquad M = \frac{B_{\text{centre}}}{I}\,a_{\text{small}}

Energy stored in an inductor and current growth in an LR circuit

Inductor energy and LR growth

U=12LI2I=ER(1−e−tR/L)τ=LRU = \frac12 LI^{2} \qquad I = \frac{E}{R}\left(1 - e^{-tR/L}\right) \qquad \tau = \frac{L}{R}

Common traps

Change of current across zero

A current that goes from −2 A to +2 A changes by 4 A. Taking the difference of the sizes, zero, or forgetting the sign gives the wrong rate.

Using the battery emf as the induced emf

When a switch opens, the back emf is L × (steady current)/(switching time). It is usually many times the battery's emf; the battery only fixes the current, E/R.

Thinking L depends on the current

L = NΦ/I is a ratio fixed by the coil's turns, size and core. A larger current gives a larger flux but the same L.

Working out the flux through the wrong loop

M is the same both ways, but only the big loop's field is known simply over the other loop. Pass the current through the big loop and find the flux through the small one.

Field at a square's centre

Each side is a finite wire at distance s/2, giving μ₀I(2 sin 45°)/(4π · s/2). Four sides make 2√2μ₀I/(πs). Using the formula for an infinite wire overcounts.

Sign of 2M in series

Coils wound the same way, carrying current the same way round, add 2M. Coils wound in opposite senses subtract 2M. The winding, often shown only in the figure, decides.

Time constant upside down

The time constant is L/R. A bigger inductance makes the current slower to grow; a bigger resistance makes it settle faster, to a smaller value.

Energy fraction taken as current fraction

Stored energy goes as I². Half the final energy needs I = I₀/√2, and half the final current stores only a quarter of the final energy.

Sign of dI/dt for a falling current

When the current is decreasing, L dI/dt is negative in E − L dI/dt − IR = 0, so the inductor's emf adds to the battery's and the current can exceed E/R for a moment.

Using μ₀ in a filled solenoid

With a core of relative permeability μr, the energy density is B²/(2μrμ₀). For the same B, a filled solenoid stores less energy per unit volume than an air-cored one.

More JEE Mains Physics formula sheets