EMI and Alternating Currents — Easy Practice (Set 1)

Quick Recap — Electromagnetic Induction

• Magnetic flux $\Phi = BA\cos\theta$ (unit weber).
• Faraday's law: induced emf $\varepsilon = -N\dfrac{d\Phi}{dt}$; Lenz's law gives its direction (it opposes the change, from energy conservation).
• Motional emf: $\varepsilon = BLv$ for a rod moving perpendicular to $B$.
• Self-inductance $L$ (unit henry): $\varepsilon = -L\dfrac{dI}{dt}$; energy stored $= \tfrac12 LI^2$.
• Eddy currents are induced in bulk conductors and are used in magnetic braking and induction heating.

Beyond-NCERT JEE Formulae

Faraday's and Lenz's laws, magnetic flux and the induced-emf basics are assumed known (they are covered in the section notes). This sheet collects the higher-yield electromagnetic-induction and AC results that JEE Main tests beyond the NCERT core.

1. Motional and rotational EMF

• Straight rod cutting field lines: $\varepsilon=Blv$ (with $B$, $l$ and $v$ mutually perpendicular; in general only the velocity component perpendicular to both the rod and $B$ contributes).
• Rod rotating about one end in a plane perpendicular to $B$ (also a disc or spoke about its centre): $\varepsilon=\tfrac12 B\omega l^2$, because the far end moves at $\omega l$ while the average speed along the rod is $\tfrac12\omega l$.
• Rod on rails of total resistance $R$: current $I=\dfrac{Blv}{R}$, retarding (Lenz) force $F=\dfrac{B^2 l^2 v}{R}$, and dissipated power $P=Fv=\dfrac{B^2 l^2 v^2}{R}$.
• Terminal speed of a rod driven by a constant force $F_0$ (or falling under gravity, $F_0=mg$): set the acceleration to zero, $v_T=\dfrac{F_0 R}{B^2 l^2}$.

When to use: any single conductor sweeping through a field — generator rods, sliding bars, rotating spokes and metal discs (the homopolar dynamo).

[JEE Tip] For a rod released on rails with speed $v_0$ and left to coast to rest, the total charge circulated is $q=\dfrac{BLx}{R}=\dfrac{mv_0}{BL}$ — independent of $R$ — and the total heat equals the whole initial kinetic energy $\tfrac12 mv_0^2$. A larger $R$ merely lets the rod slide proportionately farther.

2. Inductance, coupling and magnetic energy

• Self-inductance from flux linkage: $N\Phi=LI$, and $\varepsilon=-L\dfrac{dI}{dt}$.
• Long solenoid: $L=\mu_0 n^2 A l=\dfrac{\mu_0 N^2 A}{l}$ (with $n=N/l$). Energy stored $U=\tfrac12 LI^2$; magnetic energy density $u=\dfrac{B^2}{2\mu_0}$.
• Mutual inductance: $\varepsilon_2=-M\dfrac{dI_1}{dt}$, with $M=k\sqrt{L_1 L_2}$ and coupling $0\le k\le 1$.
• Combinations (no coupling): series inductors add, $L=L_1+L_2$; in parallel $L=\dfrac{L_1 L_2}{L_1+L_2}$.
• Two coupled coils in series: $L=L_1+L_2\pm 2M$ (plus when the fluxes aid, minus when they oppose), so $L_{aiding}-L_{opposing}=4M$.

When to use: energy-storage questions, coupled-coil and series/parallel combinations, and any "find $M$" problem.

[JEE Tip] $L$ and $M$ depend only on geometry and the core, never on the instantaneous current. Measuring a coupled pair both ways and differencing gives $M=\dfrac{L_{aiding}-L_{opposing}}{4}$ in a single step.

3. Transients: LR growth/decay and LC oscillation

• LR growth (battery switched on): $I=I_0\left(1-e^{-t/\tau}\right)$ with $I_0=\dfrac{\varepsilon}{R}$ and time constant $\tau=\dfrac{L}{R}$; at $t=\tau$ the current is $0.63 I_0$.
• LR decay (source removed): $I=I_0 e^{-t/\tau}$, falling to half in $t_{1/2}=\tau\ln 2$.
• Peak storage rate during LR growth: $\left(\dfrac{dU}{dt}\right)_{max}=\dfrac{\varepsilon^2}{4R}$, reached when $I=\tfrac12 I_0$.
• LC oscillation: $\omega=\dfrac{1}{\sqrt{LC}}$ and $f=\dfrac{1}{2\pi\sqrt{LC}}$; with $q=q_0\cos\omega t$ the current is $I=q_0\omega\sin\omega t$, so the peak current is $I_{max}=q_0\omega$.
• LC energy: $\tfrac12 LI^2+\dfrac{q^2}{2C}=\dfrac{q_0^2}{2C}=\tfrac12 LI_{max}^2$; peak capacitor voltage $V_{max}=I_{max}\sqrt{\dfrac{L}{C}}$.

When to use: switch-on/switch-off transients, "time to reach a given fraction of the final current", and energy sloshing in an LC loop.

[JEE Tip] In an LC loop the charge and current are the SHM pair (like $x$ and $v$): they are $90^\circ$ out of phase, so $\left(\dfrac{q}{q_0}\right)^2+\left(\dfrac{I}{I_{max}}\right)^2=1$. When $q=\tfrac12 q_0$ the current is $\tfrac{\sqrt3}{2}I_{max}$, not $\tfrac12 I_{max}$.

4. AC: rms, reactance, impedance and power

• RMS values for $I=I_0\sin\omega t$: $I_{rms}=\dfrac{I_0}{\sqrt2}$, the full-cycle average is zero, and the half-cycle mean is $\dfrac{2I_0}{\pi}$.
• Reactances: $X_L=\omega L$ (current lags the voltage by $90^\circ$) and $X_C=\dfrac{1}{\omega C}$ (current leads by $90^\circ$).
• Series LCR: impedance $Z=\sqrt{R^2+(X_L-X_C)^2}$, phase $\tan\phi=\dfrac{X_L-X_C}{R}$, power factor $\cos\phi=\dfrac{R}{Z}$.
• Power: average $P=V_{rms}I_{rms}\cos\phi=I_{rms}^2 R$; apparent power $V_{rms}I_{rms}$; the wattless (reactive) current $I_{rms}\sin\phi$ carries no average power.
• Non-sinusoidal source: the rms of a sum of different harmonics adds in quadrature, $V_{rms}=\sqrt{V_{1,rms}^2+V_{2,rms}^2+\cdots}$.

When to use: any driven AC circuit, heating and power questions, choke coils, and multi-harmonic (non-sinusoidal) supplies.

[JEE Tip] A pure inductor or capacitor dissipates zero average power ($\cos 90^\circ=0$), so heat appears only in $R$ and $P=I_{rms}^2 R$ holds in any circuit. RMS values at different frequencies never add arithmetically — square them, add, then take the root.

5. Series resonance, Q-factor and transformers

• Series resonance ($X_L=X_C$): $\omega_0=\dfrac{1}{\sqrt{LC}}$, so $Z=R$ is a minimum and the current $\dfrac{V_{rms}}{R}$ is a maximum.
• Quality factor: $Q=\dfrac{\omega_0 L}{R}=\dfrac{1}{\omega_0 CR}=\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$.
• Bandwidth between the half-power points: $\Delta\omega=\dfrac{R}{L}$, so $Q=\dfrac{\omega_0}{\Delta\omega}$; at a half-power point the current is $\dfrac{I_{max}}{\sqrt2}$.
• Voltage magnification: at resonance $V_L=V_C=Q\,V_{source}$, so a high-$Q$ circuit places many times the supply voltage across the L and the C.
• Transformer: ideal $\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}=\dfrac{I_p}{I_s}$ (power conserved, $V_p I_p=V_s I_s$); real efficiency $\eta=\dfrac{V_s I_s}{V_p I_p}$, and a transmission line wastes $\dfrac{P^2 R}{V^2}$, falling as $1/V^2$.

When to use: tuning circuits and radios, half-power and bandwidth questions, and transformer or power-transmission numericals.

[JEE Tip] "Maximum current" or "maximum power" is code for resonance — go straight to $X_L=X_C$. There a voltmeter across the L or the C can legitimately read $Q$ times the source voltage, because $V_L$ and $V_C$ are in antiphase and cancel at the source terminals.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Motional emf on rails: current, force, power

Q. A conducting rod of length $0.4$ m slides at $10$ m/s on smooth parallel rails closed by a $2\ \Omega$ resistor, in a uniform field of $0.5$ T perpendicular to the plane of the rails. Find the induced emf, the current, the force needed to keep the rod moving steadily, and the mechanical power supplied.

Solution. The rod sweeps field lines, so the motional emf is $$\varepsilon=Blv=(0.5)(0.4)(10)=2.$$ So $\varepsilon=2$ V and the current is $I=\dfrac{\varepsilon}{R}=\dfrac{2}{2}=1$ A. At steady speed the applied force balances the magnetic drag $F=BIl$: $$F=BIl=(0.5)(1)(0.4)=0.2.$$ So $F=0.2$ N, and the mechanical power fed in is $P=Fv=(0.2)(10)=2$ W. As a check this equals the heat dissipated, $\dfrac{\varepsilon^2}{R}=\dfrac{4}{2}=2$ W — every watt of work becomes resistive heat.

Example 2 — EMF of a rotating rod

Q. A conducting rod of length $0.5$ m rotates about one end in a plane perpendicular to a uniform field of $0.4$ T, at an angular speed of $100$ rad/s. Find the emf between its ends.

Solution. The far end moves at $\omega l$ and the pivot not at all, so the effective speed is the average $\tfrac12\omega l$; equivalently the rod sweeps area at the rate $\tfrac12\omega l^2$. The emf is $$\varepsilon=\tfrac12 B\omega l^2=\tfrac12(0.4)(100)(0.5)^2=\tfrac12(0.4)(100)(0.25).$$ So $\varepsilon=5$ V. The length enters squared, so doubling the rod would quadruple the emf. (If the rate were quoted in rev/min, convert first with $\omega=2\pi n$, where $n$ is in rev/s.)

Example 3 — LR time constant and growth

Q. A coil of inductance $0.5$ H and resistance $25\ \Omega$ is switched across a $50$ V battery of negligible internal resistance. Find the time constant, the final current, the current after one time constant, and the time to reach half the final current (take $e^{-1}=0.368$ and $\ln 2=0.693$).

Solution. The time constant and the final current are $$\tau=\dfrac{L}{R}=\dfrac{0.5}{25}=0.02,\qquad I_0=\dfrac{\varepsilon}{R}=\dfrac{50}{25}=2.$$ So $\tau=20$ ms and $I_0=2$ A. During growth $I=I_0\left(1-e^{-t/\tau}\right)$, so after one time constant $$I(\tau)=2\left(1-0.368\right)=2(0.632)=1.26.$$ So $I(\tau)=1.26$ A, which is 63% of the final value. Setting $I=\tfrac12 I_0$ gives $t=\tau\ln 2=(0.02)(0.693)=0.0139$ s, that is $13.9$ ms — shorter than one time constant, since the current is already at 63% by then.

Example 4 — Reactance, impedance and phase of a series LCR

Q. A series LCR circuit has $R=75\ \Omega$, $L=0.4$ H and $C=20\ \mu$F, and is driven by a $250$ V rms source of angular frequency $500$ rad/s. Find the two reactances, the impedance, the rms current and the phase angle.

Solution. The reactances are $$X_L=\omega L=(500)(0.4)=200,\qquad X_C=\dfrac{1}{\omega C}=\dfrac{1}{(500)(20\times10^{-6})}=100.$$ So $X_L=200\ \Omega$ and $X_C=100\ \Omega$, leaving the circuit net inductive by $100\ \Omega$. The impedance is $$Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{75^2+100^2}=\sqrt{15625}=125.$$ So $Z=125\ \Omega$ and the current is $I_{rms}=\dfrac{V_{rms}}{Z}=\dfrac{250}{125}=2$ A. The phase follows from $\cos\phi=\dfrac{R}{Z}=\dfrac{75}{125}=0.6$, giving $\phi\approx 53^\circ$ with the current lagging the applied voltage.

Example 5 — Resonant frequency and Q-factor

Q. A series LCR circuit has $L=2$ H, $C=8\ \mu$F and $R=10\ \Omega$. Find the resonant frequency, the quality factor and the bandwidth (take $\pi=3.14$).

Solution. At resonance $X_L=X_C$, so $$\omega_0=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{(2)(8\times10^{-6})}}=\dfrac{1}{\sqrt{16\times10^{-6}}}=250.$$ So $\omega_0=250$ rad/s and $f_0=\dfrac{\omega_0}{2\pi}=\dfrac{250}{6.28}=39.8$ Hz. The quality factor is $$Q=\dfrac{\omega_0 L}{R}=\dfrac{(250)(2)}{10}=50,$$ which also equals $\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$. The bandwidth is $\Delta\omega=\dfrac{R}{L}=\dfrac{10}{2}=5$ rad/s, consistent with $Q=\dfrac{\omega_0}{\Delta\omega}=\dfrac{250}{5}=50$. A large $Q$ means a sharp, narrow resonance peak.

Example 6 — Average power, power factor and wattless current

Q. A coil draws an rms current of $5$ A from a $220$ V, $50$ Hz supply and consumes an average power of $660$ W. Find the power factor, the impedance, the effective resistance and reactance, and the wattless component of the current.

Solution. The power factor comes straight from the power: $$\cos\phi=\dfrac{P}{V_{rms}I_{rms}}=\dfrac{660}{(220)(5)}=0.6.$$ The impedance is $Z=\dfrac{V_{rms}}{I_{rms}}=\dfrac{220}{5}=44\ \Omega$, and since only resistance dissipates, $R=\dfrac{P}{I_{rms}^2}=\dfrac{660}{25}=26.4\ \Omega$. The reactance follows from the impedance triangle: $$X_L=\sqrt{Z^2-R^2}=\sqrt{44^2-26.4^2}=\sqrt{1239}=35.2.$$ So $X_L=35.2\ \Omega$ and $\sin\phi=\dfrac{X_L}{Z}=0.8$. The wattless (reactive) current is $I_{rms}\sin\phi=(5)(0.8)=4$ A — it flows but carries none of the $660$ W.

Example 7 — Ideal transformer and efficiency

Q. A step-up transformer has $100$ turns in its primary and $500$ in its secondary, with $200$ V rms applied to the primary. The secondary supplies a current of $2$ A. Find the secondary voltage and the ideal primary current, and then the actual primary current if the transformer is 80% efficient.

Solution. The voltage scales with the turns ratio: $$V_s=V_p\dfrac{N_s}{N_p}=(200)\dfrac{500}{100}=1000.$$ So $V_s=1000$ V and the output power is $V_s I_s=(1000)(2)=2000$ W. For an ideal transformer power is conserved, so $$I_p=\dfrac{V_s I_s}{V_p}=\dfrac{2000}{200}=10,$$ giving $I_p=10$ A (equivalently $I_p=I_s\dfrac{N_s}{N_p}$: stepping the voltage up steps the current down). At 80% efficiency the input power is $\dfrac{2000}{0.8}=2500$ W, so the real primary current is $\dfrac{2500}{200}=12.5$ A.

Example 8 — LC oscillation: frequency, peak current and energy

Q. An ideal LC loop has $L=0.2$ H and $C=5\ \mu$F, and the capacitor is charged to $q_0=4\ \mu$C before the loop is closed. Find the angular frequency, the frequency, the peak current and the total energy (take $\pi=3.14$).

Solution. The oscillation frequency is set by $L$ and $C$ alone: $$\omega_0=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{(0.2)(5\times10^{-6})}}=\dfrac{1}{\sqrt{10^{-6}}}=1000.$$ So $\omega_0=1000$ rad/s and $f_0=\dfrac{\omega_0}{2\pi}=\dfrac{1000}{6.28}=159$ Hz. The charge plays the part of displacement, so the peak current is the analogue of $v_{max}=A\omega$: $$I_{max}=q_0\omega_0=(4\times10^{-6})(1000)=4\times10^{-3}.$$ So $I_{max}=4$ mA. The total energy, all electric at the start, is $\dfrac{q_0^2}{2C}=\dfrac{(4\times10^{-6})^2}{2(5\times10^{-6})}=1.6\times10^{-6}$ J $=1.6\ \mu$J, equal to $\tfrac12 LI_{max}^2$ at the instant the current peaks.