Induction

Magnetic Flux

Magnetic flux ($\Phi_B$) is quantity of magnetic field ($\bm{B}$) passing through a surface ($S$) perpendicular to the local area ($d\bm{A}$) of the surface:

\begin`\{aligned}` \bm{\Phi_B} = \int_S \bm{B} \cdot d\bm{A} \end`\{aligned}`

Magnetic Flux in Uniform Magnetic Field

In a uniform magnetic field, the magnetic field ($\bm{B}$) is a space invariant, where if the surface also have no curvature (such as a plane), then the magnetic flux can be expressed with the area ($\bm{A}$) of the plane:

\begin`\{aligned}` {\Phi_B} &= \bm{B} \cdot \bm{A} = |\bm{B}||\bm{A}|\cos\theta\\ \Phi_B &= BA \cos \theta \end`\{aligned}`

where $\theta$ is the angular displacement between the normal vector of the plane and the direction of magnetic field.

Inductors

A inductor is a electrical component that stores energy in a magnetic field when a current pass through.

Faraday's Law of Induction

Faraday's Electromagnetic Lab

Investigate electromagnetic induction by moving magnets through coils, changing magnetic fields, and observing the induced current and EMF in real time.

Faraday's law of induction states that the electromotive force ($\epsilon$) is equal to the negative (according to Lenz's Law) rate of change of magnetic flux with time:

\begin`\{aligned}` \epsilon = -\frac{d\Phi_B}`\{dt}` \epsilon = -\frac{\Delta \Phi_B}{\Delta t} \end`\{aligned}`

Electromotive force (EMF) had been given a misleading name, where EMF refers to the energy transfer to an electric circuit per unit charge, and is not a force. EMF of a coil can be scaled by the number of turns:

$$\epsilon = -N\frac{\Delta \Phi_B}{\Delta t}$$

info

One situation that generates EMF is a magnet passing through a solenoid, where an EMF is generated to oppose the change in flux through the coil.

Lenz's Law

Lenz's law states that the direction of the induced EMF (and hence the induced current) is such that it opposes the change in magnetic flux that produced it. This is the origin of the negative sign in Faraday's law.

The law is a consequence of conservation of energy — if the induced current reinforced the change in flux, it would create a runaway effect producing infinite energy.

Determining the Direction of Induced Current

1. Identify whether magnetic flux is increasing or decreasing.
2. If flux is increasing, the induced current creates a magnetic field to oppose the increase (opposite direction to the original field).
3. If flux is decreasing, the induced current creates a magnetic field to reinforce the original field (same direction as the original field).
4. Use the right-hand grip rule to find the current direction from the magnetic field direction.

tip

Exam Tip When answering Lenz's law questions, always state both: (a) the direction of the induced current, and (b) why it flows in that direction (to oppose the change in flux).

Motional EMF

When a straight conductor of length $l$ moves with velocity $v$ perpendicular to a uniform magnetic field $B$, an EMF is induced across its ends. This can be derived from Faraday's law.

Derivation

Consider a conducting rod of length $l$ moving at velocity $v$ perpendicular to a uniform field $B$:

• In time $\Delta t$, the rod sweeps out an area $\Delta A = l \cdot v \cdot \Delta t$
• The change in flux is $\Delta \Phi_B = B \cdot \Delta A = Blv\Delta t$
• By Faraday's law:

\begin`\{aligned}` \epsilon = -\frac{\Delta \Phi_B}{\Delta t} = -\frac{Blv\Delta t}{\Delta t} = -Blv \end`\{aligned}`

Taking the magnitude:

$$\epsilon = Blv$$

If the velocity makes an angle $\theta$ with the magnetic field:

$$\epsilon = Blv \sin \theta$$

Worked Example 1: Motional EMF

Problem: A rod of length $0.5\mathrm{ m}$ moves at $3.0\mathrm{ m s}^{-1}$ perpendicular to a magnetic field of $0.2\mathrm{ T}$. Calculate the induced EMF.

Solution:

$$\epsilon = Blv = (0.2)(0.5)(3.0) = 0.3\mathrm{ V}$$

Worked Example 2: EMF in a Rotating Coil

Problem: A rectangular coil of $200$ turns, each of area $0.01\mathrm{ m}^2$, rotates at $50\mathrm{ Hz}$ in a magnetic field of $0.5\mathrm{ T}$. Calculate the peak EMF.

Solution:

\begin`\{aligned}` \epsilon_0 &= NAB\omega = NAB \cdot 2\pi f\\ \epsilon_0 &= (200)(0.01)(0.5)(2\pi \times 50)\\ \epsilon_0 &= 314\mathrm{ V} \end`\{aligned}`

AC Generation

When a coil rotates uniformly in a uniform magnetic field, the magnetic flux through the coil varies sinusoidally, producing an alternating EMF.

Flux as a Function of Time

$$\Phi_B(t) = BA \cos(\omega t)$$

where $\omega$ is the angular velocity of rotation.

Induced EMF

\begin`\{aligned}` \epsilon &= -N\frac{d\Phi_B}`\{dt}` = -N\frac{d}`\{dt}`\left[BA\cos(\omega t)\right]\\ \epsilon &= NBA\omega \sin(\omega t) \end`\{aligned}`

The peak EMF is:

$$\epsilon_0 = NBA\omega$$

Key Features of AC

• Peak EMF ($\epsilon_0$): maximum value of the alternating EMF
• Root mean square (RMS) EMF: $\epsilon_{\mathrm{rms}} = \frac{\epsilon_0}{\sqrt{2}}$
• Frequency ($f$): number of complete cycles per second, $f = \frac{\omega}{2\pi}$
• Period ($T$): time for one complete cycle, $T = \frac{1}{f}$

tip

Exam Tip RMS values are what multimeters display. Always use RMS values when calculating average power in AC circuits: $P_{\mathrm{avg}} = \frac{\epsilon_0^2}{2R} = \frac{\epsilon_{\mathrm{rms}}^2}{R}$

Transformers

A transformer changes the voltage of an alternating current. It consists of a primary coil and a secondary coil wound around a common iron core.

Ideal Transformer Equation

For an ideal transformer (no energy losses):

$$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$

where $V_p$ and $V_s$ are the primary and secondary voltages, and $N_p$ and $N_s$ are the number of turns on the primary and secondary coils.

• Step-up transformer: $N_s \gt{} N_p$ (increases voltage)
• Step-down transformer: $N_s \lt{} N_p$ (decreases voltage)

Power in an Ideal Transformer

By conservation of energy (ideal case):

$$V_p I_p = V_s I_s$$

Therefore:

$$\frac{I_s}{I_p} = \frac{N_p}{N_s}$$

A step-up transformer increases voltage but decreases current, and vice versa.

Why Transformers Only Work with AC

Transformers require a changing magnetic flux to induce an EMF in the secondary coil. DC provides a constant current and constant flux, so no EMF is induced.

Efficiency of Real Transformers

Real transformers have losses due to:

• Eddy currents in the iron core — reduced by laminating the core
• Hysteresis losses — energy lost as the core is magnetised and demagnetised repeatedly
• Resistive heating in the coils ($I^2R$ losses)
• Flux leakage — not all magnetic flux links both coils

Efficiency:

$$\eta = \frac{V_s I_s}{V_p I_p} \times 100\%$$

Worked Example 3: Transformer

Problem: A step-down transformer has $2000$ turns on the primary and $100$ turns on the secondary. The primary voltage is $240\mathrm{ V}$ and the primary current is $2\mathrm{ A}$. Calculate the secondary voltage and current (assuming ideal).

Solution:

$$\frac{V_s}{V_p} = \frac{N_s}{N_p} \implies V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{100}{2000} = 12\mathrm{ V}$$ $$I_s = I_p \times \frac{N_p}{N_s} = 2 \times \frac{2000}{100} = 40\mathrm{ A}$$

Energy Stored in an Inductor

An inductor stores energy in its magnetic field. For an inductor with inductance $L$ carrying current $I$:

$$E = \frac{1}{2}LI^2$$

Self-Inductance

Self-inductance ($L$) is defined as:

$$\epsilon = -L\frac{\Delta I}{\Delta t}$$

where $L$ is measured in henrys (H). Self-inductance is the property of a coil that opposes changes in current through itself.

For a solenoid:

$$L = \frac{\mu_0 N^2 A}{l}$$

where $N$ is the number of turns, $A$ is the cross-sectional area, and $l$ is the length of the solenoid.

Summary Table

Quantity	Formula	Units
Magnetic flux	$\Phi_B = BA\cos\theta$	Wb (Tm$^2$)
Faraday's law	$\epsilon = -N\frac{\Delta\Phi_B}{\Delta t}$	V
Motional EMF	$\epsilon = Blv\sin\theta$	V
Peak AC EMF	$\epsilon_0 = NBA\omega$	V
Transformer ratio	$\frac{V_s}{V_p} = \frac{N_s}{N_p}$	—
Energy in inductor	$E = \frac{1}{2}LI^2$	J

tip

Exam Tip In Paper 2, induction questions often combine Faraday's law with energy conservation. Make sure you can explain why Lenz's law is necessary (conservation of energy) and derive the motional EMF expression from Faraday's law.

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Additional Worked Examples

Worked Example 4: Lenz's Law Direction Determination

Problem: A bar magnet is pushed north-pole first into a solenoid viewed from the left. Determine the direction of the induced current in the coil.

Solution:

1. The north pole approaches the left end of the solenoid, so the flux through the coil (directed to the right, towards the south pole of the magnet inside the coil) is increasing.
2. By Lenz's law, the induced current creates a magnetic field that opposes this increase. The induced field inside the coil points to the left — making the left end of the coil a north pole to repel the approaching magnet.
3. Using the right-hand grip rule (fingers curl in the direction of current, thumb points to the induced north pole), when viewed from the left end the current flows counter-clockwise.

Worked Example 5: Faraday's Law with Changing Area

Problem: A square loop of side length $0.10\mathrm{ m}$ lies perpendicular to a uniform magnetic field of $0.40\mathrm{ T}$. The loop is pulled out of the field in $0.20\mathrm{ s}$, shrinking the area inside the field from $0.010\mathrm{ m}^2$ to zero. The loop has $50$ turns. Find the average induced EMF.

Solution:

$$\Delta\Phi_B = B\,\Delta A = (0.40)(0.010 - 0) = 4.0 \times 10^{-3}\mathrm{ Wb}$$ $$\epsilon = N\frac{\Delta\Phi_B}{\Delta t} = 50 \times \frac{4.0 \times 10^{-3}}{0.20} = 1.0\mathrm{ V}$$

Note that here $B$ is constant and the flux changes because the area changes — Faraday's law applies equally to $\Delta A$ as to $\Delta B$.

Worked Example 6: EMF in a Coil Being Pulled Out of a Field

Problem: A circular coil of radius $5.0\mathrm{ cm}$ with $120$ turns is positioned so that its plane is perpendicular to a uniform field of $0.60\mathrm{ T}$. The coil is moved completely out of the field region in $0.050\mathrm{ s}$. Calculate the magnitude of the average EMF induced.

Solution:

$$A = \pi r^2 = \pi(0.050)^2 = 7.85 \times 10^{-3}\mathrm{ m}^2$$ $$\Delta\Phi_B = BA = (0.60)(7.85 \times 10^{-3}) = 4.71 \times 10^{-3}\mathrm{ Wb}$$ $$\epsilon = N\frac{\Delta\Phi_B}{\Delta t} = 120 \times \frac{4.71 \times 10^{-3}}{0.050} = 11.3\mathrm{ V}$$

Worked Example 7: Transformer Efficiency with Losses

Problem: A transformer has $N_p = 4000$ and $N_s = 200$. The input power is $1200\mathrm{ W}$ at $V_p = 240\mathrm{ V}$. The output power measured at the secondary is $1140\mathrm{ W}$. Calculate: (a) the secondary voltage, (b) the secondary current, (c) the efficiency.

Solution:

(a) $V_s = 240 \times 200/4000 = 12.0\mathrm{ V}$

(b) $I_s = 1140/12.0 = 95.0\mathrm{ A}$

(c) $\eta = 1140/1200 \times 100\% = 95.0\%$

Worked Example 8: Power Transmission — Step-Up Then Step-Down

Problem: A power station generates $500\mathrm{ kW}$ at $5000\mathrm{ V}$. The power is transmitted through cables of total resistance $10\,\Omega$.

(a) Calculate the power loss if transmitted directly at $5000\mathrm{ V}$.

(b) A step-up transformer raises the voltage to $50\,000\mathrm{ V}$ for transmission, and a step-down transformer reduces it back to $5000\mathrm{ V}$ at the consumer end. Calculate the power loss now.

Solution:

(a) Direct transmission:

$$I = \frac{P}{V} = \frac{500\,000}{5000} = 100\mathrm{ A}$$ $$P_{\mathrm{loss}} = I^2 R = (100)^2(10) = 100\,000\mathrm{ W} = 100\mathrm{ kW}$$

This is $20\%$ of the generated power wasted as heat.

(b) With step-up to $50\,000\mathrm{ V}$:

$$I' = \frac{P}{V'} = \frac{500\,000}{50\,000} = 10\mathrm{ A}, \quad P'_{\mathrm{loss}} = (10)^2(10) = 1000\mathrm{ W} = 1.0\mathrm{ kW}$$

The loss drops from $100\mathrm{ kW}$ to $1.0\mathrm{ kW}$ — a factor of 100, since $P_{\mathrm{loss}} \propto I^2$ and the current was reduced by a factor of 10.

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Eddy Currents

Definition and Mechanism

When a solid conductor is placed in a changing magnetic field, the changing flux induces circular currents within the bulk of the conductor. These are called eddy currents (or Foucault currents). They arise because the conductor itself forms closed conducting paths in which Faraday's law drives EMFs.

By Lenz's law, eddy currents produce magnetic fields that oppose the change in flux that created them. This opposition manifests as a resistive force on moving conductors or as energy dissipation in stationary conductors in oscillating fields.

Applications

Electromagnetic braking. When a conducting disc spins between the poles of a magnet, eddy currents in the disc create a magnetic field opposing the rotation. The resulting braking torque is proportional to the angular speed, so the braking force is largest at high speed and decreases smoothly to zero — providing smooth, wear-free braking. This principle is used in trains, amusement-park rides, and some exercise equipment.

Induction heating. A high-frequency alternating magnetic field induces eddy currents in a metallic object. The $I^2R$ losses within the object itself generate heat. This is the basis of induction cooktops, where the pot is heated directly while the glass surface remains cool, and of industrial induction furnaces for melting metals.

Problems

Energy loss in transformers. The iron core of a transformer is itself a conductor in a changing magnetic field. Eddy currents in the core dissipate energy as heat, reducing efficiency.

Mitigation: laminated cores. The core is constructed from thin, electrically insulated sheets (laminations) stacked together. Each lamination has a small cross-sectional area, which limits the magnitude of the eddy currents ($\epsilon \propto A$, so smaller area means smaller induced EMF per lamination). The insulation between laminations prevents currents from flowing across sheets.

Worked Example: Eddy Current Braking Force

Problem: A conducting disc of radius $R$ and thickness $t$ rotates at angular velocity $\omega$ in a uniform magnetic field $B$ perpendicular to the disc. Show that the power dissipated by eddy currents scales as $P \propto B^2\omega^2 R^4 t / \rho$, and evaluate for $R = 0.15\mathrm{ m}$, $t = 5.0\mathrm{ mm}$, $\omega = 300\mathrm{ rad s}^{-1}$, $B = 0.80\mathrm{ T}$, $\rho = 1.7 \times 10^{-8}\,\Omega\mathrm{m}$.

Solution:

Model a thin annular ring at radius $r$ of width $dr$. The EMF induced in the ring is $d\epsilon = 2\pi B\omega r^2$, and its resistance is $dR = \rho \cdot 2\pi r / (t\,dr)$. The power in the ring is:

dP = \frac{(d\epsilon)^2}`\{dR}` = \frac{2\pi B^2\omega^2 r^3\,t\,dr}{\rho}

Integrating from $r = 0$ to $R$:

$$P = \int_0^R \frac{2\pi B^2\omega^2 r^3\,t}{\rho}\,dr = \frac{\pi B^2\omega^2 t R^4}{2\rho}$$

Substituting:

$$P = \frac{\pi(0.80)^2(300)^2(5.0 \times 10^{-3})(0.15)^4}{2(1.7 \times 10^{-8})} \approx 1.4 \times 10^{7}\mathrm{ W}$$

warning

This model assumes the entire disc is immersed in the field. In practice only a localized region passes through the field gap, so actual dissipation is far lower. The key result is the scaling: $P \propto \omega^2$ and $P \propto R^4$.

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Back EMF in Motors

When the coil of a DC motor rotates in the magnetic field of the stator, it simultaneously acts as a generator. The changing flux through the rotating coil induces an EMF that opposes the supply voltage. This is called the back EMF ($\epsilon_{\mathrm{back}}$).

Net Voltage and Current

The net voltage driving current through the motor coil is:

$$V_{\mathrm{net}} = V_{\mathrm{supply}} - \epsilon_{\mathrm{back}}$$

and the current in the motor is:

$$I = \frac{V_{\mathrm{supply}} - \epsilon_{\mathrm{back}}}{R}$$

where $R$ is the total resistance of the motor windings.

Startup vs Operating Speed

• At startup ($\omega = 0$): $\epsilon_{\mathrm{back}} = 0$, so $I = V_{\mathrm{supply}}/R$. This is the maximum current — it can be very large if $R$ is small, which is why motors often use a starter resistor or electronic speed controller to limit initial current.
• At operating speed: the back EMF increases with $\omega$ (since $\epsilon_{\mathrm{back}}$ is proportional to the rate of change of flux), reducing the net voltage and hence the current. The motor reaches a steady speed when the back EMF is large enough that the current produces just enough torque to balance the load torque.

Power Delivered to the Mechanical Load

The mechanical power output of the motor is:

$$P_{\mathrm{mech}} = \epsilon_{\mathrm{back}} \cdot I$$

By energy conservation, $V_{\mathrm{supply}} \cdot I = \epsilon_{\mathrm{back}} \cdot I + I^2 R$.

Worked Example: Back EMF

Problem: A DC motor is connected to a $120\mathrm{ V}$ supply. The resistance of the armature windings is $4.0\,\Omega$. When the motor is running at full speed, the current is $5.0\mathrm{ A}$. Calculate: (a) the back EMF, (b) the mechanical power output, (c) the current at startup.

Solution:

(a) $\epsilon_{\mathrm{back}} = V_{\mathrm{supply}} - IR = 120 - (5.0)(4.0) = 100\mathrm{ V}$

(b) $P_{\mathrm{mech}} = \epsilon_{\mathrm{back}} \cdot I = (100)(5.0) = 500\mathrm{ W}$

(c) At startup, $\epsilon_{\mathrm{back}} = 0$, so $I_{\mathrm{startup}} = 120/4.0 = 30\mathrm{ A}$ — six times the operating current, confirming the need for startup protection.

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Common Pitfalls

Confusing Flux Through a Coil with Flux Density

Magnetic flux density $B$ is measured in tesla and describes field strength at a point. Magnetic flux $\Phi_B = BA\cos\theta$ is measured in weber and describes the total field through a surface. A common error is to substitute $B$ where $\Phi_B$ is required in Faraday's law, or to forget the $\cos\theta$ factor when the field is not perpendicular to the coil.

Forgetting Lenz's Law Sign (Direction of Induced Current)

Faraday's law gives the magnitude of the EMF; the negative sign indicates direction via Lenz's law. Many students calculate $\epsilon$ correctly but cannot determine which way the current flows. Always explicitly identify whether flux is increasing or decreasing, then apply the right-hand grip rule to determine the current direction.

Using DC in Transformer Calculations

Transformers require a time-varying flux. If a problem states that a transformer is connected to a DC source, no secondary voltage is produced (except for a brief transient pulse at the moment the switch is closed). Always check that the source is AC before applying the transformer equations.

Confusing Peak and RMS Values

The peak EMF $\epsilon_0$ and the RMS EMF $\epsilon_{\mathrm{rms}} = \epsilon_0/\sqrt{2}$ are different quantities. Using peak values where RMS is expected (or vice versa) leads to errors of $\sqrt{2}$ in voltage, $2$ in power, or $\sqrt{2}$ in current. Multimeters display RMS. When a problem gives "$240\mathrm{ V}$ AC" without qualification, it means RMS.

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Problem Set

Problem 1 (SL)

A rectangular coil of $100$ turns, each of dimensions $0.10\mathrm{ m} \times 0.05\mathrm{ m}$, is placed perpendicular to a uniform magnetic field of $0.30\mathrm{ T}$. The field drops to zero uniformly in $0.040\mathrm{ s}$. Calculate the average EMF induced in the coil.

Solution

$$\Delta\Phi_B = B\,\Delta A\cos\theta = (0.30)(0.10 \times 0.05)(1) - 0 = 1.5 \times 10^{-3}\mathrm{ Wb}$$$$\epsilon = N\frac{\Delta\Phi_B}{\Delta t} = 100 \times \frac{1.5 \times 10^{-3}}{0.040} = 3.75\mathrm{ V}$$

Problem 2 (SL)

A straight wire of length $0.60\mathrm{ m}$ moves at $8.0\mathrm{ m s}^{-1}$ at an angle of $30^\circ$ to a magnetic field of $0.25\mathrm{ T}$. Calculate the induced EMF.

Solution

$$\epsilon = Blv\sin\theta = (0.25)(0.60)(8.0)\sin 30^{\circ} = (0.25)(0.60)(8.0)(0.500) = 0.60\mathrm{ V}$$

Problem 3 (SL)

A step-up transformer converts $120\mathrm{ V}$ to $2400\mathrm{ V}$. The primary coil has $200$ turns. The secondary is connected to a load of $100\,\Omega$. Calculate: (a) the number of secondary turns, (b) the secondary current, (c) the primary current (ideal).

Solution

(a)

$$N_s = N_p \times \frac{V_s}{V_p} = 200 \times \frac{2400}{120} = 4000$$

(b)

$$I_s = \frac{V_s}{R_L} = \frac{2400}{100} = 24\mathrm{ A}$$

(c)

$$I_p = I_s \times \frac{N_s}{N_p} = 24 \times \frac{4000}{200} = 480\mathrm{ A}$$

Check: $V_p I_p = 120 \times 480 = 57\,600\mathrm{ W}$, $V_s I_s = 2400 \times 24 = 57\,600\mathrm{ W}$.

Problem 4 (SL)

An AC generator produces a peak EMF of $340\mathrm{ V}$ at $60\mathrm{ Hz}$. Calculate: (a) the RMS voltage, (b) the peak voltage across a $50\,\Omega$ resistor, (c) the average power dissipated.

Solution

(a) $V_{\mathrm{rms}} = 340/\sqrt{2} = 240\mathrm{ V}$

(b) Peak voltage across the resistor (negligible internal resistance): $V_0 = 340\mathrm{ V}$.

(c) $P_{\mathrm{avg}} = V_{\mathrm{rms}}^2/R = (240)^2/50 = 1152\mathrm{ W}$

Problem 5 (HL)

A solenoid of length $0.50\mathrm{ m}$, cross-sectional area $4.0 \times 10^{-3}\mathrm{ m}^2$, and $500$ turns carries a current that increases uniformly from $0$ to $3.0\mathrm{ A}$ in $0.10\mathrm{ s}$. A secondary coil of $50$ turns is wound around the centre of the solenoid. Calculate: (a) the self-inductance of the solenoid, (b) the EMF induced in the secondary coil.

Solution

(a)

$$L = \frac{\mu_0 N^2 A}{l} = \frac{(4\pi \times 10^{-7})(500)^2(4.0 \times 10^{-3})}{0.50} = 2.51 \times 10^{-3}\mathrm{ H} = 2.51\mathrm{ mH}$$

(b) The field inside the solenoid is $B = \mu_0(N/l)I$, so the flux through one secondary turn is $\Phi_B = \mu_0(N/l)IA$. The rate of change:

$$\frac{\Delta\Phi_B}{\Delta t} = \mu_0 \frac{N}{l}A \frac{\Delta I}{\Delta t} = (4\pi \times 10^{-7}) \cdot 1000 \cdot (4.0 \times 10^{-3}) \cdot 30 = 1.508 \times 10^{-4}\mathrm{ Wb s}^{-1}$$$$\epsilon_{\mathrm{secondary}} = N_s \frac{\Delta\Phi_B}{\Delta t} = 50 \times 1.508 \times 10^{-4} = 7.54\mathrm{ mV}$$

Problem 6 (HL)

A metal ring of radius $r = 0.040\mathrm{ m}$ and resistance $0.50\,\Omega$ falls vertically into a region of uniform horizontal magnetic field of $0.80\mathrm{ T}$. At the instant the ring enters the field (moving at $2.0\mathrm{ m s}^{-1}$), calculate: (a) the induced EMF, (b) the induced current, (c) the direction of the magnetic force on the ring.

Solution

(a) As the ring enters, the flux increases at rate $\Delta\Phi_B/\Delta t = B \cdot 2r \cdot v$:

$$\epsilon = B \cdot 2r \cdot v = (0.80)(0.080)(2.0) = 0.128\mathrm{ V}$$

(b)

$$I = \frac{\epsilon}{R} = \frac{0.128}{0.50} = 0.256\mathrm{ A}$$

(c) By Lenz's law, the induced current creates a field opposing the increasing downward flux, so the force on the ring is upward, opposing the fall.

Problem 7 (HL)

A power station produces $2.0\mathrm{ MW}$ at $4000\mathrm{ V}$. The transmission line has total resistance $8.0\,\Omega$. A step-up transformer increases the voltage to $80\,000\mathrm{ V}$ for transmission, and a step-down transformer at the destination returns it to $4000\mathrm{ V}$. Calculate: (a) the current in the transmission line, (b) the power loss in the line, (c) the overall efficiency of the transmission system.

Solution

(a) $I_{\mathrm{line}} = 2.0 \times 10^6 / 80\,000 = 25\mathrm{ A}$

(b) $P_{\mathrm{loss}} = (25)^2(8.0) = 5.0\mathrm{ kW}$

(c) $\eta = (2.0 \times 10^6 - 5000)/(2.0 \times 10^6) = 99.75\%$

Without transformers the current would be $500\mathrm{ A}$, giving $P_{\mathrm{loss}} = 2.0\mathrm{ MW}$ — all power would be lost.

Problem 8 (HL)

A DC motor operates from a $48\mathrm{ V}$ supply. The armature resistance is $2.0\,\Omega$. When driving a mechanical load at steady speed, the back EMF is $42\mathrm{ V}$. Calculate: (a) the current drawn, (b) the mechanical power output, (c) the power dissipated as heat, (d) the efficiency of the motor.

Solution

(a) $I = (48 - 42)/2.0 = 3.0\mathrm{ A}$

(b) $P_{\mathrm{mech}} = (42)(3.0) = 126\mathrm{ W}$

(c) $P_{\mathrm{heat}} = (3.0)^2(2.0) = 18\mathrm{ W}$ (check: $126 + 18 = 48 \times 3.0 = 144\mathrm{ W}$)

(d) $\eta = 126/144 = 87.5\%$

Problem 9 (HL)

An inductor of $0.50\mathrm{ H}$ carries a current of $4.0\mathrm{ A}$. The current is reduced to zero in $0.020\mathrm{ s}$. Calculate: (a) the energy initially stored, (b) the average EMF induced during the decay.

Solution

(a)

$$E = \frac{1}{2}LI^2 = \frac{1}{2}(0.50)(4.0)^2 = 4.0\mathrm{ J}$$

(b)

$$\epsilon = L\frac{\Delta I}{\Delta t} = (0.50)\frac{4.0 - 0}{0.020} = 100\mathrm{ V}$$

The negative sign (from Lenz's law) indicates the EMF opposes the decrease in current, but the question asks for the magnitude.

Problem 10 (SL/HL)

A coil of $150$ turns and area $2.0 \times 10^{-3}\mathrm{ m}^2$ is placed in a magnetic field that varies with time as $B(t) = 0.50\sin(120\pi t)\mathrm{ T}$. The coil is oriented with its plane perpendicular to the field. Calculate: (a) the flux as a function of time, (b) the peak EMF, (c) the RMS EMF.

Solution

(a)

$$\Phi_B(t) = NAB(t) = (150)(2.0 \times 10^{-3})(0.50\sin 120\pi t) = 0.150\sin(120\pi t)\mathrm{ Wb}$$

(b)

\epsilon(t) = -\frac{d\Phi_B}`\{dt}` = -(0.150)(120\pi)\cos(120\pi t) = -56.5\cos(120\pi t)\mathrm{ V}$$\epsilon_0 = 56.5\mathrm{ V}$$

(c)

$$\epsilon_{\mathrm{rms}} = \frac{\epsilon_0}{\sqrt{2}} = \frac{56.5}{\sqrt{2}} = 40.0\mathrm{ V}$$

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tip

Diagnostic Test Ready to test your understanding of Induction? The diagnostic test contains the hardest questions within the IB specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Induction with other physics topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.