Magnetism and Electromagnetism — Diagnostic Tests

Unit Tests

UT-1: Magnetic Fields and Electromagnets

Question: (a) Describe the magnetic field pattern around a bar magnet. (b) Explain how to magnetise a steel nail using a solenoid. (c) An electromagnet consists of a coil of 200 turns with a current of 3 A. Calculate the magnetic field strength (in ampere-turns) and describe two ways to increase the strength. (d) Explain the difference between permanent and temporary magnets.

Solution:

(a) Magnetic field lines emerge from the north pole and curve around to enter the south pole, forming closed loops. The field is strongest near the poles (lines closest together) and weakest far away. The field direction is from north to south outside the magnet. Field lines never cross.

(b) Wrap an insulated wire around the nail many times to form a solenoid. Connect the coil to a DC power supply. The current creates a magnetic field that aligns the magnetic domains in the steel nail. After the current is removed, the domains remain aligned (steel retains magnetism). The nail is now a permanent magnet.

(c) Magnetic field strength (ampere-turns) $= NI = 200 \times 3 = 600$ At.

Two ways to increase strength: (1) Increase the current. (2) Increase the number of turns. (3) Use a soft iron core (which concentrates the magnetic field lines). (4) Reduce the air gap.

(d) Permanent magnets (e.g., steel) retain their magnetism after the magnetising field is removed because their magnetic domains remain aligned. Temporary (soft) magnets (e.g., iron) lose their magnetism when the external field is removed because their domains return to random orientations. Electromagnets use soft iron cores because they can be switched on and off.

UT-2: The Motor Effect

Question: (a) State the motor effect and Fleming’s left-hand rule. (b) A wire carrying a current of 5 A is placed in a magnetic field of flux density 0.3 T. The wire is 20 cm long and is at 90 degrees to the field. Calculate the force on the wire. (c) Explain how a DC motor works, including the role of the commutator. (d) Calculate the force if the wire is at 30 degrees to the magnetic field.

Solution:

(a) Motor effect: A current-carrying wire in a magnetic field experiences a force perpendicular to both the current direction and the field direction.

Fleming’s left-hand rule: Point your thumb (thuMb) in the direction of Motion (force), first finger (First) in the direction of the magnetic Field, and second finger (seCond) in the direction of the Current.

(b) $F = BIL = 0.3 \times 5 \times 0.2 = 0.3$ N.

(c) A DC motor has a coil of wire on an axle in a magnetic field. When current flows, the motor effect creates a force on each side of the coil (in opposite directions due to opposite current directions), producing a turning effect (torque). The commutator is a split-ring that reverses the direction of current in the coil every half-turn, ensuring the forces continue to rotate the coil in the same direction. Brushes maintain electrical contact with the spinning commutator.

(d) $F = BIL \sin \theta = 0.3 \times 5 \times 0.2 \times \sin 30^\circ = 0.3 \times 0.5 = 0.15$ N. The force is halved because only the component of the current perpendicular to the field contributes.

UT-3: Electromagnetic Induction

Question: (a) State Faraday’s law of electromagnetic induction. (b) A coil of 500 turns is placed in a magnetic field. The magnetic flux through the coil changes from 0.02 Wb to 0.08 Wb in 0.1 seconds. Calculate the average EMF induced. (c) Explain how a generator (dynamo) produces an alternating current. (d) State two factors that affect the magnitude of the induced EMF.

Solution:

(a) The induced EMF in a coil is directly proportional to the rate of change of magnetic flux linkage through the coil: $\varepsilon = -N\frac{\Delta \Phi}{\Delta t}$.

(b) $\Delta \Phi = 0.08 - 0.02 = 0.06$ Wb. $\Delta t = 0.1$ s. $\varepsilon = N\frac{\Delta \Phi}{\Delta t} = 500 \times \frac{0.06}{0.1} = 500 \times 0.6 = 300$ V.

(c) A generator has a coil that rotates in a magnetic field. As the coil rotates, the magnetic flux through it continuously changes (increasing, then decreasing, then increasing in the opposite direction). By Faraday’s law, this changing flux induces an EMF that alternates in direction, producing an alternating current. Slip rings (instead of a commutator) maintain continuous contact without reversing the current, so the output is AC.

(d) Two factors: (1) Speed of rotation — faster rotation changes the flux more rapidly, inducing larger EMF. (2) Strength of the magnetic field — stronger field produces greater flux change for the same rotation. (3) Number of turns on the coil — more turns increase the total flux linkage. (4) Area of the coil — larger area captures more flux.

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Integration Tests

IT-1: Transformers and Power Transmission (with Electricity)

Question: A power station generates electricity at 25 kV. A step-up transformer increases this to 400 kV for transmission. The transformer has 2000 turns on the primary coil. (a) Calculate the number of turns on the secondary coil. (b) The transmission line has a total resistance of 10 $\Omega$ and carries 100 A. Calculate the power loss in the transmission line. (c) If the electricity were transmitted at 25 kV with the same power, calculate the current and power loss. (d) Explain why high-voltage transmission is more efficient.

Solution:

(a) $\frac{V_s}{V_p} = \frac{N_s}{N_p}$. $\frac{400,000}{25,000} = \frac{N_s}{2000}$. $N_s = 2000 \times 16 = 32,000$ turns.

(b) Power loss $= I^2R = 100^2 \times 10 = 100,000$ W $= 100$ kW.

(c) Power transmitted $= 400,000 \times 100 = 40,000,000$ W $= 40$ MW. At 25 kV: $I = 40,000,000/25,000 = 1600$ A. Power loss $= 1600^2 \times 10 = 25,600,000$ W $= 25,600$ kW $= 25.6$ MW.

(d) Power loss in transmission lines $= I^2R$. By increasing the voltage (using a step-up transformer), the current is reduced for the same power ($P = VI$). Since power loss depends on $I^2$Reducing the current by a factor of 16 (from 1600 A to 100 A) reduces the power loss by a factor of $16^2 = 256$. This is why high-voltage transmission is far more efficient: losses at 400 kV (100 kW) are negligible compared to losses at 25 kV (25,600 kW).

IT-2: Electromagnetic Devices in Daily Life (with Energy)

Question: (a) Explain how an electric bell works using an electromagnet. (b) A relay uses an electromagnet to switch a 230 V circuit using a 12 V circuit. Explain the advantage of this arrangement. (c) An induction hob transfers energy to a metal pan using electromagnetic induction. Explain the process. (d) A metal detector generates an alternating magnetic field. Explain how it detects buried metal objects.

Solution:

(a) When the bell switch is pressed, current flows through the electromagnet coil, creating a magnetic field. The electromagnet attracts the iron armature, which pulls the hammer to strike the gong. As the armature moves, it breaks the circuit (opens a contact), the electromagnet demagnetises, and a spring returns the armature to its original position. The circuit closes again and the cycle repeats, producing a continuous ringing sound.

(b) A relay allows a low-voltage, low-current circuit (12 V, safe to touch) to control a high-voltage, high-current circuit (230 V, dangerous). This provides: (1) Safety — the user interacts only with the low-voltage circuit. (2) Isolation — the control and power circuits are electrically isolated. (3) Convenience — a small switch can control a large load (e.g., a motor or heater).

(c) The induction hob has a coil beneath the ceramic surface that carries high-frequency alternating current. This creates a rapidly changing magnetic field above the hob. When a ferromagnetic pan is placed on the hob, the changing magnetic field induces eddy currents (circulating currents) in the bottom of the pan. These eddy currents flow through the pan’s electrical resistance, generating heat (by the Joule effect, $P = I^2R$). The pan heats up directly, while the hob surface remains relatively cool.

(d) The metal detector’s coil generates an alternating magnetic field that penetrates the ground. When this field encounters a metal object, it induces eddy currents in the object. These eddy currents create their own magnetic field, which opposes the original field (Lenz’s law). The detector’s receiving coil senses this change in the total magnetic field. The change triggers an alert (audio or visual). Different metals produce different responses, allowing some detectors to discriminate between types of metal.

IT-3: Electromagnetic Induction and Lenz’s Law (with Forces)

Question: A rectangular coil of 50 turns, width 0.1 m and length 0.15 m, is pulled out of a uniform magnetic field of flux density 0.5 T at a constant speed of 2 m/s. (a) Calculate the EMF induced in the coil as it leaves the field. (b) The coil has a total resistance of 5 $\Omega$. Calculate the induced current. (c) Using Lenz’s law, determine the direction of the induced current relative to the motion. (d) Calculate the force required to maintain the constant speed, explaining why this force is needed.

Solution:

(a) As the coil leaves the field, the rate of change of area = \text{width \times \text{speed = 0.1 \times 2 = 0.2 \text{ m^2/\text{s. Rate of change of flux $= B \times \frac{\Delta A}{\Delta t} = 0.5 \times 0.2 = 0.1$ Wb/s. EMF $= N \times \frac{\Delta \Phi}{\Delta t} = 50 \times 0.1 = 5$ V.

(b) Current = \text{EMF/R = 5/5 = 1 A.

(c) By Lenz’s law, the induced current opposes the change causing it. The coil is being pulled out of the field, reducing the flux through it. The induced current creates a magnetic field that tries to maintain the flux (i.e., in the same direction as the original field). By the right-hand grip rule, this means the induced current flows clockwise (as viewed from the north pole of the magnet). This current-carrying coil in the magnetic field experiences a force that opposes the pulling motion (Fleming’s left-hand rule confirms the force is to the left, opposing the rightward pull).

(d) The induced current (1 A) in the magnetic field (0.5 T) in a conductor of length 0.15 m experiences a force: $F = BIL = 0.5 \times 1 \times 0.15 = 0.075$ N per side. For 50 turns: $F = 50 \times 0.075 = 3.75$ N (per active side).

Alternatively, from energy conservation: P = \text{EMF \times I = 5 \times 1 = 5 W. Force $= P/v = 5/2 = 2.5$ N.

This force opposes the motion (by Lenz’s law). An equal and opposite external force of 2.5 N must be applied to maintain the constant speed. The work done by this force is converted to electrical energy in the coil (which is then dissipated as heat in the resistance). This demonstrates the conservation of energy: mechanical work $\to$ electrical energy $\to$ thermal energy.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

• Confusing terminology or concepts that appear similar but have distinct meanings.
• Overlooking key assumptions or boundary conditions that limit applicability.