Ch 1 · 20 Magnetic Fields · A Level Physics 9702
Topic 1 of 2 in Magnetic Fields and Alternating Currents — A Level Physics 9702 — 94 lessons.
Introduction
Magnetic fields push on moving charges. That one fact explains electric motors, the curved tracks of particles in detectors, the Hall probe in your phone's compass, and, when the field changes, the electricity from every generator. (Syllabus topic 20: sections 20.1–20.5.)
Key ideas — 20.1 Magnetic fields
A magnetic field is a field of force produced by moving charges (currents) or by permanent magnets. It is a region where a moving charge, a current or a magnetic pole experiences a force.
🧑🏫 Teacher's explanation
Imagine you are walking through a crowd. If the crowd is dense, you feel a lot of pressure pushing you around; if the crowd is sparse, you barely notice them. A magnetic field is like that crowd. It is a region of space where a magnetic charge (a pole) or a moving electric charge (a current) feels a push or a pull. We represent this 'invisible crowd' using field lines. Where the lines are packed tightly together, the 'pressure' or field strength is high; where they are far apart, it is weak.
📖 Definitions
🔍 Worked example — Easy
A straight wire of length 0.50 m carries a current of 4.0 A. It is placed in a uniform magnetic field of 0.20 T at an angle of 90^° to the field lines. Calculate the force acting on the wire.
🔍 Worked example — Standard
A proton (charge q = 1.6 × 10^-19 C) moves at 3.0 × 10^6 m/s perpendicular to a uniform magnetic field of 0.50 T. Calculate the force on the proton.
🔍 Worked example — Exam-style
A rectangular coil of area 0.020 m^2 is placed in a magnetic field of 0.45 T. The field lines make an angle of 30^° with the plane of the coil. Calculate the magnetic flux Phi.
⚠️ Common errors
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🎯 Exam question
Key ideas — 20.2 Force on a current-carrying conductor
A conductor carrying current I, with length L in a field of flux density B, at angle θ to the field, feels a force
🧑🏫 Teacher's explanation
Imagine you are walking through a crowd. If you walk straight through, you might bump into people occasionally. But if you try to walk sideways through a line of people, you experience a much stronger 'push' against your direction of travel. In physics, a magnetic field is like that crowd. A current-carrying wire is essentially a stream of moving charges. When these charges move through a magnetic field, the field exerts a force on each individual charge. Because there are billions of charges in a wire, these tiny individual forces add up to a macroscopic force on the whole wire.
📖 Definitions
🔍 Worked example — Easy
A straight wire of length 0.50 m carries a current of 4.0 A. It is placed in a uniform magnetic field of 0.20 T at an angle of 90^° to the field lines. Calculate the force acting on the wire.
🔍 Worked example — Standard
A wire of length 2.0 m carries a current of 1.5 A and experiences a force of 0.60 N when placed in a uniform magnetic field. Calculate the magnetic flux density B if the wire is perpendicular to the field.
🔍 Worked example — Exam-style
An electron moves at 3.0 × 10^6 m· s^-1 perpendicular to a magnetic field of 0.50 T. The charge of an electron is 1.6 × 10^-19 C. Calculate the force on the electron.
⚠️ Common errors
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🎯 Exam question
Key ideas — 20.3 (a) Force on a moving charge
A charge Q moving with velocity v at angle θ to a field B feels a force
🧑🏫 Teacher's explanation
Imagine you are running through a heavy rainstorm. If you run straight, you only hit raindrops hitting your chest. If you run sideways (perpendicular) to the rain, you hit many more raindrops on your side. A moving charge in a magnetic field is similar. The magnetic field doesn't pull on the charge just because it's there; it only exerts a force if the charge is moving across the field lines. The faster you move, the more 'collisions' with the field you experience, and the stronger the force.
📖 Definitions
🔍 Worked example — Easy
An electron with charge 1.6 × 10^-19 C moves at 3.0 × 10^6 m/s perpendicular to a uniform magnetic flux density of 0.50 T. Calculate the magnitude of the force acting on the electron.
🔍 Worked example — Standard
A proton (Q = 1.6 × 10^-19 C) moves at 5.0 × 10^5 m/s at an angle of 30^° to a magnetic field of 0.80 T. Determine the magnitude of the force.
🔍 Worked example — Exam-style
A particle with charge Q moves in a uniform magnetic field B perpendicular to its velocity v. The particle follows a circular path of radius r = 0.20 m. If the magnetic field is 0.40 T and the velocity is 2.0 × 10^4 m/s, calculate the charge Q of the particle.
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🎯 Exam question
Key ideas — 20.3 (b) Circular motion of charged particles
A charged particle moving perpendicular to a uniform magnetic field feels a force of constant magnitude perpendicular to its velocity, so it moves in a circle at constant speed. The magnetic force provides the centripetal force:
🧑🏫 Teacher's explanation
Imagine you are running through a crowded hallway, and suddenly a strong wind starts blowing sideways against you. If the wind is always pushing you exactly perpendicular to the direction you are running, you won't speed up or slow down, but you will be forced to turn. This is exactly what happens to a charged particle in a magnetic field.
📖 Definitions
🔍 Worked example — Standard
A proton (charge Q = 1.6 × 10^-19 C, mass m = 1.67 × 10^-27 kg) moves at 5.0 × 10^5 m/s perpendicular to a magnetic field of 0.20 T. Calculate the radius of the orbit.
🔍 Worked example — Exam-style
An alpha particle (charge Q = 3.2 × 10^-19 C, mass m = 6.64 × 10^-27 kg) enters a magnetic field B perpendicularly. If the radius of its path is 0.050 m and its velocity is 1.5 × 10^7 m/s, calculate the magnetic flux density B.
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🎯 Exam question
Key ideas — 20.3 (c) Velocity selection
Charged particles pass through a region with an electric field and a magnetic field at right angles to each other and to the beam. The electric force QE and the magnetic force BQv act in opposite directions. Only particles with
🧑🏫 Teacher's explanation
Imagine you are walking through a heavy wind (the electric field) while someone is trying to push you sideways with a fan (the magnetic field). If you walk at just the right speed, the wind's push and the fan's push cancel each other out perfectly, and you walk in a straight line. If you walk too slow, the fan wins and pushes you off course. If you walk too fast, the wind's resistance or the fan's force changes the balance.
📖 Definitions
🔍 Worked example — Easy
A proton (Q = 1.6 × 10^-19 C) passes through perpendicular electric and magnetic fields. The electric field strength is 2.0 × 10^4 V/m and the magnetic flux density is 5.0 × 10^-3 T. Calculate the velocity required for the proton to pass undeflected.
🔍 Worked example — Standard
An electron moves through a region where E = 4.5 × 10^3 V/m and B = 1.5 × 10^-4 T. Determine the velocity of the electron if it is to pass through a narrow slit undeflected.
🔍 Worked example — Exam-style
In a velocity selector, the electric field is 8.0 × 10^4 V/m and the magnetic field is 4.0 × 10^-3 T. A particle with charge q is moving at v = 2.0 × 10^7 m/s. Determine if the particle will be deflected and, if so, by which force (electric or magnetic) if the electric force is acting downwards and the magnetic force is acting upwards.
⚠️ Common errors
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🎯 Exam question
Key ideas — 20.3 (d) The Hall effect
When a current flows through a thin slice of conductor (or semiconductor) in a magnetic field perpendicular to it, the moving charge carriers are pushed to one side by the magnetic force. Charge builds up on that side, producing a p.d. across the slice: the Hall voltage V_H. It grows until the electric force from it balances the magnetic force on each carrier.
🧑🏫 Teacher's explanation
Imagine a busy hallway where everyone is walking straight forward. Suddenly, a strong wind blows from the side. Everyone is pushed toward one wall. Because people (charges) are now crowding against that wall, there is a 'pressure' difference between that wall and the other side. In a conductor, this 'pressure' is the Hall voltage.
📖 Definitions
🔍 Worked example — Exam-style
A conductor of thickness t is placed in a magnetic field B. The current is I and the charge carrier density is n. If the thickness is halved and the current is doubled, how does the Hall voltage V_H change?
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🎯 Exam question
Key ideas — 20.4 (a) Fields due to currents
A ferrous (iron) core inside a solenoid greatly increases the field strength (the iron becomes magnetised and adds its field).
🧑🏫 Teacher's explanation
Imagine a current-carrying wire as a spinning carousel of charges. Just as the spinning carousel creates a wind that circles around it, the moving charges in a wire create a magnetic field that circles around the wire. The strength of this 'magnetic wind' depends on how much current is flowing and how close you are to the wire.
📖 Definitions
🔍 Worked example — Easy
A straight wire carries a current of 5.0 A. Describe the shape and direction of the magnetic field lines around the wire.
🔍 Worked example — Standard
A student places a long solenoid into a magnetic field. The solenoid is wound such that the field inside is uniform. If a ferrous core is inserted into the center of the solenoid, what happens to the magnetic field strength inside?
🔍 Worked example — Exam-style
Sketch the magnetic field pattern produced by a flat circular coil. Describe the field at the center of the coil.
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🎯 Exam question
(a) State the effect on the magnetic field strength inside the solenoid when a soft iron rod is inserted into the center. [1] (b) Sketch the magnetic field pattern produced by the solenoid, showing the field lines both inside and outside the coil. [3] [4 marks]
Key ideas — 20.4 (b) Forces between currents
Each current-carrying conductor sits in the magnetic field of the other, so each feels a force BIL. Using the right-hand grip rule for the field and Fleming's left-hand rule for the force:
🧑🏫 Teacher's explanation
Imagine two people walking through a crowd. If they both walk in the same direction, they might naturally cluster together. In physics, current-carrying wires act like these people. Each wire creates a magnetic field around itself. When a second wire is placed in that field, it feels a push or a pull.
📖 Definitions
🔍 Worked example — Easy
Two long, straight, parallel wires carry currents of 5.0 A and 8.0 A respectively. The wires are separated by a distance of 0.10 m. Calculate the force per unit length acting between them.
🔍 Worked example — Standard
Two parallel wires carry currents of 12 A and 15 A in the same direction. They are 0.05 m apart. Calculate the force per meter between them.
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🎯 Exam question
Key ideas — 20.5 Electromagnetic induction
Magnetic flux Phi through an area is the product of the magnetic flux density and the cross-sectional area perpendicular to the field:
🧑🏫 Teacher's explanation
Imagine you are holding a net while walking through a rainstorm. The amount of water caught in your net depends on how big the net is and how hard the rain is falling. In physics, magnetic flux Phi is like that amount of rain caught; it depends on the strength of the magnetic field B and the area A of your net. However, if you tilt the net, you catch less rain because the rain is no longer hitting the net head-on.
📖 Definitions
🔍 Worked example — Easy
A rectangular coil of dimensions 0.20 m by 0.15 m is placed in a uniform magnetic field of 0.40 T. Calculate the magnetic flux Phi through the coil if the field is perpendicular to the plane of the coil.
🔍 Worked example — Standard
A solenoid has 500 turns. The magnetic flux through each turn changes from 2.0 × 10^-3 Wb to 6.0 × 10^-3 Wb in 0.05 s. Calculate the average induced e.m.f.
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🎯 Exam question
Worked example
Example 1: force on a wire. A 0.20 m wire carrying 3.0 A lies at 30° to a field of 0.50 T. F = 0.50 × 3.0 × 0.20 × sin 30° = 0.15 N.
🔍 Worked example — Easy
A straight conductor of length 0.20 m carries a current of 3.0 A. It is placed in a uniform magnetic field of 0.50 T at an angle of 30^° to the field lines. Calculate the force acting on the wire.
🔍 Worked example — Standard
A proton (m = 1.67 × 10^-27 kg, q = 1.60 × 10^-19 C) moves at 2.0 × 10^6 m s^-1 perpendicular to a uniform magnetic field of 0.10 T. Determine the radius of the circular path taken by the proton.
🔍 Worked example — Exam-style
A semiconductor slice of thickness t = 0.10 mm has a charge carrier density n = 1.0 × 10^23 m^-3. It carries a current of 20 mA in a uniform magnetic field of 0.30 T. Calculate the Hall voltage V_H.
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Common mistakes
Practice
Answers: (1) B is the force per unit current per unit length on a wire at right angles to the field; 1 T is the flux density giving a force of 1 N on 1 m of wire carrying 1 A at right angles to the field. (2) The force BQv is constant in size and always perpendicular to the velocity, so it provides a centripetal force: BQv = mv^2/r, r = mv/BQ. (3) v = E/B = 4.0 × 10^5 m s⁻¹. (4) V_H = BI/(ntq): a semiconductor has a much smaller n, so the Hall voltage is much larger and easier to measure. (5) Faraday: induced e.m.f. equals the rate of change of magnetic flux linkage. Lenz: the induced e.m.f. acts to oppose the change producing it.