AP Physics · topic 12
Magnetic Fields and Electromagnetism
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
Mapping an electromagnet without mistaking a field line for a path
A long solenoid is tested at its centre. Current and magnetic flux-density measurements are supplied. A compass at the right-hand end points away from the coil when current is positive. For a long air-core solenoid use B = μ₀nI, with μ₀ = 4π×10⁻⁷ T m A⁻¹. Ignore Earth's field after background subtraction. Supplied data (Current / A; Flux density / mT): (-2, -2), (-1, -1), (0, 0), (1, 1), (2, 2). (a) Identify the right-end pole for positive current, estimate turns per metre, and predict the field after reversing a 1.5 A current. [3 marks] (b) Design a fair comparison of air and iron cores and explain a limitation of field-line maps. [3 marks]
Solenoid field after background subtraction
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Flux density / mT
Horizontal axis (x): Current / A
Read the exact data as a table
| Current / A | Flux density / mT |
|---|---|
| -2 | -2 |
| -1 | -1 |
| 0 | 0 |
| 1 | 1 |
| 2 | 2 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Identify the right-end pole for positive current, estimate turns per metre, and predict the field after reversing a 1.5 A current.
Reveal the worked solution
- Outside a magnet, a compass's north-seeking end follows the field direction from north to south. Pointing away from the right end identifies that end as north.
- The measured gradient is 1.0×10⁻³ T A⁻¹ = μ₀n, so n = 0.001/(4π×10⁻⁷)≈796 m⁻¹.
- Reversing current reverses field direction. At −1.5 A the linear calibration predicts −1.5 mT; its magnitude is unchanged.
- Control current rather than supply voltage, because resistance and heating may change. Keep probe orientation and centre position fixed, repeat measurements and subtract the ambient field.
- Measure a current-up and current-down sequence to detect hysteresis or remanence in the iron core; the air-core proportional model need not describe a saturating core.
- Field lines represent local directions and relative density. They are not physical wires or trajectories that isolated magnetic poles follow; a complete magnetic field map forms loops.
Answer (a) Right end is north; n≈796 turns m⁻¹; reversed field is −1.5 mT. (b) Vary core material at fixed coil geometry and current; measure signed field at fixed locations and check for residual magnetism.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A velocity selector followed by a magnetic analyser
Positive singly charged ions first cross perpendicular uniform electric and magnetic fields E = 2.0×10⁴ N C⁻¹ and Bselector = 0.10 T, oriented so their forces oppose. Undeflected ions then enter a separate perpendicular analyser field 0.20 T. Charge is e = 1.60×10⁻¹⁹ C. The supplied calibration gives analyser radius against mass for ions at the selected speed. Supplied data (Ion mass / 10⁻²⁷ kg; Orbit radius / m): (2, 0.0125), (4, 0.025), (6, 0.0375), (8, 0.05). (a) Find selected speed and the mass of an ion following a 0.0375 m analyser radius. Explain the origin of the circular path. [3 marks] (b) Predict the effect of a doubly charged ion of the same mass at the selected speed, and describe the analyser's sign discrimination. [3 marks]
Mass analyser calibration
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Orbit radius / m
Horizontal axis (x): Ion mass / 10⁻²⁷ kg
Read the exact data as a table
| Ion mass / 10⁻²⁷ kg | Orbit radius / m |
|---|---|
| 2 | 0.0125 |
| 4 | 0.025 |
| 6 | 0.0375 |
| 8 | 0.05 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find selected speed and the mass of an ion following a 0.0375 m analyser radius. Explain the origin of the circular path.
Reveal the worked solution
- In the selector, qE = qvB gives v = E/B = 2.0×10⁵ m s⁻¹; charge cancels, so this stage selects speed rather than mass.
- In the analyser qvB = mv²/r, so m = qBr/v = (1.60×10⁻¹⁹)(0.20)(0.0375)/(2.0×10⁵) = 6.0×10⁻²⁷ kg.
- The magnetic force is perpendicular to velocity, changing its direction and providing the circular resultant without doing work on the ion.
- The selected speed E/B does not depend on charge magnitude when both selector forces scale with q.
- In the analyser r = mv/(|q|B), so doubling charge halves radius at fixed mass, speed and field. The apparatus measures mass-to-charge ratio, not mass independently.
- Changing charge sign reverses the magnetic force direction. Field direction and incoming velocity must be specified to name the clockwise or counterclockwise sense; radius alone gives no sign.
Answer (a) Selected speed = 2.0×10⁵ m s⁻¹; ion mass = 6.0×10⁻²⁷ kg. (b) Its radius halves to 0.01875 m; an opposite-sign charge curves in the opposite direction.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.