IB Diploma Physics · unit D
Fields
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
A selection from this pathway. Open a topic below for its complete practice and challenge set.
A satellite raised to twice the orbital radius
A spherical planet has radius R = 6.0×10⁶ m and surface gravitational field g₀ = 10 N kg⁻¹. A 100 kg satellite follows a circular orbit at radius 2R measured from the planet's centre. Ignore rotation and atmosphere. Use GM = g₀R² and the supplied inverse-square field data. Supplied model: Use g = GM/r², circular acceleration v²/r and orbital period T = 2πr/v. Supplied data (Radius / R; Field magnitude / N kg⁻¹): (1, 10), (2, 2.5), (3, 1.111111), (4, 0.625). (a) Find the satellite's weight, orbital speed and orbital period. Explain what supplies the centripetal resultant. [3 marks] (b) The radius is increased to 4R. Predict ratios of field, speed and period, and explain why astronauts can feel weightless despite nonzero gravity. [3 marks]
Planetary gravitational field
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Field magnitude / N kg⁻¹
Horizontal axis (x): Radius / R
Read the exact data as a table
| Radius / R | Field magnitude / N kg⁻¹ |
|---|---|
| 1 | 10 |
| 2 | 2.5 |
| 3 | 1.111111 |
| 4 | 0.625 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find the satellite's weight, orbital speed and orbital period. Explain what supplies the centripetal resultant.
Reveal the worked solution
- At 2R the field is g₀/4 = 2.5 N kg⁻¹, so the gravitational force is mg = 250 N. The satellite's mass remains 100 kg.
- Gravity alone supplies mv²/r: v = √(gr) = √(2.5×1.2×10⁷)≈5477 m s⁻¹.
- Period is circumference/speed, T = 2πr/v≈1.38×10⁴ s. 'Centripetal force' names the inward resultant; it is not an additional interaction alongside gravity.
- For a circular orbit g∝r⁻², v∝r⁻¹/² and T∝r³/². Doubling r therefore gives the stated ratios.
- Astronauts and spacecraft accelerate together in free fall. Little support force acts on the astronauts, so their apparent weight can be near zero.
- The gravitational field is not zero in orbit, and constant speed still requires acceleration because velocity direction changes.
Answer (a) Weight = 250 N; orbital speed ≈5.48 km s⁻¹; period ≈1.38×10⁴ s. (b) Field becomes 1/4, speed 1/√2 and period 2√2 times the initial orbital values.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Finding the location where a field vanishes
Two fixed point charges lie on the x-axis: +4.0 nC at x = 0 and +1.0 nC at x = 0.30 m. Use k = 9.0×10⁹ N m² C⁻² and V = kQ/r with zero potential at infinity. The supplied graph gives the field magnitude of the +4.0 nC charge alone. Supplied model: For point charges use field magnitude E = k|Q|/r² and force F = qE, adding field vectors with directions. Supplied data (Distance from left charge / m; Field magnitude / N C⁻¹): (0.1, 3600), (0.2, 900), (0.3, 400). (a) Locate the zero-resultant-field point between the charges and find the electric potential there. [3 marks] (b) Find force on a −2.0 nC test charge at x = 0.10 m, and describe useful field-line features without treating them as literal objects. [3 marks]
Field of the left charge alone
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Field magnitude / N C⁻¹
Horizontal axis (x): Distance from left charge / m
Read the exact data as a table
| Distance from left charge / m | Field magnitude / N C⁻¹ |
|---|---|
| 0.1 | 3600 |
| 0.2 | 900 |
| 0.3 | 400 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Locate the zero-resultant-field point between the charges and find the electric potential there.
Reveal the worked solution
- Between two positive charges their fields oppose. Equality requires 4/x² = 1/(0.30−x)², giving x/(0.30−x) = 2.
- Solve x = 0.20 m, with distances 0.20 and 0.10 m. Field directions, not just magnitudes, must be checked.
- Potential adds as a scalar: V = 9.0×10⁹[(4.0×10⁻⁹)/0.20+(1.0×10⁻⁹)/0.10] = 270 V. Zero field does not require zero potential.
- At x = 0.10 m the left charge contributes +3600 N C⁻¹, while the right contributes −9.0/(0.20²) = −225 N C⁻¹.
- Net field is +3375 N C⁻¹. F = qE = (−2.0×10⁻⁹)(3375) = −6.75×10⁻⁶ N; a negative charge feels force opposite the field.
- Lines point away from positive charges, never cross and are denser where field is stronger in a consistent drawing. They represent a continuous vector field, not discrete channels carrying test charges.
Answer (a) Zero field at x = 0.20 m; potential there = +270 V. (b) Resultant field = 3375 N C⁻¹ east; force = 6.75×10⁻⁶ N west.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.