Magnetic fields: A Level Physics questions with answers
Free Magnetic fields questions for A Level Physics (9702), part of Fields & motion. Each question has the answer and the marking points an examiner would look for.
Protons are accelerated from rest through a potential difference of 10 kV and then enter a uniform magnetic field of flux density 0.20 T at right angles to the field.
(a)
Calculate the speed of the protons.
[2]
(b)
Calculate the radius of the path in the field.
[2]
(c)
Explain why the protons move in a circle.
[2]
(d)
Electrons with the same speed enter the field. State two differences in their path.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
1.38 × 106 m s−1
eV = ½mv2C1 v = 1.38 × 106 m s−1A1
2
(b)
0.0722 m
Bqv = mv2/r C1 r = 0.0722 m A1
2
(c)
force ⟂ velocity, constant magnitude
the magnetic force is always perpendicular to the velocity B1 it is constant in magnitude and provides the centripetal force B1
2
(d)
curve opposite way; much smaller radius
curve in the opposite direction B1 radius much smaller (by me/mp) B1
2
Question 2Structured · 9 marksA* challenge
Electrons are accelerated from rest through a p.d. of 200 V and then enter a uniform magnetic field of flux density 0.1 T at right angles to the field.
(a)
Calculate the speed of the electrons.
[2]
(b)
Show that the path is circular and calculate its radius.
[3]
(c)
An electric field is now applied so that the electrons pass undeflected. Calculate the electric field strength and state its direction relative to the magnetic force.
[2]
(d)
The accelerating p.d. is doubled. Determine the factor by which the radius changes.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
8.38 × 106 m s−1
eV = ½mv2C1 8.38 × 106A1
2
(b)
4.77 × 10−4 m
force perpendicular to velocity, constant magnitude B1 Bqv = mv2/r C1 4.77 × 10−4A1
3
(c)
8.38 × 105 V m−1, electric force opposite to magnetic force
E = Bv C1 8.38 × 105 with direction A1
2
(d)
√2 (1.41)
r ∝ v ∝ √V C1 √2 A1
2
Question 3Structured · 10 marks
A proton moving at 6.0 × 105 m s−1 enters a uniform magnetic field of flux density 0.20 T at right angles to the field.
(a)
Define the tesla.
[2]
(b)
Explain why the proton moves in a circle.
[2]
(c)
Calculate the radius of the path. (mass of proton = 1.67 × 10−27 kg)
[3]
(d)
Show that the time for one orbit does not depend on the speed of the proton, and calculate it.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
B = F/(IL); 1 N on 1 m of wire carrying 1 A perpendicular
unit of magnetic flux density B1 a field in which a straight wire of length 1 m carrying 1 A at right angles to the field experiences a force of 1 N B1
2
(b)
force always perpendicular to velocity; provides centripetal force, speed constant
the magnetic force is always perpendicular to the velocity B1 so it does no work and provides the centripetal force (constant speed) B1
2
(c)
0.0313 m
Bqv = mv2/r C1 r = mv/(Bq) C1 0.0313 m A1
3
(d)
T = 2πm/(Bq) = 3.28 × 10−7 s
T = 2πr/v C1 = 2πm/(Bq), independent of v A1 3.28 × 10−7 s A1
3
Question 4Structured · 8 marks
Electrons are accelerated from rest through a p.d. of 500 V and enter a region of uniform magnetic field of flux density 3.0 mT at right angles to the field.
(a)
Calculate the speed of the electrons.
[2]
(b)
Use your answer to (a) to calculate the radius of their path.
[2]
(c)
Explain why the kinetic energy of the electrons is constant in the magnetic field.
[2]
(d)
State and explain the effect on the radius of doubling the accelerating p.d.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
1.33 × 107 m s−1
eV = ½mv2C1 1.33 × 107 m s−1A1
2
(b)
0.0252 m
r = mv/(Be) C1 0.0252 m A1
2
(c)
force perpendicular to velocity; no work done
the magnetic force is always perpendicular to the velocity B1 so no work is done on the electrons B1
2
(d)
× √2; v ∝ √V and r ∝ v
v ∝ √V so v increases by √2 B1 r ∝ v so the radius increases by a factor of √2 B1
2
Question 5Structured · 10 marks
A straight wire of length 8.0 cm carrying a current of 4.0 A lies at right angles to a uniform magnetic field of flux density 0.080 T.
(a)
Calculate the force on the wire.
[2]
(b)
The wire is turned until it makes 30° with the field. Calculate the new force.
[2]
(c)
Describe how a current balance could be used to measure B.
[3]
(d)
Explain why two parallel wires carrying currents in the same direction attract.
[2]
(e)
State Lenz’s law.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
0.0256 N
F = BIL C1 0.0256 N A1
2
(b)
0.0128 N
F = BIL sin θ C1 0.0128 N A1
2
(c)
wire on balance in field; measure force/mass change for known I and L; B = F/IL
place the wire on a balance (or pivoted frame) in the field B1 measure the change in reading (force) for a known current B1 B = F/(IL) from the gradient of F against I B1
3
(d)
each in field of other; force by left-hand rule towards other
each wire lies in the magnetic field of the other B1 Fleming’s left-hand rule gives a force on each towards the other B1
2
(e)
induced e.m.f. opposes the change producing it
the direction of the induced e.m.f. is such as to oppose the change that produces it B1
1
Question 6Structured · 5 marks
An electron travelling at 4.0 × 106 m s−1 enters a uniform magnetic field of flux density 0.05 T at right angles to the field.
(a)
Explain why the electron moves in a circular path.
[2]
(b)
Calculate the radius of the path.
[2]
(c)
State the effect on the radius of doubling the speed.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
force perpendicular to velocity, constant magnitude
magnetic force is always perpendicular to velocity B1 constant magnitude so provides centripetal force B1
Riyaz Shaikh has mentored IGCSE, IB and A Level students for 15+ years, one-on-one, online and in Mumbai. The hardest questions on this site are the ones he uses to find a student's weak points.
Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in