PhysicsCBSE Class 12 · Moving Charges and Magnetism

Motion of a Charge in a Magnetic Field Class 12: Circle and Helix

Q: Why does a magnetic field not change the speed of a charged particle?

The magnetic force is always perpendicular to the velocity. A force perpendicular to motion does no work, so the kinetic energy, and hence the speed, stay constant; only the direction changes.

Q: Why is the period of circular motion independent of speed?

The radius r = mv/qB is proportional to v, so the circumference 2πr grows in the same ratio as the speed. The time 2πr/v then cancels v and leaves T = 2πm/qB.

Q: When does a charged particle follow a helical path?

When its velocity makes an angle other than 0°, 90° or 180° with the field. The component along B carries it forward uniformly while the perpendicular component makes it circle.

Q: Does a stationary charge feel a magnetic force?

No. The magnetic force qv × B is zero when v = 0. Only an electric field can act on a charge at rest.

Q: Is the velocity selector in the CBSE 2026-27 syllabus?

No. The syllabus includes the force on a moving charge in uniform electric and magnetic fields, but the velocity selector and the cyclotron are not to be assessed.

Updated 15 October 2026 · Fact-checked

A charge q moving with velocity v in a magnetic field B feels F = q(v × B), always perpendicular to v, so it does no work and the speed stays constant. With v ⊥ B the path is a circle of radius mv/qB and period 2πm/qB; at any other angle it is a helix.

Force F is always at right angles to velocity v, so it only turns the charge: speed stays constant and the path is a circle of radius mv/qB.

Illustration drawn for QuestionBuddy.

Understand Motion of a Charge in a Magnetic Field

An electron beam in a glass tube bends into a circle when a magnet is brought near, yet the electrons never speed up or slow down. The magnetic force changes only the direction of motion.

Lorentz force. A charge q moving with velocity $v$ through electric field $E$ and magnetic field $B$ feels $F = q (E + v \times B)$ . The magnetic part, $F_{B} = q (v \times B)$ , has magnitude $q v B sin θ$ ( $θ$ = angle between $v$ and $B$ ) and points perpendicular to both $v$ and $B$ . Key facts:

It is zero if the charge is at rest or moves parallel or antiparallel to $B$ .
It is largest ( $q v B$ ) when $v ⊥ B$ .
For a negative charge it is opposite to the direction given by $v \times B$ .
Because it is always perpendicular to $v$ , it does no work: kinetic energy and speed stay the same.
The SI unit of B, the tesla, follows: 1 T is the field in which a charge of 1 C moving at 1 m/s perpendicular to the field feels 1 N.

**Circular motion ( $v ⊥ B$ ).** The force has constant size qvB and is always perpendicular to the velocity, so it acts as the centripetal force: $q v B = \frac{m v ^{2}}{r}$ , giving $r = \frac{m v}{q B} = \frac{p}{q B}$ . The time for one revolution is $T = \frac{2 π r}{v} = \frac{2 π m}{q B}$ , and the frequency is $ν = \frac{q B}{2 π m}$ , with angular frequency $ω = \frac{q B}{m}$ . Notice that T does not depend on the speed or the radius: a faster particle moves on a bigger circle and takes the same time. Writing $p = 2 m K$ , a particle of kinetic energy K has $r = \frac{2 m K}{q B}$ ; if it was accelerated from rest through a potential difference V, $K = q V$ .

**Helical motion ( $v$ at an angle $θ$ to $B$ ).** Split $v$ into $v_{∥} = v cos θ$ along $B$ and $v_{⊥} = v sin θ$ across it. The parallel part feels no force and stays constant; the perpendicular part gives circular motion of radius $r = \frac{m v _{⊥}}{q B}$ and period $T = \frac{2 π m}{q B}$ . Together the path is a helix. The distance moved along $B$ in one revolution is the pitch: $p = v_{∥} T = \frac{2 π m v c o s θ}{q B}$ .

Electric and magnetic fields together. The electric force $q E$ acts whether or not the charge moves, and it does work (it can change the speed). The magnetic force only bends the path. In a problem with both fields, find each force as a vector and add them.

Direction rule. Use the right-hand rule for $v \times B$ : fingers along $v$ , curl them towards $B$ , and the thumb gives the force on a positive charge. Equivalently, use Fleming's left-hand rule with the current direction taken along the motion of a positive charge.

Motion of a Charge in a Magnetic Field: formulas and key facts

Lorentz force: $F = q (E + v \times B)$; Magnetic part: $F = q v B sin θ$ , perpendicular to both v and B.
Radius of circular path: $r = \frac{m v}{q B} = \frac{p}{q B} = \frac{2 m K}{q B}$; For v ⊥ B.
Period and frequency: $T = \frac{2 π m}{q B}, ν = \frac{q B}{2 π m}, ω = \frac{q B}{m}$; Independent of speed and radius.
Particle accelerated through V: $r = \frac{1}{B} \frac{2 mV}{q}$; Uses $K = q V$ .
Helix: radius and pitch: $r = \frac{m v s i n θ}{q B}, p = \frac{2 π m v c o s θ}{q B}$; θ is the angle between v and B.
Work done by magnetic force: $W = 0$; F ⊥ v, so speed and kinetic energy stay constant.

How to solve Motion of a Charge in a Magnetic Field questions

Use this method for any problem on a charge moving in a magnetic (and electric) field.

1Write q (with sign), m, v and B in SI units; if the particle was accelerated through V, first find $v = \frac{2 q V}{m}$ or use $K = q V$ .
2Find the angle θ between $v$ and $B$ . θ = 0° or 180°: straight line. θ = 90°: circle. Otherwise: helix.
3For a circle, equate $q v B = \frac{m v ^{2}}{r}$ to get r; get T from $\frac{2 π m}{q B}$ .
4For a helix, split v into $v sin θ$ (sets r) and $v cos θ$ (sets the pitch $v cos θ \times T$ ).
5For direction, use $v \times B$ (right-hand rule), and reverse it for a negative charge.
6For ratio questions, write r or T as a formula in m, q, v (or K, or V) and cancel what is common.
7If an electric field is also present, add $q E$ to $q v \times B$ as vectors.

Motion of a Charge in a Magnetic Field in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Separating ions by mass

Suppose singly charged neon ions of mass 20 u and 22 u are accelerated from rest through 1000 V and then enter a uniform magnetic field of 0.2 T at right angles. Each ion travels half a circle before hitting a detector plate. Take 1 u = $1.66 \times 1 0^{- 27}$ kg.

(a) Find the radius of the path of each ion. (b) How far apart do the two kinds of ion land on the detector? (c) Why does a heavier ion land further away?

Try it first, then show the solutionSolution

$r = \frac{1}{B} \frac{2 mV}{q}$ . For 20 u: $r = \frac{1}{0.2} \frac{2 \times 3.32 \times 1 0 ^{- 26} \times 1000}{1.6 \times 1 0 ^{- 19}} \approx 0.102$ m.
For 22 u: $r = 0.102 \times 1.1 \approx 0.107$ m.
After half a circle each lands a distance 2r from the entry point, so the separation is $2 (0.1068 - 0.1019) \approx 0.0099$ m, about 1 cm.
With the same charge and accelerating voltage, $r \propto m$ : the heavier ion has more momentum and bends less.

Answer: (a) about 10.2 cm and 10.7 cm; (b) about 1 cm apart; (c) r is proportional to √m for the same q and V.

Memory hook: Heavier ions are harder to bend: r grows as √m.

Period independent of speed

Imagine a lab demonstration in which an electron beam moving at $2 \times 1 0^{7}$ m/s enters a uniform field of 1 mT at right angles and traces a glowing circle.

(a) Find the radius of the circle. (b) Find the time for one revolution. (c) If the electron speed is doubled, what happens to the radius and to the period?

Try it first, then show the solutionSolution

$r = \frac{m v}{e B} = \frac{9.1 \times 1 0 ^{- 31} \times 2 \times 1 0 ^{7}}{1.6 \times 1 0 ^{- 19} \times 1 0 ^{- 3}} \approx 0.114$ m.
$T = \frac{2 π m}{e B} = \frac{2 π \times 9.1 \times 1 0 ^{- 31}}{1.6 \times 1 0 ^{- 22}} \approx 3.57 \times 1 0^{- 8}$ s.
Doubling v doubles r (to about 0.23 m), but T stays the same, since it does not contain v.

Answer: (a) about 11.4 cm; (b) about $3.57 \times 1 0^{- 8}$ s; (c) radius doubles, period unchanged.

Memory hook: Faster means bigger circle, same lap time.

Tricks to remember Motion of a Charge in a Magnetic Field

Short memory aids. Say each one aloud once, then use it on a question the same day.

r = mv/qB: 'my velocity over quite Big'
Momentum on top, charge × field below.
Use it when: Writing the radius quickly.
Period has no v
T = 2πm/qB: no speed, no radius in it.
Use it when: Conceptual MCQs on period and frequency.
Magnetic force: turns, never pushes
It steers the particle but never changes its speed; only E can speed it up.
Use it when: Work and kinetic energy questions.
Helix = circle + slide
Sin part makes the circle, cos part slides it along B.
Use it when: Radius and pitch of a helix.

Motion of a Charge in a Magnetic Field: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Force and radius for a proton · 2 marks · Very short answerBoard level

A proton moving at $3 \times 1 0^{6}$ m/s enters a uniform magnetic field of 0.5 T at right angles. Find (a) the magnetic force on it and (b) the radius of its path. ( $m_{p} = 1.67 \times 1 0^{- 27}$ kg)

Try it first, then show the solutionSolution

(a) $F = q v B = 1.6 \times 1 0^{- 19} \times 3 \times 1 0^{6} \times 0.5 = 2.4 \times 1 0^{- 13}$ N. (1)
(b) $r = \frac{m v}{q B} = \frac{1.67 \times 1 0 ^{- 27} \times 3 \times 1 0 ^{6}}{1.6 \times 1 0 ^{- 19} \times 0.5} \approx 0.063$ m. (1)

Answer: (a) $2.4 \times 1 0^{- 13}$ N; (b) about 6.3 cm.

Example 2: Derivation: radius and period of circular motion · 3 marks · Short answerBoard level

A particle of mass m and charge q enters a uniform magnetic field B with velocity v perpendicular to the field. Show that it moves in a circle, and derive its radius and time period. Show that the period does not depend on the speed.

Try it first, then show the solutionSolution

The force $q v \times B$ has constant size qvB and is always perpendicular to $v$ ; it changes only the direction of motion, so the particle moves in a circle with this force as the centripetal force. (1)
$q v B = \frac{m v ^{2}}{r}$ gives $r = \frac{m v}{q B}$ . (1)
$T = \frac{2 π r}{v} = \frac{2 π m}{q B}$ , which contains no v (or r): a faster particle moves on a proportionally larger circle in the same time. (1)

Answer: $r = \frac{m v}{q B}$ , $T = \frac{2 π m}{q B}$ , independent of speed.

Example 3: Electron accelerated and then bent · 3 marks · Short answerHarder board level

An electron is accelerated from rest through 100 V and enters a uniform field of $1.0 \times 1 0^{- 3}$ T at right angles. Find (a) its speed, (b) the radius of its path and (c) its frequency of revolution.

Try it first, then show the solutionSolution

(a) $\frac{1}{2} m v^{2} = e V$ : $v = \frac{2 \times 1.6 \times 1 0 ^{- 19} \times 100}{9.1 \times 1 0 ^{- 31}} \approx 5.93 \times 1 0^{6}$ m/s. (1)
(b) $r = \frac{m v}{e B} = \frac{9.1 \times 1 0 ^{- 31} \times 5.93 \times 1 0 ^{6}}{1.6 \times 1 0 ^{- 19} \times 1 0 ^{- 3}} \approx 0.034$ m. (1)
(c) $ν = \frac{e B}{2 π m} = \frac{1.6 \times 1 0 ^{- 22}}{2 π \times 9.1 \times 1 0 ^{- 31}} \approx 2.8 \times 1 0^{7}$ Hz. (1)

Answer: (a) about $5.9 \times 1 0^{6}$ m/s; (b) about 3.4 cm; (c) about $2.8 \times 1 0^{7}$ Hz.

Example 4: Helical path and pitch · 3 marks · Short answerHarder board level

An electron moving at $2 \times 1 0^{6}$ m/s enters a uniform magnetic field of 0.01 T, its velocity making 60° with the field. Find (a) the radius of the helix, (b) the time period, and (c) the pitch.

Try it first, then show the solutionSolution

$v_{⊥} = v sin 60° \approx 1.73 \times 1 0^{6}$ m/s; $v_{∥} = v cos 60° = 1 \times 1 0^{6}$ m/s.
(a) $r = \frac{m v _{⊥}}{e B} = \frac{9.1 \times 1 0 ^{- 31} \times 1.73 \times 1 0 ^{6}}{1.6 \times 1 0 ^{- 21}} \approx 9.9 \times 1 0^{- 4}$ m. (1)
(b) $T = \frac{2 π m}{e B} = \frac{2 π \times 9.1 \times 1 0 ^{- 31}}{1.6 \times 1 0 ^{- 21}} \approx 3.57 \times 1 0^{- 9}$ s. (1)
(c) Pitch $= v_{∥} T = 1 0^{6} \times 3.57 \times 1 0^{- 9} \approx 3.6 \times 1 0^{- 3}$ m. (1)

Answer: (a) about 0.99 mm; (b) about $3.57 \times 1 0^{- 9}$ s; (c) about 3.6 mm.

Example 5: Comparing proton, deuteron and alpha particle · 5 marks · Long answerHOTS / competency

A proton, a deuteron and an alpha particle enter the same uniform magnetic field at right angles.
(a) Find the ratio of the radii of their paths if they have the same kinetic energy. (2)
(b) Find the ratio if they are accelerated from rest through the same potential difference. (2)
(c) Which of them takes the longest time for one revolution? (1)
(Masses 1 : 2 : 4 in units of the proton mass; charges e, e, 2e.)

Try it first, then show the solutionSolution

(a) $r = \frac{2 m K}{q B} \propto \frac{m}{q}$ for the same K and B. (1)
$r_{p} : r_{d} : r_{α} = \frac{1}{1} : \frac{2}{1} : \frac{4}{2} = 1 : 2 : 1$ . (1)
(b) $K = q V$ , so $r = \frac{1}{B} \frac{2 mV}{q} \propto \frac{m}{q}$ . (1)
$r_{p} : r_{d} : r_{α} = 1 : 2 : \frac{4}{2} = 1 : 2 : 2$ . (1)
(c) $T = \frac{2 π m}{q B} \propto \frac{m}{q}$ : ratios 1 : 2 : 2, so the deuteron and alpha particle take equally long, twice the proton's period. (1)

Answer: (a) 1 : √2 : 1; (b) 1 : √2 : √2; (c) the deuteron and alpha particle tie, each with twice the proton's period.

Example 6: Force in combined fields · 4 marks · Case-basedHarder board level

Read the passage and answer.

A charge of $+ 1$ μC moves with velocity $v = 2 \times 1 0^{3} \hat{i}$ m/s in a region with $B = 0.5 \hat{j}$ T and $E = 500 \hat{j}$ V/m.
(a) Find the magnetic force. (1)
(b) Find the electric force. (1)
(c) Find the total force and its magnitude. (1)
(d) Which of the two forces can change the charge's kinetic energy? Why? (1)

Try it first, then show the solutionSolution

(a) $v \times B = 2 \times 1 0^{3} \times 0.5 (\hat{i} \times \hat{j}) = 1 0^{3} \hat{k}$ ; $F_{B} = 1 0^{- 6} \times 1 0^{3} \hat{k} = 1 0^{- 3} \hat{k}$ N. (1)
(b) $F_{E} = q E = 1 0^{- 6} \times 500 \hat{j} = 5 \times 1 0^{- 4} \hat{j}$ N. (1)
(c) $F = (5 \times 1 0^{- 4} \hat{j} + 1 0^{- 3} \hat{k})$ N; magnitude $(5 \times 1 0^{- 4})^{2} + (1 0^{- 3})^{2} \approx 1.12 \times 1 0^{- 3}$ N. (1)
(d) Only the electric force: the magnetic force is always perpendicular to the velocity and does no work. (1)

Answer: (a) $1 0^{- 3} \hat{k}$ N; (b) $5 \times 1 0^{- 4} \hat{j}$ N; (c) about $1.12 \times 1 0^{- 3}$ N; (d) the electric force.

Where students lose marks in Motion of a Charge in a Magnetic Field

The slips examiners see most often, and what to write instead.

Mistake: Forgetting to reverse the force direction for an electron.
Using v × B directly for a negative charge.
Do this: Find v × B, then reverse it for a negative charge.
Usually costs: 1 mark for direction.
Mistake: Using the full speed for the radius of a helix.
Ignoring the angle between v and B.
Wrong working
r = mv/(qB)
Right working
r = mv sin θ/(qB); pitch uses v cos θ
Do this: Split v into components first.
Usually costs: 1 to 2 marks.
Mistake: Saying the magnetic force increases the speed of the particle.
Thinking any force speeds a body up.
Do this: A force perpendicular to velocity changes only direction; W = 0.
Usually costs: 1 mark in reasoning questions.
Mistake: Not converting the accelerating voltage into speed or kinetic energy.
Putting V directly into r = mv/qB.
Do this: Use $\frac{1}{2} m v^{2} = q V$ first, or $r = \frac{1}{B} \frac{2 mV}{q}$ .
Usually costs: Most of the numerical.
Mistake: Writing T ∝ v or T ∝ r.
Assuming a bigger circle takes longer.
Do this: T = 2πm/(qB): the larger circle is matched by the larger speed.
Usually costs: 1 mark in MCQs.

Exam tips

Write the Lorentz force in vector form and state that the magnetic part does no work; both earn marks in 2-mark answers.
In derivations, say explicitly that qvB provides the centripetal force.
For ratio questions, write the formula with what is kept the same (v, K or V) before substituting.
Board papers have set circular and helical motion of charges as 2-, 3- and 5-mark items (for example 2023 55/1/1 Q19, 2024 55/1/1 Q24 and 2025 55/1/1 Q32).
The velocity selector and the cyclotron are not in the 2026-27 syllabus; do not spend time on them.

Practice questions on Motion of a Charge in a Magnetic Field

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The work done by the magnetic force on a charged particle moving in a uniform magnetic field is
(a) qvB (b) qvBr (c) zero (d) depends on the angle
Hint
Angle between force and velocity.
Answer
(c) zero.
Q2 · 1 mark · Very short answerBoard level
A charged particle moves parallel to a uniform magnetic field. What force acts on it and what is its path?
Hint
sin 0° = 0.
Answer
No magnetic force (F = qvB sin 0° = 0); it moves in a straight line with constant velocity.
Q3 · 2 marks · Very short answerHarder board level
What magnetic field is needed to make an electron moving at $5 \times 1 0^{6}$ m/s travel in a circle of radius 10 cm?
Hint
B = mv/(er).
Answer
$B = \frac{m v}{er} = \frac{9.1 \times 1 0 ^{- 31} \times 5 \times 1 0 ^{6}}{1.6 \times 1 0 ^{- 19} \times 0.1} \approx 2.8 \times 1 0^{- 4}$ T.
Q4 · 2 marks · Very short answerHarder board level
An electron and a proton enter the same magnetic field at right angles. Find the ratio of the radii of their paths if they have (a) the same speed, (b) the same kinetic energy. ( $m_{p} \approx 1836 m_{e}$ )
Hint
r ∝ m for same v; r ∝ √m for same K.
Answer
(a) $\frac{r _{e}}{r _{p}} = \frac{m _{e}}{m _{p}} \approx \frac{1}{1836}$ . (b) $\frac{r _{e}}{r _{p}} = \frac{m _{e}}{m _{p}} \approx \frac{1}{43}$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The kinetic energy of a charged particle moving in a uniform magnetic field does not change.
Reason (R): The magnetic force on a moving charge is always perpendicular to its velocity.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Power = F · v.
Answer
(a).
Q6 · 3 marks · Short answerHOTS / competency
A proton moving at $1 0^{6}$ m/s enters, at right angles, a region 5 cm wide containing a uniform field of 0.1 T perpendicular to its motion. Through what angle is it deflected when it leaves the region?
Hint
Find r; the angle turned satisfies sin θ = d/r when d < r.
Answer
$r = \frac{m v}{e B} = \frac{1.67 \times 1 0 ^{- 27} \times 1 0 ^{6}}{1.6 \times 1 0 ^{- 20}} \approx 0.104$ m. The proton moves on an arc of this circle; when it has gone 5 cm across the region, $sin θ = \frac{0.05}{0.104} \approx 0.479$ , so it is deflected by about 28.6°.
Q7 · 2 marks · Very short answerHOTS / competency
A charged particle moves in a circle in a uniform magnetic field. (a) If the field is doubled, what happens to the period and the radius? (b) Why can the same field be used to keep particles of different speeds in step with one another?
Hint
T = 2πm/qB, r = mv/qB.
Answer
(a) Both the period and the radius halve. (b) The period does not depend on speed, so particles of the same q/m but different speeds all complete their circles in the same time.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Expression for the Lorentz force?Show answer
$F = q (E + v \times B)$ .
2.Radius of a charge moving perpendicular to B?Show answer
$r = \frac{m v}{q B}$ .
3.Period of revolution?Show answer
$T = \frac{2 π m}{q B}$ , independent of speed.
4.Path when v makes an angle with B?Show answer
A helix.
5.Pitch of a helix?Show answer
$v cos θ \times \frac{2 π m}{q B}$ .
6.Work done by the magnetic force?Show answer
Zero.

Motion of a Charge in a Magnetic Field: frequently asked questions

Why does a magnetic field not change the speed of a charged particle?

The magnetic force is always perpendicular to the velocity. A force perpendicular to motion does no work, so the kinetic energy, and hence the speed, stay constant; only the direction changes.

Why is the period of circular motion independent of speed?

The radius r = mv/qB is proportional to v, so the circumference 2πr grows in the same ratio as the speed. The time 2πr/v then cancels v and leaves T = 2πm/qB.

When does a charged particle follow a helical path?

When its velocity makes an angle other than 0°, 90° or 180° with the field. The component along B carries it forward uniformly while the perpendicular component makes it circle.

Does a stationary charge feel a magnetic force?

No. The magnetic force qv × B is zero when v = 0. Only an electric field can act on a charge at rest.

Is the velocity selector in the CBSE 2026-27 syllabus?

No. The syllabus includes the force on a moving charge in uniform electric and magnetic fields, but the velocity selector and the cyclotron are not to be assessed.