Motion of a Charged Particle in a Magnetic Field

Q: Why does the magnetic force do no work on a charged particle?

The magnetic force qv×B is always perpendicular to v, so W = F·v dt = 0. A magnetic field bends the particle's path but cannot change its speed or kinetic energy.

Q: What is the radius of circular motion of a charged particle in a magnetic field?

For v perpendicular to B, r = mv/qB = p/qB = √(2mK)/qB. Faster or heavier particles orbit larger circles; stronger fields give tighter orbits.

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

T = 2πm/(qB) depends only on m/q and B. A faster particle covers a longer circumference at proportionally higher speed, so the revolution time is constant. This 'isochronism' is why cyclotrons work.

Q: What is helical motion and when does it occur?

When v makes an angle θ with B, parallel component v cos θ carries the particle along the field lines while perpendicular v sin θ makes it circle. The path is a helix with radius mv sin θ/(qB) and pitch 2πmv cos θ/(qB).

Q: What is a velocity selector?

A velocity selector uses crossed E and B perpendicular to a charged beam. Only particles with v = E/B pass undeflected; others are pushed out. It is used in mass spectrometers to pick out ions of a single speed.

Q: What is the cyclotron frequency?

f_c = qB/(2πm), the rate at which a charged particle circles in a magnetic field. This is the frequency at which the accelerating voltage must be applied in a cyclotron.

Q: What is the maximum energy a cyclotron can give to an ion?

If the dee radius is R, K_max = q²B²R²/(2m). Larger B and R give higher energies, but relativistic mass increase limits standard cyclotrons.

Q: Why can't a cyclotron accelerate electrons?

Electrons reach relativistic speeds quickly; γm grows and T = 2πm/(qB) increases, so they fall out of sync with the fixed-frequency oscillator. Synchrocyclotrons vary frequency to compensate.

Q: Does a stationary charge experience any force in a magnetic field?

No. qv×B vanishes when v = 0. A magnetic field only exerts force on charges in motion.

PhysicsMagnetic Effects of Current and MagnetismFor JEE aspirants

A charged particle moving through a magnetic field feels the magnetic (Lorentz) force $F = q (v \times B)$ , which is always perpendicular to $v$ . Because the force is perpendicular to velocity, it does no work: the particle's speed and kinetic energy stay constant, only its direction changes. The three canonical cases for JEE Main and NEET are: (1) $v ⊥ B$ giving pure circular motion of radius $r = m v / q B$ ; (2) $v$ at an angle $θ$ to $B$ giving helical motion; (3) $v$ in crossed $E$ and $B$ fields, the basis of the velocity selector and the cyclotron. All are tested every year.

Key Formulas - Quick Reference

Lorentz force (combined $E$ and $B$ ): $F = q E + q (v \times B)$
Magnetic force alone: $F = q (v \times B)$ , magnitude $F = q v B sin θ$
Radius of circular motion ( $v ⊥ B$ ): $r = \frac{m v}{q B} = \frac{p}{q B} = \frac{2 m K}{q B}$
Time period: $T = \frac{2 π m}{q B}$ (independent of speed)
Cyclotron frequency: $f = \frac{q B}{2 π m}$ ; angular frequency $ω = \frac{q B}{m}$
Helical motion pitch: $p = v_{∥} T = \frac{2 π m v cos θ}{q B}$
Velocity selector (crossed $E, B$ ): selected speed $v = \frac{E}{B}$
Cyclotron maximum kinetic energy: $K_{m a x} = \frac{q ^{2} B ^{2} R ^{2}}{2 m}$

Colours in figures: magnetic field

B

velocity

v

force

F

electric field

E

1The Lorentz Force

The total electromagnetic force on a charge $q$ moving with velocity $v$ in electric field $E$ and magnetic field $B$ is the Lorentz force:

F = q E + q (v \times B)

The magnetic part $F_{m} = q (v \times B)$ has these key properties:

Perpendicular to velocity: $F_{m} ⊥ v$ always, so it changes direction, not speed.
Perpendicular to field: $F_{m} ⊥ B$ always, so motion parallel to $B$ is unaffected.
Does no work: $P = F_{m} \cdot v = 0$ , so kinetic energy is conserved.
Vanishes when $v ∥ B$ or when $v = 0$ : stationary charges and charges moving along the field feel no magnetic force.
Maximum when $v ⊥ B$ : $F_{m a x} = q v B$ .

Electric vs magnetic force - the key contrasts

Property	Electric force $q E$	Magnetic force $q v \times B$
Direction relative to field	Along $E$ (or opposite)	Perpendicular to $B$
Depends on velocity?	No	Yes (both magnitude and direction)
Acts on stationary charge?	Yes	No
Does work?	Yes	Never (on the particle)
Frame dependence	Frame-dependent	Frame-dependent; but total Lorentz force is frame-independent

Solved Example 1

A charged particle of mass

5 mg

and charge

q = + 2 μ C

has velocity

v = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}

m/s. Find the magnetic force and acceleration in field

B = 3 \hat{i} - 2 \hat{k}

Wb/m².

Solution:

$F = q (v \times B) = 2 \times 1 0^{- 6} [(2 \hat{i} - 3 \hat{j} + 4 \hat{k}) \times (3 \hat{i} - 2 \hat{k})]$

Expanding the cross product: $2 \hat{i} \times 3 \hat{i} = 0$ , $2 \hat{i} \times (- 2 \hat{k}) = 4 \hat{j}$ ; $(- 3 \hat{j}) \times 3 \hat{i} = 9 \hat{k}$ , $(- 3 \hat{j}) \times (- 2 \hat{k}) = 6 \hat{i}$ ; $4 \hat{k} \times 3 \hat{i} = 12 \hat{j}$ , $4 \hat{k} \times (- 2 \hat{k}) = 0$ .

Sum: $6 \hat{i} + 16 \hat{j} + 9 \hat{k}$ . So $F = 2 \times 1 0^{- 6} (6 \hat{i} + 16 \hat{j} + 9 \hat{k})$ N.

$a = F / m = \frac{2 \times 1 0 ^{- 6}}{5 \times 1 0 ^{- 6}} (6 \hat{i} + 14 \hat{j} + 9 \hat{k}) = 0.4 (6 \hat{i} + 16 \hat{j} + 9 \hat{k})$ m/s².

2Motion Perpendicular to $B$ : Circular Motion

If a particle enters a uniform $B$ with $v ⊥ B$ , the magnetic force $q v B$ acts as a centripetal force perpendicular to $v$ . Result: uniform circular motion in the plane perpendicular to $B$ .

Figure 1A positive charge in a uniform field into the page.

F = q v \times B

is always perpendicular to

v

and points to the centre, so speed stays constant and the path is a circle traversed anticlockwise, with

r = \frac{m v}{q B}

and

T = \frac{2 π m}{q B}

(independent of speed).

Applying Newton's second law with centripetal acceleration:

$q v B = \frac{m v ^{2}}{r} ⟹ r = \frac{m v}{q B} = \frac{p}{q B}$

where $p = m v$ is the momentum. In terms of kinetic energy $K = p^{2} / (2 m)$ : $r = \frac{2 m K}{q B}$ .

Time period (one full revolution):

$T = \frac{2 π r}{v} = \frac{2 π m}{q B}; f = \frac{q B}{2 π m}; ω = \frac{q B}{m}$

Note that $T$ , $f$ , and $ω$ are all independent of $v$ and $r$ . Fast particles orbit larger circles in the same time as slow ones - this is the principle behind the cyclotron.

Figure 2Particles entering the same field (into the page) with the same speed. The positive charge curves one way, the negative charge curves the opposite way, and a neutral particle is undeflected. For equal

∣ q ∣

and

m

the two radii are equal.

Solved Example 2

A proton, alpha particle, and deuteron move in circular paths with the same kinetic energy in the same magnetic field. Compare their radii and time periods.

Solution:

$r = \frac{2 m K}{q B}$ . Masses: $m_{p} : m_{α} : m_{D} = 1 : 4 : 2$ . Charges: $q_{p} : q_{α} : q_{D} = 1 : 2 : 1$ .

$r_{p} : r_{α} : r_{D} = \frac{1}{1} : \frac{4}{2} : \frac{2}{1} = 1 : 1 : 2$ .

$T = \frac{2 π m}{q B}$ : $T_{p} : T_{α} : T_{D} = \frac{1}{1} : \frac{4}{2} : \frac{2}{1} = 1 : 2 : 2$ .

Solved Example 3

A positive charge

q

of mass

m

moving with speed

v

enters a region of uniform

B

(into page) that lies to the right of a straight boundary

P Q

. Its velocity makes angle

θ

with the boundary. Find (i) time spent in the field, (ii) distance travelled, (iii) impulse from the field. There is no field on the left of

P Q

Figure 3Solved Example 3: the charge enters the field region at

A

making angle

θ

with the boundary

P Q

. It follows an arc that subtends

2 θ

at the centre

O

and leaves at

A^{'}

, again at angle

θ

P Q

. The velocity component along

P Q

is unchanged; the component perpendicular to

P Q

is reversed, giving impulse

2 m v sin θ

Solution:

The particle traces a circular arc of angle $2 θ$ at the centre, then exits the field on the same side.

(i) Time: arc angle $= 2 θ$ , angular speed $ω = q B / m$ , so $t = 2 θ / ω = \frac{2 m θ}{q B}$ .

(ii) Distance: $s = r (2 θ) = \frac{2 m v θ}{q B}$ .

(iii) Impulse: $∣Δ p ∣ = 2 m v sin θ$ (the velocity component perpendicular to $P Q$ is reversed).

3Motion at an Angle to $B$ : Helical Motion

If $v$ makes angle $θ$ with $B$ , decompose:

$v_{∥} = v cos θ$ : unaffected by $B$ ; particle moves uniformly along field lines.
$v_{⊥} = v sin θ$ : produces circular motion in the plane perpendicular to $B$ , radius $r = \frac{m v sin θ}{q B}$ .

The combined motion is a helix around the field lines.

Figure 4When

v

makes angle

θ

with

B

v_{⊥} = v sin θ

produces circular motion of radius

r = \frac{m v sin θ}{q B}

, while

v_{∥} = v cos θ

is unaffected. The result is a helix about the field direction with pitch

p = v_{∥} T = \frac{2 π m v cos θ}{q B}

Pitch = axial distance per full revolution:

$p = v_{∥} T = v cos θ \cdot \frac{2 π m}{q B} = \frac{2 π m v cos θ}{q B}$

Solved Example 4

Protons of speed

4 \times 1 0^{5}

m/s enter a uniform

B = 0.3

T at angle

60°

to the field. Find the radius and pitch of the helix. (

m_{p} = 1.67 \times 1 0^{- 27}

kg)

Solution:

$v_{⊥} = v sin 60° = 4 \times 1 0^{5} \times \frac{3}{2} = 23 \times 1 0^{5}$ m/s.

$r = \frac{m v _{⊥}}{q B} = \frac{1.67 \times 1 0 ^{- 27} \times 2 3 \times 1 0 ^{5}}{1.6 \times 1 0 ^{- 19} \times 0.3} \approx 1.2 \times 1 0^{- 2}$ m = $1.2$ cm.

$T = \frac{2 π m}{q B} = \frac{2 π \times 1.67 \times 1 0 ^{- 27}}{1.6 \times 1 0 ^{- 19} \times 0.3} \approx 2.18 \times 1 0^{- 7}$ s.

Pitch $= v cos 60° \cdot T = (4 \times 1 0^{5}) (0.5) (2.18 \times 1 0^{- 7}) \approx 4.35$ cm.

4Combined Electric and Magnetic Fields

4.1Velocity selector (crossed $E$ and $B$ )

If $E$ , $B$ , and $v$ are mutually perpendicular, the electric force $q E$ and magnetic force $q v \times B$ can be arranged to cancel. Setting them equal in magnitude and opposite in direction:

q E = q v B ⟹ v = \frac{E}{B}

Only particles with this specific speed pass through undeflected; faster or slower particles are deflected out. This device is a velocity selector, used in mass spectrometers and similar experiments to pick out ions of a single speed.

Figure 5Velocity selector:

E

(plates) and

B

(into page) are crossed. For a positive charge moving right,

q E

acts down and

q v B

acts up. Only

v = E / B

passes straight through the exit slit; faster particles bend towards the magnetic force, slower ones towards the electric force.

4.2 $E ∥ B$ , both along $v$

Magnetic force is zero. Only electric force acts, along the field. Particle accelerates/decelerates in a straight line, gaining or losing kinetic energy.

4.3 $E ∥ B$ , $v$ at angle to them

Component of $v$ parallel to $B$ is accelerated linearly by $E$ ; perpendicular component circles. Result: a helix whose pitch grows with each turn.

5The CyclotronBeyond syllabus

The cyclotron is a particle accelerator that uses the fact that $T$ (time to complete a semicircle) is independent of speed. Two hollow D-shaped conductors ("dees") sit in a strong uniform $B$ ; between the dees, an oscillating $E$ across a gap re-accelerates the particle each half-turn.

Figure 6Cyclotron (top view). The ion source

S

sits in the gap near the centre between dees

D_{1}

and

D_{2}

. Inside a dee there is no electric field, so the ion moves in a semicircle under

B

alone. Each time it crosses the gap, the alternating voltage accelerates it, so the next semicircle is larger (

r \propto v

). The time for each semicircle,

π m / q B

, does not change, which is why a fixed oscillator frequency

f = \frac{q B}{2 π m}

stays in step.

5.1Cyclotron frequency

Because $T = 2 π m / (q B)$ is independent of speed, the applied oscillator must run at the fixed cyclotron frequency:

f_{c} = \frac{q B}{2 π m}

5.2Maximum energy

The maximum orbit radius equals the radius $R$ of the dees. Setting $r = R$ in $r = m v / q B$ :

$v_{m a x} = \frac{q B R}{m} ⟹ K_{m a x} = \frac{1}{2} m v_{m a x}^{2} = \frac{q ^{2} B ^{2} R ^{2}}{2 m}$

5.3Limitations of the cyclotron

Not for electrons: electrons quickly reach relativistic speeds where $m \to γ m$ increases and $T$ is no longer constant; they fall out of sync with the oscillator.
Not for neutrons or neutral particles: requires a charge.
Relativistic limit: for heavy ions the increase in $m$ at very high energies also spoils synchronism. The synchrocyclotron and synchrotron address this by varying $f$ or $B$ .

Solved Example 5

An alpha particle (

m = 6.8 \times 1 0^{- 27}

kg,

q = 3.2 \times 1 0^{- 19}

C) moves in a circle of radius

0.45

m in

B = 1.2

T. Find its speed, cyclotron frequency, and kinetic energy.

Solution:

$v = \frac{q B r}{m} = \frac{3.2 \times 1 0 ^{- 19} \times 1.2 \times 0.45}{6.8 \times 1 0 ^{- 27}} = 2.54 \times 1 0^{7}$ m/s.

$f = \frac{q B}{2 π m} = \frac{3.2 \times 1 0 ^{- 19} \times 1.2}{2 π \times 6.8 \times 1 0 ^{- 27}} \approx 9.0 \times 1 0^{6}$ Hz = $9$ MHz.

$K = \frac{1}{2} m v^{2} = \frac{1}{2} (6.8 \times 1 0^{- 27}) (2.54 \times 1 0^{7})^{2} \approx 2.19 \times 1 0^{- 12}$ J $\approx 13.7$ MeV.

Common Mistakes to Avoid

Watch out

Thinking the magnetic force can change kinetic energy: it cannot; $F ⊥ v$ always, so $W = 0$ . Speed changes only via electric forces or collisions.
Confusing $r$ and $T$ dependence on speed: $r = m v / q B$ scales with $v$ ; $T = 2 π m / q B$ does not depend on $v$ . Faster particle = bigger orbit at same $T$ .
Applying $r = m v / q B$ blindly for helical motion: use $v_{⊥} = v sin θ$ , not $v$ itself.
Wrong direction of the magnetic force: use right-hand rule for $v \times B$ , then flip if the charge is negative.
Assuming $v = E / B$ works with anything: the velocity selector formula holds only when $v$ , $E$ , $B$ are mutually perpendicular.
Ignoring the sign of charge in force expressions: $F = q (v \times B)$ includes the sign of $q$ . Positive and negative charges deflect in opposite directions.
Forgetting that cyclotron won't accelerate electrons: relativistic mass increase spoils the constant- $T$ trick.

Frequently Asked Questions

Q1. Why does the magnetic force do no work on a charged particle?

The magnetic force $F = q (v \times B)$ is always perpendicular to $v$ . Work $W = F \cdot d r = F \cdot v d t = 0$ because a vector dot itself into a perpendicular vector is zero. Hence a magnetic field can bend a particle's path but cannot change its speed or kinetic energy.

Q2. What is the radius of circular motion of a charged particle in a magnetic field?

If $v ⊥ B$ , the radius is $r = \frac{m v}{q B} = \frac{p}{q B} = \frac{2 m K}{q B}$ . Faster or heavier particles orbit larger circles; stronger fields give tighter orbits.

Q3. Why is the time period of circular motion independent of speed?

$T = 2 π m / (q B)$ depends only on the particle's mass-to-charge ratio and the field. A faster particle covers a longer circumference at proportionally higher speed, so the time per revolution stays the same. This "isochronism" is the physics that makes the cyclotron work.

Q4. What is helical motion and when does it occur?

When $v$ makes an angle $θ$ with $B$ , the parallel component $v cos θ$ carries the particle along the field lines while the perpendicular component $v sin θ$ makes it circle. The resulting path is a helix with radius $r = \frac{m v sin θ}{q B}$ and pitch $p = \frac{2 π m v cos θ}{q B}$ .

Q5. What is a velocity selector?

A velocity selector uses crossed electric and magnetic fields perpendicular to a charged beam. Only particles with speed $v = E / B$ experience balanced electric and magnetic forces and pass through undeflected. Faster or slower particles are pushed out. It is used in mass spectrometers to pick out ions of a single speed.

Q6. What is the cyclotron frequency?

The cyclotron (angular) frequency is $ω_{c} = q B / m$ , or $f_{c} = q B / (2 π m)$ . It is the rate at which a charged particle circles in a magnetic field, and equals the frequency at which the accelerating voltage must be applied in a cyclotron.

Q7. What is the maximum energy a cyclotron can give to an ion?

If the dee radius is $R$ , the largest orbit fits at $r = R$ giving $v_{m a x} = q B R / m$ and $K_{m a x} = q^{2} B^{2} R^{2} / (2 m)$ . Larger fields and bigger dees yield higher energies, but relativistic mass increase eventually limits standard cyclotrons.

Q8. Why can't a cyclotron accelerate electrons?

Electrons have small mass, so they reach relativistic speeds quickly. As $v \to c$ , effective mass grows as $γ m$ , and the time period $T = 2 π m / (q B)$ increases. The particle falls out of sync with the fixed-frequency oscillator, so acceleration stops. Synchrocyclotrons vary the frequency to compensate.

Q9. Does a stationary charge experience any force in a magnetic field?

No. The magnetic force $F = q (v \times B)$ vanishes if $v = 0$ . A magnetic field only pushes on charges in motion. If the charge is also in an electric field, of course, it feels $q E$ whether it moves or not.

Previous year questions on Motion of a Charged Particle in a Magnetic Field

18 questions from past papers, each with a step-by-step solution.

Show all 18 questions

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1The Lorentz Force

Electric vs magnetic force - the key contrasts

2Motion Perpendicular to B: Circular Motion

3Motion at an Angle to B: Helical Motion

4Combined Electric and Magnetic Fields

4.1Velocity selector (crossed E and B)

4.2E∥B, both along v

4.3E∥B, v at angle to them

5The CyclotronBeyond syllabus

5.1Cyclotron frequency

5.2Maximum energy

5.3Limitations of the cyclotron

Common Mistakes to Avoid

Frequently Asked Questions

Q1. Why does the magnetic force do no work on a charged particle?

Q2. What is the radius of circular motion of a charged particle in a magnetic field?

Q3. Why is the time period of circular motion independent of speed?

Q4. What is helical motion and when does it occur?

Q5. What is a velocity selector?

Q6. What is the cyclotron frequency?

Q7. What is the maximum energy a cyclotron can give to an ion?

Q8. Why can't a cyclotron accelerate electrons?

Q9. Does a stationary charge experience any force in a magnetic field?

Previous year questions on Motion of a Charged Particle in a Magnetic Field

Ready to master Magnetic Effects of Current and Magnetism?

2Motion Perpendicular to $B$ : Circular Motion

3Motion at an Angle to $B$ : Helical Motion

4.1Velocity selector (crossed $E$ and $B$ )

4.2 $E ∥ B$ , both along $v$

4.3 $E ∥ B$ , $v$ at angle to them