Charged Particle Trajectory in Spatially Varying Magnetic Fields: Solving equations of motion and finding maximum displacement/penetration of a charged particle projected into a space-varying magnetic field

1. Introduction

When a charged particle enters a magnetic field, it experiences a magnetic force. If the magnetic field changes from one point to another, the path of the particle is no longer a simple circle of constant radius.

F = q(v × B)

The magnitude of magnetic force is:

F = |q|vB sinθ

where θ is the angle between the velocity and magnetic field.

Important: A magnetic field changes the direction of velocity but does not change the speed of a charged particle because magnetic force is always perpendicular to velocity.

2. Motion in a Uniform Magnetic Field

For a charged particle moving perpendicular to a uniform magnetic field:

qvB = mv²/r

Therefore:

r = mv/(|q|B)

This is called the cyclotron radius or Larmor radius.

The angular velocity is:

ω = |q|B/m

The time period is:

T = 2πm/(|q|B)

3. Spatially Varying Magnetic Field

Suppose the magnetic field depends on position:

B = B(x)

Then the instantaneous radius becomes:

r(x)=mv/(|q|B(x))

Therefore:

• Large B → smaller radius
• Small B → larger radius
• Increasing B → particle bends more strongly
• Decreasing B → particle bends less strongly

The trajectory must be studied using the local value of B at every point. The particle does not generally follow one circle throughout its motion.

4. Equation of Motion

The general equation of motion is:

m dv/dt = q(v × B)

If:

B = B(x) k̂

and the particle moves in the x-y plane, then:

m dv_x/dt = qB(x)v_y

m dv_y/dt = -qB(x)v_x

The exact signs depend upon the chosen coordinate system and charge sign.

5. Speed Remains Constant

The magnetic force is perpendicular to velocity:

F · v = 0

Therefore:

d(½mv²)/dt = 0

Hence:

v = constant

Even though the magnetic field changes with position, an ideal static magnetic field does no work on the particle.

6. Maximum Displacement / Penetration

A common competitive-exam situation is a particle entering a region where the magnetic field becomes stronger with distance.

If the field increases sufficiently rapidly, the particle bends more and more strongly and may turn back before reaching the end of the region.

The turning or maximum penetration point is obtained by combining:

r(x)=mv/(|q|B(x))

with the geometrical condition imposed by the boundary or trajectory.

Key idea: There is no single universal formula for maximum penetration in every non-uniform magnetic field. The field profile B(x), initial direction, geometry and boundary conditions must be used.

7. Important Special Case: B(x) = B₀ + kx

Suppose:

B(x)=B₀+kx

Then:

r(x)=mv/[|q|(B₀+kx)]

As x increases:

B ↑

Therefore:

r ↓

Thus the path becomes progressively more curved.

8. Magnetic Field Gradient and Drift

In a slowly varying magnetic field, a charged particle may experience a gradient-B drift.

For perpendicular velocity:

v_∇B = (mv_⊥²)/(2qB³) (B × ∇B)

The exact direction depends on the sign of q.

For school-level CBSE questions, the main focus is generally the magnetic force, changing radius, curvature and qualitative trajectory. Gradient-B drift is mainly useful for advanced competitive problems.

9. Slowly Varying Magnetic Field

If the magnetic field changes slowly over one cyclotron orbit, the motion can be considered approximately circular locally.

The local radius is:

r = mv_⊥/(|q|B)

The magnetic moment is approximately conserved:

μ = mv_⊥²/(2B)

Thus, if B increases:

v_⊥²/B ≈ constant

This is an advanced competitive-exam result.

10. Magnetic Mirror Effect

When a charged particle moves toward a region of increasing magnetic field, its perpendicular velocity component can increase while its parallel component decreases.

Using conservation of magnetic moment:

mv_⊥²/(2B)=constant

If the parallel velocity becomes zero:

v_∥=0

the particle turns back.

This is called magnetic mirroring.

11. Solved Example 1 — Local Radius

An electron moves with speed 2 × 10⁶ m/s in a magnetic field whose magnitude at a certain point is 0.5 T. Find its instantaneous radius.

Given:

• m = 9.1 × 10^-31 kg
• |q| = 1.6 × 10^-19 C
• v = 2 × 10⁶ m/s
• B = 0.5 T

r=mv/(|q|B)

r= (9.1×10^-31)(2×10⁶) / [(1.6×10^-19)(0.5)]

r ≈ 2.28 × 10^-5 m

12. Solved Example 2 — Effect of Increasing B

A charged particle enters a region where the magnetic field increases from 0.2 T to 0.8 T. If its speed remains constant, compare the radii of curvature.

r∝1/B

Therefore:

r₁/r₂=B₂/B₁

r₁/r₂=0.8/0.2=4

r₁ : r₂ = 4 : 1

Thus the radius becomes four times smaller.

13. Solved Example 3 — Magnetic Mirror

A charged particle has perpendicular velocity v_⊥1 in a magnetic field B₁. At another point the magnetic field becomes B₂. Assuming adiabatic motion, find v_⊥2.

Using:

v_⊥²/B = constant

Therefore:

v_⊥2²/B₂ = v_⊥1²/B₁

Hence:

v_⊥2 = v_⊥1 √(B₂/B₁)

14. Solved Example 4 — Turning Condition

A particle enters a magnetic mirror with speed v at pitch angle α. The magnetic field changes from B₀ to B_m. Find the condition for reflection.

At the initial point:

v_⊥0=v sinα

and:

v_∥0=v cosα

Conservation of magnetic moment gives:

v_⊥m²/B_m = v²sin²α/B₀

At the mirror point:

v_∥m=0

Therefore:

v_⊥m=v

Hence:

B_m = B₀/sin²α

This is the mirror condition.

15. Important Formula Sheet

Quantity	Formula
Magnetic force	F=q(v×B)
Force magnitude	F=|q|vB sinθ
Radius	r=mv⊥/(|q|B)
Angular frequency	ω=|q|B/m
Time period	T=2πm/(|q|B)
Position-dependent radius	r(x)=mv⊥/(|q|B(x))
Magnetic moment	μ=mv⊥²/(2B)
Adiabatic condition	v⊥²/B ≈ constant
Mirror condition	Bm=B₀/sin²α

Section A — 1 Mark MCQs
10 Questions

Q1. A magnetic field does work on a charged particle when the field is static and magnetic force acts alone: 1 Mark

A) Always

B) Never

C) Only for positive charge

D) Only for negative charge

Answer: B

Q2. The radius of a charged particle in a magnetic field is proportional to: 1 Mark

A) B

B) 1/B

C) B²

D) 1/B²

Answer: B

Q3. If magnetic field doubles, the radius of circular motion becomes: 1 Mark

A) Double

B) Four times

C) Half

D) Unchanged

Answer: C

Q4. The magnetic force is perpendicular to: 1 Mark

A) Magnetic field only

B) Velocity only

C) Both velocity and magnetic field

D) Neither

Answer: C

Q5. For perpendicular entry into B, the radius is: 1 Mark

A) mv/qB

B) qB/mv

C) qv/mB

D) mB/qv

Answer: A

Q6. If B increases with x, the local radius generally: 1 Mark

A) Increases

B) Decreases

C) Remains constant

D) Becomes infinite

Answer: B

Q7. Cyclotron angular frequency is: 1 Mark

A) qB/m

B) m/qB

C) q/mB

D) B/qm

Answer: A

Q8. Magnetic mirror reflection occurs in a region of: 1 Mark

A) Decreasing B

B) Increasing B

C) Zero B

D) Constant electric field

Answer: B

Q9. If B changes spatially, the particle's speed in a static magnetic field remains: 1 Mark

A) Increasing

B) Decreasing

C) Constant

D) Zero

Answer: C

Q10. The local curvature radius is inversely proportional to: 1 Mark

A) Mass

B) Speed

C) Charge magnitude

D) Magnetic field

Answer: D

Section B — 2 Marks
10 Questions

Q11. Write the expression for the radius of a charged particle moving perpendicular to a magnetic field. 2 Marks

r=mv/(|q|B)

Q12. Why does a magnetic field not change the speed of a charged particle? 2 Marks

Magnetic force is perpendicular to velocity, so it does no work. Therefore kinetic energy and speed remain constant.

Q13. A particle enters a region where B changes from 1 T to 2 T. Compare its local radii. 2 Marks

r∝1/B

r₁:r₂=2:1

Q14. State the cyclotron frequency of a charged particle. 2 Marks

f=|q|B/(2πm)

Q15. What happens to the trajectory when B increases with distance? 2 Marks

The radius of curvature decreases because r∝1/B. Hence the trajectory bends more strongly.

Q16. What is magnetic moment of a charged particle's circular orbit? 2 Marks

μ=mv_⊥²/(2B)

Q17. What is meant by maximum penetration? 2 Marks

It is the maximum distance reached by the particle inside a magnetic-field region before its trajectory turns back or becomes unable to penetrate further.

Q18. A particle has radius 10 cm in B=0.5 T. What happens to radius if B becomes 1 T? 2 Marks

Since r∝1/B, doubling B halves the radius. New radius = 5 cm.

Q19. Define magnetic mirror effect. 2 Marks

A charged particle can be reflected when it moves into a region of sufficiently strong magnetic field. This phenomenon is called magnetic mirroring.

Q20. Why is the trajectory not a single circle in a non-uniform magnetic field? 2 Marks

Because the radius depends on local magnetic field:

r(x)=mv/(|q|B(x))

As B changes with position, the radius also changes.

Section C — 3 Marks
10 CBSE + Competitive Questions

Q21. Derive the radius of circular motion of a charged particle in a uniform magnetic field. 3 Marks

Magnetic force:

F=qvB

Centripetal force:

F=mv²/r

Equating:

qvB=mv²/r

Therefore:

r=mv/(|q|B)

Q22. A proton enters a magnetic field of 0.4 T perpendicular to the field with speed 4×10⁶ m/s. Calculate its radius. 3 Marks

Using:

r=mv/(qB)

Taking: m=1.67×10^-27 kg, q=1.6×10^-19 C.

r= (1.67×10^-27)(4×10⁶) / [(1.6×10^-19)(0.4)]

r≈0.104 m ≈10.4 cm

Q23. The magnetic field varies as B(x)=B₀+kx. Explain how the radius changes with x. 3 Marks

r(x)=mv/[|q|(B₀+kx)]

If k>0:

B increases with x

Therefore:

r decreases with x

The particle bends increasingly sharply as it penetrates deeper into the region.

Q24. Show that a magnetic field cannot change the kinetic energy of a charged particle. 3 Marks

Magnetic force:

F=q(v×B)

Power:

P=F·v

Therefore:

P=q(v×B)·v=0

Hence:

Kinetic energy remains constant.

Q25. A charged particle has radius 20 cm in B=0.2 T. Find the radius when the magnetic field becomes 0.8 T, assuming speed remains constant. 3 Marks

r₁B₁=r₂B₂

Therefore:

r₂=r₁B₁/B₂

r₂=20×0.2/0.8

r₂=5 cm

Q26. Derive the mirror condition for a charged particle entering a stronger magnetic-field region. 3 Marks

Magnetic moment is conserved:

mv_⊥²/(2B)=constant

At the starting point:

v_⊥0=v sinα

At the mirror point:

v_∥=0

Hence:

B_msin²α=B₀

Therefore:

B_m=B₀/sin²α

Q27. A particle enters a magnetic mirror with pitch angle 30°. If the initial magnetic field is B₀, find the magnetic field required for reflection. 3 Marks

B_m=B₀/sin²30°

Since:

sin30°=1/2

Therefore:

B_m=B₀/(1/4)

B_m=4B₀

Q28. A particle has perpendicular velocity v⊥ in magnetic field B. The field becomes 4B slowly. Find its new perpendicular velocity under adiabatic conditions. 3 Marks

Using:

v_⊥²/B=constant

Therefore:

v'_⊥²/(4B)=v_⊥²/B

Hence:

v'_⊥=2v_⊥

New perpendicular velocity = 2v_⊥

Q29. Explain qualitatively why a particle may have a maximum penetration distance in a magnetic field that increases with distance. 3 Marks

As the particle moves deeper:

B(x) ↑

Hence:

r(x)=mv/(|q|B(x)) ↓

Therefore the path bends increasingly strongly. Depending on the geometry and field profile, the particle can eventually turn back.

The point where the particle reverses direction represents its maximum penetration distance.

Q30. A charged particle enters a region where B changes from B₀ to 9B₀ slowly. If its initial perpendicular velocity is v⊥, find its final perpendicular velocity and compare the local radii. 3 Marks

Using adiabatic conservation:

v'_⊥²/(9B₀)=v_⊥²/B₀

Therefore:

v'_⊥=3v_⊥

Radius:

r=mv_⊥/(qB)

Thus:

r'/r=(3v⊥/9B₀)/(v⊥/B₀)

v'_⊥=3v_⊥
r'=r/3

Final Quick Revision

F=q(v×B)

r=mv_⊥/(|q|B)

ω=|q|B/m

T=2πm/(|q|B)

B↑ ⇒ r↓

B↓ ⇒ r↑

μ=mv_⊥²/(2B)

v_⊥²/B≈constant

B_m=B₀/sin²α

Competitive Exam Trick:

For a spatially varying magnetic field, first identify B at the particle's position. Then calculate the local radius:

r(x)=mv/(|q|B(x))

Do not use one constant radius unless B is uniform.

Charged Particle Trajectory in Spatially Varying Magnetic Fields
CBSE + Competitive Physics | Concept + Derivation + 30 Questions