Intermediate

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Magnetic Force on Moving Charges & Lorentz Force

Master Magnetic Force on Moving Charges and Lorentz Force for JEE Main, JEE Advanced, NEET & Board Exams. Covers magnetic force formula, right-hand rule, Lorentz force, circular and helical motion in magnetic fields, velocity selector, force on current-carrying conductors and radius from accelerating potential with solved examples.

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When a charged particle moves through a magnetic field, it experiences a force — the magnetic force. Unlike the electric force (which acts along the field direction), the magnetic force acts perpendicular to both the velocity and the field. The combined effect of electric and magnetic forces on a moving charge is called the Lorentz Force. Understanding this force is the gateway to the entire chapter of Moving Charges and Magnetism — it explains how charged particles move in fields, how cyclotrons work, how galvanometers deflect, and how motors generate torque. For JEE and NEET, this is a very high-weightage topic with questions appearing every year on force directions, circular motion in fields, and velocity selectors.

1. Magnetic Force on a Moving Charge

A charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences a force:

${\vec{F}}_{B} = q (\vec{v} \times \vec{B})$

The magnitude of this force is:

$F_{B} = q v B \sin θ$

where $θ$ is the angle between $\vec{v}$ and $\vec{B}$ .

Key Properties of Magnetic Force

Property	Details
Direction	Perpendicular to both $\vec{v}$ and $\vec{B}$ (given by right-hand rule or cross product)
Maximum force	$F = q v B$ when $θ = 90 °$ (velocity $⊥$ field)
Zero force	$F = 0$ when $θ = 0 °$ or $180 °$ (velocity parallel or antiparallel to field)
Work done	Always zero — force is always $⊥$ to velocity, so $W = \vec{F} \cdot \vec{v} = 0$
Speed change	Does NOT change speed or kinetic energy — only changes direction
Nature	Non-conservative (cannot define a potential energy for magnetic force)
On stationary charge	$F = 0$ — magnetic force requires motion ( $v = 0 \Rightarrow F = 0$ )

SI Unit of Magnetic Field

From $F = q v B \sin θ$ , the SI unit of $B$ is the Tesla (T):

$1 T = \frac{1 N}{1 C \cdot 1 m/s} = 1 {N·A}^{- 1} {·m}^{- 1}$

Another common unit: $1 Tesla = 10^{4} Gauss (G)$ . Earth's magnetic field $\approx 10^{- 5}$ T = 0.1 G (very weak).

2. Direction of Magnetic Force — Right-Hand Rule

The direction of ${\vec{F}}_{B} = q (\vec{v} \times \vec{B})$ is determined by the right-hand rule for the cross product:

Point the fingers of the right hand along $\vec{v}$ (velocity direction).
Curl them towards $\vec{B}$ (magnetic field direction).
The thumb points in the direction of $\vec{v} \times \vec{B}$ .
If the charge $q$ is positive: force is in the direction of $\vec{v} \times \vec{B}$ .
If the charge $q$ is negative: force is in the direction opposite to $\vec{v} \times \vec{B}$ .

Fleming's Left-Hand Rule (for positive charge)

Stretch the thumb, index finger, and middle finger of the left hand mutually perpendicular:

Index finger → direction of $\vec{B}$ (field)
Middle finger → direction of $\vec{v}$ (velocity / conventional current)
Thumb → direction of force $\vec{F}$

Standard Direction Conventions

Symbol	Meaning	Represents
$⊙$ (dot)	Out of the page / plane	Arrow tip coming towards you
$\otimes$ (cross)	Into the page / plane	Arrow tail going away from you

Quick Force Direction Examples

Charge	Velocity $\vec{v}$	Field $\vec{B}$	Force $\vec{F}$
$+ q$	$+ \hat{x}$ (right)	$+ \hat{y}$ (up)	$+ \hat{z}$ (out of page)
$- q$	$+ \hat{x}$ (right)	$+ \hat{y}$ (up)	$- \hat{z}$ (into page)
$+ q$	$+ \hat{x}$ (right)	$\otimes$ (into page, $- \hat{z}$ )	$- \hat{y}$ (downward)
$+ q$	Parallel to $\vec{B}$	Any	$\vec{0}$ (zero)

3. The Lorentz Force

When a charge $q$ moves with velocity $\vec{v}$ in a region where both electric field $\vec{E}$ and magnetic field $\vec{B}$ are present, the total force on the charge is the Lorentz Force:

$\vec{F} = q (\vec{E} + \vec{v} \times \vec{B})$

The two components:

Electric force ${\vec{F}}_{E} = q \vec{E}$ : acts along $\vec{E}$ , does work, changes speed.
Magnetic force ${\vec{F}}_{B} = q (\vec{v} \times \vec{B})$ : acts $⊥$ to $\vec{v}$ , does no work, changes direction only.

Property	Electric Force ${\vec{F}}_{E}$	Magnetic Force ${\vec{F}}_{B}$
Direction	Along $\vec{E}$ (or opposite for $- q$ )	$⊥$ to both $\vec{v}$ and $\vec{B}$
Acts on	Stationary and moving charges	Only moving charges ( $v \neq 0$ )
Work done	Can be non-zero; changes KE	Always zero; KE unchanged
Changes speed?	Yes	No — only direction
Depends on velocity?	No	Yes — $F_{B} \propto v$

4. Motion of a Charged Particle in a Uniform Magnetic Field

Case 1 — Velocity Perpendicular to $\vec{B}$ ( $θ = 90 °$ )

The magnetic force acts as a centripetal force, causing the particle to move in a circle in the plane perpendicular to $\vec{B}$ .

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

Other important quantities:

Radius: $r = \frac{m v}{q B} = \frac{p}{q B}$ where $p = m v$ is momentum.
Angular velocity (cyclotron frequency): $ω = \frac{q B}{m}$
Time period: $T = \frac{2 π m}{q B}$ — independent of speed and radius!
Frequency: $f = \frac{1}{T} = \frac{q B}{2 π m}$ — also independent of speed.

The fact that $T$ and $f$ are independent of velocity is the operating principle of the cyclotron.

Case 2 — Velocity Parallel to $\vec{B}$ ( $θ = 0 °$ or $180 °$ )

$F_{B} = q v B \sin 0 ° = 0$ . The particle experiences no magnetic force and continues in a straight line with constant velocity.

Case 3 — Velocity at Angle $θ$ to $\vec{B}$ (General Case)

Resolve velocity into two components:

$v_{∥} = v \cos θ$ — parallel to $\vec{B}$ : unaffected, moves in straight line.
$v_{⊥} = v \sin θ$ — perpendicular to $\vec{B}$ : causes circular motion.

The combination of uniform linear motion along $\vec{B}$ and circular motion perpendicular to $\vec{B}$ results in a helix (helical path).

Radius of helix: $r = \frac{m v \sin θ}{q B}$
Pitch of helix (distance advanced per revolution): $p = v_{∥} \cdot T = \frac{2 π m v \cos θ}{q B}$

Angle $θ$ between $\vec{v}$ and $\vec{B}$	Path of Particle
$0 °$ or $180 °$	Straight line (no force)
$90 °$	Circle (in plane perpendicular to $\vec{B}$ )
$0 ° < θ < 90 °$ (or $90 ° < θ < 180 °$ )	Helix (axis along $\vec{B}$ )

5. Velocity Selector (Filter)

A velocity selector is a device that uses crossed electric and magnetic fields to select particles of a specific velocity, regardless of their charge or mass.

When $\vec{E}$ and $\vec{B}$ are perpendicular to each other and to the particle's velocity:

Electric force: $F_{E} = q E$ (upward, say)
Magnetic force: $F_{B} = q v B$ (downward, say)

For the particle to pass through undeflected, $F_{E} = F_{B}$ :

$q E = q v B$ $\Rightarrow$ $v = \frac{E}{B}$

Only particles with speed $v = E / B$ pass through undeviated. Faster particles are deflected one way, slower particles the other way. The selected speed is independent of charge and mass.

Applications: Mass spectrometers, electron guns, ion beam equipment.

6. Force on a Current-Carrying Conductor in a Magnetic Field

A current-carrying conductor in a magnetic field experiences a force because the moving charge carriers (electrons) experience the magnetic force, and this force is transmitted to the conductor as a whole.

For a straight conductor of length $L$ carrying current $I$ in a field $\vec{B}$ :

$\vec{F} = I (\vec{L} \times \vec{B})$ or $F = B I L \sin θ$

where $θ$ is the angle between the current direction $\vec{L}$ and $\vec{B}$ .

Maximum force: $F = B I L$ when current $⊥$ field ( $θ = 90 °$ ).
Zero force: when current is parallel to field ( $θ = 0 °$ or $180 °$ ).
Direction: given by Fleming's Left-Hand Rule or $\vec{L} \times \vec{B}$ .

Relation Between Current Force and Charge Force

Consider $n$ charge carriers per unit volume, each with charge $q$ and drift velocity $v_{d}$ in a conductor of cross-section $A$ and length $L$ :

$F = (n A L) q v_{d} B = (n A q v_{d}) L B = I L B$

This confirms that $F = B I L$ is the macroscopic consequence of the microscopic $F = q v_{d} B$ on individual charge carriers.

Force on an Arbitrary Curved Conductor

For a curved conductor of arbitrary shape carrying current $I$ in a uniform field $\vec{B}$ , the net force is the same as that on a straight conductor connecting its two endpoints:

$\vec{F} = I ({\vec{L}}_{e f f} \times \vec{B})$

where ${\vec{L}}_{e f f}$ is the vector from one end to the other (straight-line distance). For a closed loop in a uniform field, the net force is zero.

7. Radius and Energy of Charged Particles

For a charged particle accelerated through a potential difference $V$ and then entering a magnetic field $B$ perpendicularly:

$\frac{1}{2} m v^{2} = q V$ $\Rightarrow$ $v = \sqrt{\frac{2 q V}{m}}$

Radius of circular path in the magnetic field:

$r = \frac{m v}{q B} = \frac{1}{B} \sqrt{\frac{2 m V}{q}}$

Comparing Radii of Different Particles (Same $V$ and $B$ )

Particle	Charge $q$	Mass $m$	$r \propto \sqrt{m / q}$
Proton ( $p$ )	$e$	$m_{p}$	$\sqrt{m_{p} / e}$
Deuteron ( $d$ )	$e$	$2 m_{p}$	$\sqrt{2} r_{p}$
Alpha ( $α$ )	$2 e$	$4 m_{p}$	$\sqrt{2} r_{p}$
Electron ( $e^{-}$ )	$e$	$m_{e} \approx m_{p} / 1836$	$r_{e} ≪ r_{p}$

Note: Deuteron and alpha particle have the same radius when accelerated through the same potential difference, since $\sqrt{m / q}$ is equal for both ( $\sqrt{2 m_{p} / e}$ ). This is a frequently tested NEET/JEE result.

Solving Magnetic Force Problems — Master Strategy

Question Type	Immediate Approach	Key Formula
Find force on moving charge	Find $θ$ between $\vec{v}$ and $\vec{B}$ ; use RHR for direction	$F = q v B \sin θ$
Radius of circular path	Set $q v B = m v^{2} / r$ ; solve for $r$	$r = m v / q B$
Time period in magnetic field	Remember: independent of $v$ and $r$	$T = 2 π m / q B$
Particle accelerated through $V$	Use energy conservation first, then radius formula	$r = \frac{1}{B} \sqrt{\frac{2 m V}{q}}$
Velocity selector condition	Equate electric and magnetic forces	$v = E / B$
Force on current in $\vec{B}$	Find angle between $I$ direction and $\vec{B}$	$F = B I L \sin θ$
Helical path parameters	Resolve $v$ into $v_{⊥}$ and $v_{∥}$	$r = m v \sin θ / q B$ ; pitch $= 2 π m v \cos θ / q B$

JEE/NEET Power Tips — Avoid These Common Mistakes

Magnetic force does NO work. Since ${\vec{F}}_{B} ⊥ \vec{v}$ always, $W = \vec{F} \cdot \vec{v} = 0$ . Therefore kinetic energy and speed are unchanged — only direction changes. Never say "the magnetic field accelerates the particle" in terms of speed.
Time period $T = 2 π m / q B$ is independent of speed and radius. This is the crucial fact behind cyclotron operation. Doubling the speed doubles the radius but the time for one revolution stays the same.
For negative charges, reverse the force direction. Apply the right-hand rule to get $\vec{v} \times \vec{B}$ , then reverse it for a negative charge. Many students forget the sign reversal for electrons.
Deuteron and alpha particle have equal radii when accelerated through the same potential difference — since $\sqrt{m / q}$ is the same ( $\sqrt{2 m_{p} / e}$ ) for both. This appears regularly in NEET MCQs.
Force on a closed current loop in a uniform $\vec{B}$ is zero — the forces on different parts cancel. However, there can be a net torque. Do not confuse force with torque.
A stationary charge feels no magnetic force — $F = q v B \sin θ = 0$ when $v = 0$ . Magnetic force requires motion. But the same stationary charge can feel an electric force.
In a velocity selector, the selected velocity $v = E / B$ is independent of charge $q$ and mass $m$ — it depends only on the ratio of the fields.

Practice Questions (JEE / NEET Level)

Q1: A proton moving with velocity $3 \times 10^{6}$ m/s enters a magnetic field of 0.4 T perpendicular to the field. The radius of the circular path is: (mass of proton $= 1.67 \times 10^{- 27}$ kg, $e = 1.6 \times 10^{- 19}$ C)

A) 0.068 m
B) 0.078 m
C) 0.048 m
D) 0.098 m

Answer: B) 0.078 m.

Explanation:

The radius of a charged particle moving perpendicular to a magnetic field is given by $r = \frac{m v}{q B}$ .

Substitute the given values:
$r = \frac{1.67 \times 10^{- 27} \times 3 \times 10^{6}}{1.6 \times 10^{- 19} \times 0.4}$

Simplify the numerator and the denominator:
$r = \frac{5.01 \times 10^{- 21}}{0.64 \times 10^{- 19}}$
$r = \frac{5.01 \times 10^{- 21}}{6.4 \times 10^{- 20}}$

Calculate the final radius:
$r = 0.07828 m \approx 0.078 m$ .

Q2: A charged particle moves in a magnetic field. Which of the following quantities remains unchanged?

A) Momentum
B) Speed
C) Velocity
D) Kinetic energy and speed

Answer: D) Kinetic energy and speed.

Explanation:

The magnetic force is always perpendicular to the velocity of the particle. Therefore, the work done by the magnetic force is zero ( $W = \vec{F} \cdot \vec{d} = 0$ ).

Since no work is done, the kinetic energy remains constant, which means the speed (the magnitude of velocity) also remains constant. However, the magnetic force continuously changes the direction of the particle, so the velocity (a vector) and momentum ( $\vec{p} = m \vec{v}$ ) both change.

Q3: A proton, a deuteron, and an alpha particle are accelerated through the same potential difference and then enter a magnetic field perpendicularly. The ratio of their radii $r_{p} : r_{d} : r_{α}$ is:

A) $1 : \sqrt{2} : \sqrt{2}$
B) $1 : 2 : 1$
C) $1 : \sqrt{2} : 1$
D) $\sqrt{2} : 1 : 1$

Answer: A) $1 : \sqrt{2} : \sqrt{2}$ .

Explanation:

The radius in terms of accelerating potential $V$ is given by $r = \frac{1}{B} \sqrt{\frac{2 m V}{q}}$ .
Since $B$ and $V$ are constant for all three particles, $r \propto \sqrt{\frac{m}{q}}$ .

For the Proton ( $m = m_{p}, q = e$ ):
$r_{p} \propto \sqrt{\frac{m_{p}}{e}}$

For the Deuteron ( $m = 2 m_{p}, q = e$ ):
$r_{d} \propto \sqrt{\frac{2 m_{p}}{e}} = \sqrt{2} r_{p}$

For the Alpha particle ( $m = 4 m_{p}, q = 2 e$ ):
$r_{α} \propto \sqrt{\frac{4 m_{p}}{2 e}} = \sqrt{\frac{2 m_{p}}{e}} = \sqrt{2} r_{p}$

Taking the ratio:
$r_{p} : r_{d} : r_{α} = 1 : \sqrt{2} : \sqrt{2}$ .

Q4: In a velocity selector, the electric field is $4 \times 10^{4}$ V/m and the magnetic field is 0.02 T. The velocity of the particle that passes through undeflected is:

A) $2 \times 10^{3}$ m/s
B) $2 \times 10^{6}$ m/s
C) $8 \times 10^{2}$ m/s
D) $2 \times 10^{4}$ m/s

Answer: B) $2 \times 10^{6}$ m/s.

Explanation:

For a charged particle to pass through undeflected, the electric force must exactly balance the magnetic force: $q E = q v B$ .

Solving for velocity:
$v = \frac{E}{B}$
$v = \frac{4 \times 10^{4}}{0.02} = 2 \times 10^{6} m/s$ .

Q5: A straight wire of length 0.5 m carries a current of 4 A. It is placed in a uniform magnetic field of 2 T at an angle of 30° to the field. The force on the wire is:

A) 4 N
B) 2 N
C) 1 N
D) $\sqrt{3}$ N

Answer: B) 2 N.

Explanation:

The magnetic force on a current-carrying wire is calculated using $F = B I L \sin θ$ .

Substitute the given values:
$F = 2 \times 4 \times 0.5 \times \sin (30^{\circ})$
$F = 4 \times 0.5$
$F = 2 N$ .

Q6: An electron enters a magnetic field of 0.1 T with a velocity of $4 \times 10^{6}$ m/s at an angle of 30° to the field. The pitch of the helical path is: ( $m_{e} = 9.1 \times 10^{- 31}$ kg, $e = 1.6 \times 10^{- 19}$ C)

A) 1.24 mm
B) 2.43 mm
C) 6.94 mm
D) 3.47 mm

Answer: A) 1.24 mm.

Explanation:

The pitch of a helical path is the distance traveled along the magnetic field during one complete revolution. It is calculated by $Pitch = v_{∥} \times T$ , where $v_{∥}$ is the parallel component of velocity and $T$ is the time period.

First, find the parallel component of velocity:
$v_{∥} = v \cos (30^{\circ}) = 4 \times 10^{6} \times \frac{\sqrt{3}}{2}$
$v_{∥} \approx 3.464 \times 10^{6} m/s$

Next, find the time period of revolution:
$T = \frac{2 π m}{q B} = \frac{2 π \times 9.1 \times 10^{- 31}}{1.6 \times 10^{- 19} \times 0.1}$
$T \approx \frac{57.18 \times 10^{- 31}}{1.6 \times 10^{- 20}} \approx 3.57 \times 10^{- 10} s$

Finally, calculate the pitch:
$Pitch = 3.464 \times 10^{6} \times 3.57 \times 10^{- 10}$
$Pitch \approx 1.237 \times 10^{- 3} m \approx 1.24 mm$ .

Q7 (JEE Advanced type): A particle of charge $q$ and mass $m$ moves in a circle of radius $R$ in a uniform magnetic field $B$ . The work done by the magnetic force in one complete revolution is:

A) $q v B R$
B) $2 π q v B R$
C) $π q v B R$
D) Zero

Answer: D) Zero.

Explanation:

The magnetic force is always perpendicular to the instantaneous velocity (and thus the displacement) of the particle.

Since the work done is the dot product of force and displacement ( $W = \int \vec{F} \cdot d \vec{s} = \int F d s \cos (90^{\circ}) = 0$ ), the total work done over one complete revolution (or any path whatsoever) is exactly zero. This reflects the fundamental property that a steady magnetic field cannot do work on a charged particle.

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Est. Read Time28 minutes

CategoryMoving Charges And Magnetism

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Magnetic Force on Moving Charges & Lorentz Force

1. Magnetic Force on a Moving Charge

Key Properties of Magnetic Force

SI Unit of Magnetic Field

2. Direction of Magnetic Force — Right-Hand Rule

Fleming's Left-Hand Rule (for positive charge)

Standard Direction Conventions

Quick Force Direction Examples

3. The Lorentz Force

4. Motion of a Charged Particle in a Uniform Magnetic Field

Case 1 — Velocity Perpendicular to $\vec{B}$ ( $θ = 90 °$ )

Case 2 — Velocity Parallel to $\vec{B}$ ( $θ = 0 °$ or $180 °$ )

Case 3 — Velocity at Angle $θ$ to $\vec{B}$ (General Case)

5. Velocity Selector (Filter)

6. Force on a Current-Carrying Conductor in a Magnetic Field

Relation Between Current Force and Charge Force

Force on an Arbitrary Curved Conductor

7. Radius and Energy of Charged Particles

Comparing Radii of Different Particles (Same $V$ and $B$ )

Solving Magnetic Force Problems — Master Strategy

Practice Questions (JEE / NEET Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

We Value Your Privacy

Edvaya Target

Edvaya Aspire

Take a Free Mock Test

Target Subjects

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Subject Mastery

Magnetic Force on Moving Charges & Lorentz Force

1. Magnetic Force on a Moving Charge

Key Properties of Magnetic Force

SI Unit of Magnetic Field

2. Direction of Magnetic Force — Right-Hand Rule

Fleming's Left-Hand Rule (for positive charge)

Standard Direction Conventions

Quick Force Direction Examples

3. The Lorentz Force

4. Motion of a Charged Particle in a Uniform Magnetic Field

Case 1 — Velocity Perpendicular to B→ (θ=90°)

Case 2 — Velocity Parallel to B→ (θ=0° or 180°)

Case 3 — Velocity at Angle θ to B→ (General Case)

5. Velocity Selector (Filter)

6. Force on a Current-Carrying Conductor in a Magnetic Field

Relation Between Current Force and Charge Force

Force on an Arbitrary Curved Conductor

7. Radius and Energy of Charged Particles

Comparing Radii of Different Particles (Same V and B)

Solving Magnetic Force Problems — Master Strategy

Practice Questions (JEE / NEET Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

We Value Your Privacy

Case 1 — Velocity Perpendicular to $\vec{B}$ ( $θ = 90 °$ )

Case 2 — Velocity Parallel to $\vec{B}$ ( $θ = 0 °$ or $180 °$ )

Case 3 — Velocity at Angle $θ$ to $\vec{B}$ (General Case)

Comparing Radii of Different Particles (Same $V$ and $B$ )