A proton with kinetic energy $1.3384\times10^{-14}\ \text{J}$, moving horizontally from north to south, enters a uniform magnetic field $B$ of $2.0\ \text{mT}$ directed eastward. Calculate : (a) the speed of the proton, (b) the magnitude of acceleration of the proton, (c) the radius of the path traced by the proton. [Take $(q/m)$ for proton $= 1.0\times10^{8}\ \text{C/kg}$]

A proton with kinetic energy 1.3384×10⁻¹⁴J, moving horizontally from north to south, enters a uniform magnetic

Question

1.3384\times10^{-14}\ \text{J}

(a) the speed of the proton,

(b) the magnitude of acceleration of the proton,

(q/m)

for proton

= 1.0\times10^{8}\ \text{C/kg}

]

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Concept understanding — Charged Particle in Magnetic Field

Charged Particle in a Magnetic Field

When a charged particle moves through a magnetic field, the field grabs it sideways. Unlike an electric field, which can speed a charge up or slow it down, a magnetic field only bends the path — it never changes the particle's speed. Understanding why leads directly to circular and helical motion, the basis of cyclotrons, mass spectrometers and the aurora.

The force: always sideways

A particle of charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ feels the magnetic (Lorentz) force:

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

Because of the cross product, $\vec{F}$ is perpendicular to both $\vec{v}$ and $\vec{B}$ . Its magnitude is

F = |q|\,v\,B\sin\theta

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

Important

Since $\vec{F} \perp \vec{v}$ , the force does no work: $\vec{F}\cdot\vec{v} = 0$ . Therefore the kinetic energy and the speed stay constant — the field only changes the direction of motion, never the magnitude.

Case 1: velocity perpendicular to the field → a circle

If $\vec{v} \perp \vec{B}$ ( $\theta = 90^\circ$ ), the force $F = qvB$ stays constant in size and always points toward one central point. That is exactly the condition for uniform circular motion, with the magnetic force acting as the centripetal force:

qvB = \frac{mv^2}{r}

Solving for the radius:

r = \frac{mv}{qB}

The time period of one revolution is

T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}

Note

The period $T$ (and the frequency $f = qB/2\pi m$ , the cyclotron frequency) does not depend on the speed or the radius. A faster particle traces a bigger circle but takes exactly the same time per loop. This speed-independence is what makes the cyclotron work.

Case 2: velocity at an angle → a helix

If $\vec{v}$ makes an angle $\theta$ with $\vec{B}$ , split it into two parts:

Perpendicular component $v_\perp = v\sin\theta$ — feels the magnetic force and drives circular motion of radius $r = \dfrac{mv_\perp}{qB}$ .
Parallel component $v_\parallel = v\cos\theta$ — feels no force (since $\vec{v_\parallel}\times\vec{B}=0$ ) and carries the particle steadily along the field line.

Combining a circle with a steady drift gives a helix. The distance advanced along the field in one full turn is the pitch:

p = v_\parallel\,T = v\cos\theta \cdot \frac{2\pi m}{qB}

A quick example

An electron ( $m = 9.1\times10^{-31}\ \text{kg}$ , $q = 1.6\times10^{-19}\ \text{C}$ ) enters a $0.02\ \text{T}$ field at $10^{6}\ \text{m/s}$ , perpendicular to $\vec{B}$ :

$r = \frac{mv}{qB} = \frac{(9.1\times10^{-31})(10^{6})}{(1.6\times10^{-19})(0.02)} \approx 2.8\times10^{-4}\ \text{m}$ …

Why this formula?

Charged Particle in a Magnetic Field — Why the Key Formulas Hold

Let's build this from first principles. The core idea is that a magnetic field exerts a force only on a moving charge, and that force is always perpendicular to both the velocity and the field.

1. The Fundamental Force Law: Lorentz Force

The starting point is the Lorentz force for a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ :

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

Why this form?

Cross product $\vec{v} \times \vec{B}$ means the force is perpendicular to both $\vec{v}$ and $\vec{B}$ .
Magnitude: $F_m = |q| v B \sin\theta$ , where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$ .
Direction: given by the right-hand rule (for positive $q$ ).

Key insight: Because $\vec{F}_m \perp \vec{v}$ , the magnetic force does no work — it changes only the direction of velocity, not its speed.

2. Circular Motion in a Uniform Magnetic Field

Consider a charge $q$ moving with speed $v$ perpendicular to a uniform $\vec{B}$ (so $\theta = 90^\circ$ , $\sin\theta = 1$ ).

Step 1: Force provides centripetal acceleration

The magnetic force is the only radial force:

F_m = q v B

This must equal the centripetal force required for circular motion:

F_c = \frac{m v^2}{r}

Step 2: Equate and solve for $r$

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

Cancel one $v$ (assuming $v \neq 0$ ):

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

Thus:

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

This is the radius of the circular path (cyclotron radius).

Why this makes sense:

Larger mass $m$ → harder to turn → larger $r$
Larger charge $q$ or stronger $B$ → stronger force → tighter turn → smaller $r$
Faster speed $v$ → more momentum → larger $r$

3. Angular Frequency (Cyclotron Frequency)

From the circular motion relation:

\omega = \frac{v}{r}

Substitute $r = \frac{m v}{q B}$ :

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

Thus:

\boxed{\omega_c = \frac{q B}{m}}

Why this is remarkable:

$\omega_c$ is independent of speed $v$ — all particles with same $q/m$ have the same angular frequency, regardless of how fast they move.
This is the principle behind cyclotrons (particle accelerators).

4. General Motion: Helical Path …

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A charged particle entering a magnetic field perpendicular to its velocity executes uniform circular motion. From the given kinetic energy we find $v = 4.09 \times 10^6 \, \text{m/s}$ , the centripetal acceleration is $a = 8.18 \times 10^{11} \, \text{m/s}^2$ , and the radius of the circular path is $r \approx 20.4 \, \text{m}$ ( $\approx 20\ \text{m}$ ).

When a charged particle moves through a magnetic field, it experiences a Lorentz force perpendicular to both its velocity and the field. Because this force is always perpendicular to the velocity, it does no work—the particle's speed remains constant while the direction continuously changes. The result is uniform circular motion, with the magnetic force providing exactly the centripetal force needed to maintain the circular path.

The proton moves south (horizontal), the field points east (horizontal), so the magnetic force $\vec{F} = q\vec{v} \times \vec{B}$ points vertically—either up or down depending on the sign. The magnitude of this force is $F = qvB$ , and this equals the centripetal force $mv^2/r$ .

(a) Speed of the proton

The kinetic energy relates to speed through:

KE = \frac{1}{2}mv^2

Solving for $v$ :

v = \sqrt{\frac{2 \cdot KE}{m}}

The mass follows from the given charge-to-mass ratio and the proton charge $q = 1.6 \times 10^{-19}\ \text{C}$ :

m = \frac{q}{(q/m)} = \frac{1.6 \times 10^{-19}}{1.0 \times 10^{8}} = 1.6 \times 10^{-27}\ \text{kg}

Substituting the values:

v = \sqrt{\frac{2 \times 1.3384 \times 10^{-14}}{1.6 \times 10^{-27}}} = \sqrt{\frac{2.6768 \times 10^{-14}}{1.6 \times 10^{-27}}} = \sqrt{1.673 \times 10^{13}}

v = 4.09 \times 10^{6}\ \text{m/s} \;(\approx 4.1 \times 10^{6}\ \text{m/s})

(b) Magnitude of acceleration

The magnetic force on the proton is:

F = qvB

This force is perpendicular to the velocity, causing centripetal acceleration:

a = \frac{F}{m} = \frac{qvB}{m} = \frac{q}{m} \cdot vB

Substituting:

$q/m = 1.0 \times 10^8 \, \text{C/kg}$
$v = 4.09 \times 10^6 \, \text{m/s}$
$B = 2.0 \times 10^{-3} \, \text{T}$

$a = 1.0 \times 10^8 \times 4.09 \times 10^6 \times 2.0 \times 10^{-3} = 8.18 \times 10^{11}\ \text{m/s}^2$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 23 on this concept.

CBSE 2026Set V11 markMCQ
$Q. The path traced by a charged particle moving perpendicular to a uniform magnetic field is :$
(a) circle
(b) straight line
(c) helix
(d) ellipse
›Reveal solutionSolution

(a) circle …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The distance travelled by a charged particle in one rotation along the magnetic field is called$
(a) pitch
(b) angular frequency
(c) radius of helix
(d) angular displacement
›Reveal solutionSolution

A charged particle entering a magnetic field with velocity components both along and perpendicular to B moves in a helix; the axial distance covered in one full turn is called the pitch.

The component of velocity perpendicular to B (v_perp) causes circular motion (radius r = m v_perp / qB), while the component along B (v_parallel) is unaffected and produces uniform linear motion along the field direction. The combination is a helix. I …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. A positively charged particle enters in a perpendicular uniform magnetic field, its path will be:$
(a) Elliptical
(b) Parabolic
(c) Linear
(d) Circular
›Reveal solutionSolution

A charged particle moving perpendicular to a uniform magnetic field traces a circle because the magnetic force is always perpendicular to velocity.

When a positive charge $q$ moves with speed $v$ perpendicular to a uniform field $B$ , it experiences a force $F = qvB$ directed perpendicular to $v$ (by $\vec F = q\vec v \times \vec B$ ). This force acts as a centripetal force, constantly changing the direction of veloci …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If a charged particle enters perpendicularly into a uniform magnetic field, then which of the following statements is true?$
(a) Both energy and momentum remain constant.
(b) Energy remains constant, but momentum changes.
(c) Both energy and momentum change.
(d) Energy changes but momentum remains constant.
›Reveal solutionSolution

The magnetic force is always perpendicular to the velocity, so it can never do work on the charge — kinetic energy (and hence speed) stays exactly constant. But the force continuously deflects the particle into a circular path, constantly changing the direction of its momentum vector even while its magnitude is unchanged.

The magnetic force and work done

The force on a charge $q$ moving with velocity $\vec v$ in a magnetic field $\vec B$ is the Lorentz (magnetic) force:

$\vec F=q\,\vec v\times\vec B$

By the definition of the cross product, $\vec F$ is always perpendicular to $\vec v$ . The (infinitesimal) work done by this force over a displacement $d\vec s = \vec v\,dt$ is

$dW=\vec F\cdot d\vec s = (q\,\vec v\times\vec B)\cdot(\vec v\,dt)=0$

because $\vec v\times\vec B$ is perpendicular to $\vec v$ , so its dot product with $\vec v$ is zero. Since $dW=0$ at every instant, the total work done by the magnetic force is always zero.

By the work–energy theorem, since no work is done, the kinetic energy — and hence the speed $|\vec v|$ and hence the magnitude of momentum $|\vec p|=m|\vec v|$ — of the particle remains constant.

Why momentum itself still changes

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. A charged particle enter in a magnetic field perpendicular to the magnetic lines of forces. The path of the charged particle is :$
(a) circular
(b) ellipse
(c) straight line
(d) helical
›Reveal solutionSolution

A charge moving perpendicular to a uniform magnetic field traces a circle, because the magnetic force always acts as a centripetal force.

The magnetic force on a moving charge is $\vec{F} = q\vec{v}\times\vec{B}$ . When $\vec{v} \perp \vec{B}$ , this force has constant magnitude $qvB$ and is always directed perpendicular to $\vec{v}$ — i.e., it always points toward a fixed centre. A force of constant magnitude always perpendicular to velocity is exactly the condition for unifo …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. When a charged particle moves in a uniform magnetic field in a direction perpendicular to the field, then the path of the particle will be -$
(a) Parabolic
(b) Circular
(c) Straight line
(d) Helical
›Reveal solutionSolution

A uniform magnetic force acting always perpendicular to the velocity provides centripetal force, so the particle moves in a circle.

When a charged particle of charge q moves with velocity v perpendicular to a uniform magnetic field B, it experiences a magnetic force:

$\vec{F} = q\vec{v} \times \vec{B}$

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. A charged particle enters at 30° to the magnetic field. Its path becomes :$
(a) circular
(b) helical
(c) elliptical
(d) straight line
›Reveal solutionSolution
Entering at an oblique angle (neither 0° nor 90°) to B gives a helix — the velocity component along B is unaffected, while the perpendicular component causes circular motion.

When a charged particle enters a uniform magnetic field at an angle θ (here 30°) to $\vec B$ , resolve its velocity into two components:

$v_\parallel = v\cos\theta$ , along $\vec B$ : experiences no magnetic force ( $\vec F = q\vec v\times\vec B = 0$ for this component), so the particle drifts uniformly along the field direction.

$v_\perp = v\sin\theta$ , perpendicular to $\vec B$ : experiences a force $qv_\perp B$ that is always perpendicular to this velocity component, producing uniform circular motion in the plane perpendicular to $\vec B$ . …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. A charged particle enters with some speed at 30 degrees to the magnetic field. Its path will be :$
(a) helical
(b) circular
(c) parabolic
(d) None of these
›Reveal solutionSolution

A charge entering at an angle (other than 0 or 90 degrees) to a magnetic field follows a helical path.

Resolve the velocity into a component parallel to B and one perpendicular to B. The parallel component feels no force and gives steady straight-line drift; the perpendicular component feels qvB and gives uniform circular motion. Their superposition is a helix (a circle th …

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CBSE 2022Set ANNUAL1 markMCQ
$\vec{F}$
(a) $\dfrac{q}{\vec{v}\times\vec{B}}$
(b) $\dfrac{\vec{v}\times\vec{B}}{q}$
(c) $q(\vec{v}\times\vec{B})$
(d) $\vec{v}\times\vec{q}\times\vec{B}$
›Reveal solutionSolution
The magnetic force on a charge is $\vec F = q(\vec v \times \vec B)$ : option (C).

A charge $q$ moving with velocity $\vec v$ in a magnetic field $\vec B$ experiences the magnetic part of the Lorentz force:

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

Its magnitude is $F = qvB\sin\theta$ , where $\theta$ is the angle between $\vec v$ and $\vec B$ . …
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CBSE 2022Set ANNUAL1 markMCQ
$Q. A proton enters into a uniform magnetic field perpendicularly to it. The path of the proton would be:$
(a) Elliptical
(b) Circular
(c) Parabolic
(d) Linear
›Reveal solutionSolution

A charge entering $\perp$ to a uniform $\vec B$ moves in a circle, since the magnetic force provides constant centripetal force: option (B).

The magnetic force $\vec F = q(\vec v \times \vec B)$ is always perpendicular to the velocity, so it does no work and the speed stays constant.

When the proton enters perpendicular to $\vec B$ , this constant-magnitude force acts as a centripetal force, bending the path into a circle of radius …

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CBSE 2020Set 55/2/11 markMCQ
$Q. An electron is released from rest in a region of uniform electric and magnetic fields acting parallel to each other. The electron will (A) move in a straight line. (B) move in a circle. (C) remain stationary. (D) move in a helical path.$
›Reveal solutionSolution

When electric and magnetic fields are parallel and a charged particle starts from rest, only the electric field exerts a force initially; the particle accelerates along the field direction in a straight line.

Understanding Forces on a Charged Particle

The motion of a charged particle in electromagnetic fields depends on two forces: the electric force $\vec{F}_E = q\vec{E}$ and the magnetic force $\vec{F}_B = q(\vec{v} \times \vec{B})$ . The key insight here is that the magnetic force has a peculiar property—it acts only when the particle has a velocity component perpendicular to the magnetic field.

When the electric and magnetic fields are parallel to each other, we need to think carefully about what happens at each instant.

Step-by-Step Analysis

1. Initial condition: the electron at rest

The electron starts from rest, so $\vec{v} = 0$ initially. The magnetic force is $\vec{F}_B = q(\vec{v} \times \vec{B}) = 0$ because the velocity is zero. Only the electric force acts:
$\vec{F} = -e\vec{E}$
The electron experiences a force opposite to the electric field direction (since its charge is negative).

2. The electron begins to accelerate

Under the electric force alone, the electron accelerates in a direction parallel to $\vec{E}$ (but opposite in sense). Since $\vec{E} \parallel \vec{B}$ , the velocity that develops is also parallel to $\vec{B}$ .

3. What about the magnetic force as the electron moves?

Once the electron has velocity $\vec{v}$ , we check the magnetic force:
$\vec{F}_B = -e(\vec{v} \times \vec{B})$
But here's the crucial point: $\vec{v}$ is parallel to $\vec{B}$ (both along the same line). The cross product of two parallel vectors is zero:
$\vec{v} \times \vec{B} = 0 \quad \text{when } \vec{v} \parallel \vec{B}$
So the magnetic force remains zero throughout the motion.

Tip

The magnetic force $q(\vec{v} \times \vec{B})$ vanishes whenever velocity is parallel (or antiparallel) to the magnetic field. This is why particles can stream freely along magnetic field lines.

4. The resulting motion …
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CBSE 2020Set 55/3/11 markMCQ
$45°$
›Reveal solutionSolution
A charged particle entering a magnetic field at an angle splits its velocity into parallel (unaffected) and perpendicular (circular motion) components; the combination produces helical motion. The answer is (D).

When a charged particle moves through a magnetic field, the Lorentz force acts on it. The key insight is that the magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ is always perpendicular to both the velocity and the field. This means the force can only act on the component of velocity that is perpendicular to $\vec{B}$ — the parallel component remains completely unaffected.

Think of it this way: the magnetic field "sees" only the perpendicular part of the velocity and bends it into circular motion, while the parallel part continues undisturbed, carrying the particle forward along the field direction. The result is a corkscrew path.

Let me break down what happens to our proton entering at $45°$ :

Decompose the velocity into components If the proton enters with speed $v$ at $45°$ to $\vec{B}$ , we can write:

$v_\parallel = v \cos 45° = \frac{v}{\sqrt{2}}$ $v_\perp = v \sin 45° = \frac{v}{\sqrt{2}}$
The parallel component is along $\vec{B}$ , the perpendicular component lies in the plane normal to $\vec{B}$ .

Analyze the motion due to each component

The magnetic force is $\vec{F} = q\vec{v} \times \vec{B}$ . Since $\vec{v}_\parallel$ is parallel to $\vec{B}$ , their cross product vanishes — no force acts on this component. The proton continues with constant velocity $v_\parallel$ along the field direction.

The perpendicular component $v_\perp$ experiences the full magnetic force, which is always perpendicular to $v_\perp$ itself. This is the classic setup for uniform circular motion in the plane perpendicular to $\vec{B}$ .

Combine the two motions

In the plane perpendicular to $\vec{B}$ : circular motion with radius $r = \frac{m v_\perp}{qB}$ .

Along the direction of $\vec{B}$ : uniform motion with velocity $v_\parallel$ . …
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Q.A proton with kinetic energy 1.3384×10−14 J1.3384\times10^{-14}\ \text{J}1.3384×10−14 J, moving horizontally from north to south, enters a uniform magnetic field BBB of 2.0 mT2.0\ \text{mT}2.0 mT directed eastward. Calculate :

(a) Speed of the proton

(b) Magnitude of acceleration

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Q.A proton with kinetic energy $1.3384\times10^{-14}\ \text{J}$ , moving horizontally from north to south, enters a uniform magnetic field $B$ of $2.0\ \text{mT}$ directed eastward. Calculate :

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