Verify that the cyclotron frequency $\omega = eB/m$ has the correct dimensions of $[T]^{-1}$.

> [!TLDR] > The cyclotron frequency $\omega = eB/m$ has dimensions of $[T]^{-1}$ because the Lorentz force law $F = qvB$ gives $[eB] = [M][T]^{-1}$ and dividing by mass $[M]$ leaves $[T]^{-1}$. The key insight is that dimensions must match on both sides of any physical equation. For cyclotron frequency, we're checking that $\omega = eB/m$ indeed gives inverse time — the unit of frequency. Let's work through this systematically. 1. **Start with what we know about dimensions.** Frequency $\omega$ has dimensions of $[T]^{-1}$ — that's what we need to verify. The right side is $eB/m$, so we need the dimensions of $e$, $B$, and $m$. 2. **Mass is straightforward.** Mass $m$ has dimension $[M]$. 3. **For charge $e$ and magnetic field $B$, we need a physical relation.** The Lorentz force gives us the link: a charge $q$ moving with velocity $v$ in a magnetic field $B$ experiences force $F = qvB$. 4. **Write this dimensionally.** Force has dimensions $[M][L][T]^{-2}$. Velocity has $[L][T]^{-1}$. So: $$[F] = [q][v][B]$$ $$[M][L][T]^{-2} = [e][L][T]^{-1}[B]$$ 5. **Solve for $[eB]$.** Multiply both sides by $[T]$: $$[M][L][T]^{-1} = [e][L][B]$$ Cancel $[L]$: $$[M][T]^{-1} = [e][B]$$ > [!NOTE] > This is a neat result: the product $eB$ has dimensions $[M][T]^{-1}$ — mass per unit time. 6. **Now divide by mass.** The cyclotron frequency is: $$\omega = \frac{eB}{m}$$ Dimensionally: $$[\omega] = \frac{[eB]}{[m]} = \frac{[M][T]^{-1}}{[M]} = [T]^{-1}$$ This matches exactly what we expect for frequency. …

Verify that the cyclotron frequency ω = eB/m has the correct dimensions of [T]⁻¹.

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\omega = eB/m

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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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The cyclotron frequency $\omega = eB/m$ has dimensions of $[T]^{-1}$ because the Lorentz force law $F = qvB$ gives $[eB] = [M][T]^{-1}$ and dividing by mass $[M]$ leaves $[T]^{-1}$ .

The key insight is that dimensions must match on both sides of any physical equation. For cyclotron frequency, we're checking that $\omega = eB/m$ indeed gives inverse time — the unit of frequency.

Let's work through this systematically.

Start with what we know about dimensions. Frequency $\omega$ has dimensions of $[T]^{-1}$ — that's what we need to verify. The right side is $eB/m$ , so we need the dimensions of $e$ , $B$ , and $m$ .
Mass is straightforward. Mass $m$ has dimension $[M]$ .
For charge $e$ and magnetic field $B$ , we need a physical relation. The Lorentz force gives us the link: a charge $q$ moving with velocity $v$ in a magnetic field $B$ experiences force $F = qvB$ .
Write this dimensionally. Force has dimensions $[M][L][T]^{-2}$ . Velocity has $[L][T]^{-1}$ . So:

[F] = [q][v][B]

[M][L][T]^{-2} = [e][L][T]^{-1}[B]

Solve for $[eB]$ . Multiply both sides by $[T]$ :

[M][L][T]^{-1} = [e][L][B]

Cancel $[L]$ :

[M][T]^{-1} = [e][B]

Note

This is a neat result: the product $eB$ has dimensions $[M][T]^{-1}$ — mass per unit time.

Now divide by mass. The cyclotron frequency is:

\omega = \frac{eB}{m}

Dimensionally:

[\omega] = \frac{[eB]}{[m]} = \frac{[M][T]^{-1}}{[M]} = [T]^{-1}

This matches exactly what we expect for frequency. …

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

KEAM 2025Set pha-2025-0424F4 marksMCQ
$Q. When a proton moves in a uniform magnetic field such that its velocity has a component along the direction of magnetic field, its trajectory will be a (A) circle (B) straight line (C) helix (D) parabola (E) ellipse$
›Reveal solutionSolution

The parallel velocity component moves the proton uniformly along $\vec{B}$ while the perpendicular component makes it circle; together these produce a helix.

The magnetic force $\vec{F}=q\vec{v}\times\vec{B}$ acts only on the velocity component perpendicular to $\vec{B}$ , causing circular motion in that plane. The component of velocity along $\vec{B}$ experiences no force and stays constant, giving uniform …
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KEAM 2025Set pha-2025-0429F4 marksMCQ
$Q. If an electron moves with a velocity v in a magnetic field B, the magnetic force on the electron is maximum when the angle between v and B is (A) 30º (B) 180º (C) 60º (D) 90º (E) 0º$
›Reveal solutionSolution

The Lorentz magnetic force magnitude is $F = qvB\sin\theta$ , which is greatest when the velocity is perpendicular to the field, i.e. $\theta = 90^\circ$ .

The force on a charge moving in a magnetic field is

$\vec F = q\,\vec v \times \vec B, \qquad |\vec F| = qvB\sin\theta.$ …
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KEAM 2024Set eng-2024-06054 marksMCQ
$B$
›Reveal solutionSolution
The perpendicular velocity component makes a circle while the parallel component moves uniformly along $B$ — the combination is a helix.

Resolve $v$ into components parallel and perpendicular to $B$ :

The perpendicular component gives uniform circular motion (magnetic force $qv_\perp B$ ). …
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KEAM 2024Set pha-2024-06104 marksMCQ
$3 \mu C$
›Reveal solutionSolution

Compute the Lorentz force $q\vec v\times\vec B$ .

Given $q = 3\,\mu C = 3\times10^{-6}$ C, $\vec v = (\hat i + \hat j)\times10^5$ m/s, $\vec B = 5\hat j$ T.

$\vec v\times\vec B = 10^5(\hat i+\hat j)\times 5\hat j = 5\times10^5\,(\hat i\times\hat j + \hat j\times\hat j) = 5\times10^5\,\hat k.$ …
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KEAM 2023Set eng-2023-P1-A14 marksMCQ
$\left(10^{-4}\hat{k}\right)\mathrm{T}$
›Reveal solutionSolution
With F perpendicular to B, v = F/(qB) = (5 x 10^-12)/(10^-9 x 10^-4) = 50 m/s.

Concept and Intuition

The magnetic force is F = q v x B. Here B is along z and the force lies in the x-y plane, so the velocity component producing the force is perpendicular to B; the magnitude relation reduces to F = q v B.

Step-by-Step Solution

Magnitude of force: |F| = sqrt(4^2 + 3^2) x 10^-12 = 5 x 10^-12 N.

B = 10^-4 T, q = 10^-9 C, and v is perpendicular to B. …
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KEAM 2022Set eng-2022-P1-A14 marksMCQ
$v$
›Reveal solutionSolution
Both the electron and proton move perpendicular to $\vec{B}$ , so each follows a circular path.

Concept and Intuition

A charged particle entering a uniform magnetic field with velocity perpendicular to the field feels a force $\vec{F} = q\vec{v}\times\vec{B}$ that is always perpendicular to the velocity and constant in magnitude. This is a centripetal force, producing uniform circular motion. Both the electron and proton are charged and moving perpendicular to $\vec{B}$ .

Step-by-Step Solution

Force magnitude $F = qvB$ (since $v \perp B$ ), always perpendicular to $v$ .

A constant perpendicular force gives circular motion of radius $r = mv/qB$ . …
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←Previous questionQ10 — A charged particle would continue to move with a constant velocity in …→Next questionQ12 — Show that a force that does no work must be a velocity dependent force…

Q.Verify that the cyclotron frequency $\omega = eB/m$ has the correct dimensions of $[T]^{-1}$ .

Concept understanding — Charged Particle in Magnetic Field

Charged Particle in a Magnetic Field

The force: always sideways

Case 1: velocity perpendicular to the field → a circle

Case 2: velocity at an angle → a helix

A quick example

Why this formula?

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

1. The Fundamental Force Law: Lorentz Force

Why this form?

2. Circular Motion in a Uniform Magnetic Field

Step 1: Force provides centripetal acceleration

Step 2: Equate and solve for $r$

Why this makes sense:

3. Angular Frequency (Cyclotron Frequency)

Why this is remarkable:

4. General Motion: Helical Path …

Method: Verifying a Formula's Dimensions by Re-Deriving It From a Known Force Law

Steps

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.Verify that the cyclotron frequency ω=eB/m\omega = eB/mω=eB/m has the correct dimensions of [T]−1[T]^{-1}[T]−1.

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Q.Verify that the cyclotron frequency $\omega = eB/m$ has the correct dimensions of $[T]^{-1}$ .

Unlock everything free for 14 days