A charged particle of charge $e$ and mass $m$ is moving in an electric field $\vec{E}$ and magnetic field $\vec{B}$. Construct dimensionless quantities and quantities of dimension $[T]^{-1}$.

A charged particle of charge e and mass m is moving in an electric field vecE and magnetic field vecB. Constru

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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 key idea is to use dimensional analysis to combine $e$ , $m$ , $E$ , and $B$ into quantities with dimensions of $[T]^{-1}$ (frequency). The cyclotron frequency $\omega_c = eB/m$ is one such quantity, and another is $\sqrt{eE/md}$ where $d$ is a length scale — but without a length scale, only $eB/m$ is purely $[T]^{-1}$ from the given parameters.

Why Magnetic Force Balance?

When a charged particle moves through electric and magnetic fields, two forces act on it: the electric force $\vec{F}_E = e\vec{E}$ and the magnetic force $\vec{F}_B = e(\vec{v} \times \vec{B})$ . The magnetic force depends on velocity, which introduces a natural timescale into the problem.

The most famous frequency that emerges is the cyclotron frequency — the rate at which a charged particle gyrates in a uniform magnetic field. This comes from equating the magnetic force to the centripetal force: $evB = m v^2 / r$ , giving $\omega = v/r = eB/m$ . Notice that this frequency depends only on $e$ , $B$ , and $m$ — not on $v$ or $r$ individually.

But the problem also includes an electric field $E$ . Can we construct another frequency involving $E$ ? Let's see.

Step-by-step construction

1. Identify the dimensions of each quantity

First, let's write down the dimensions of everything we have:

Charge $e$ : $[e] = [\text{current} \times \text{time}] = [I T]$
Mass $m$ : $[m] = [M]$
Electric field $E$ : $[E] = [\text{force/charge}] = [M L T^{-2}]/[I T] = [M L T^{-3} I^{-1}]$
Magnetic field $B$ : $[B] = [\text{force/(charge × velocity)}] = [M L T^{-2}]/[I T \cdot L T^{-1}] = [M T^{-2} I^{-1}]$

We want quantities with dimension $[T]^{-1}$ , i.e., frequency.

2. The obvious candidate: $eB/m$

Take the combination $eB/m$ :

$[eB] = [I T] \cdot [M T^{-2} I^{-1}] = [M T^{-1}]$
Dividing by $m$ : $[eB/m] = [M T^{-1}]/[M] = [T^{-1}]$

So $\omega_c = \frac{eB}{m}$ is a frequency. This is the cyclotron frequency — the angular frequency of circular motion in a pure magnetic field.

\omega_c = \frac{eB}{m}

3. Can we get another frequency involving $E$ ?

Try $eE/m$ :

$[eE] = [I T] \cdot [M L T^{-3} I^{-1}] = [M L T^{-2}]$
$[eE/m] = [M L T^{-2}]/[M] = [L T^{-2}]$

This has dimensions of acceleration, not frequency. To get $[T^{-1}]$ , we'd need to divide by a velocity or multiply by a time — but we don't have those as independent parameters.

What about $\sqrt{eE/md}$ ? If we introduce a length $d$ , then:

$[eE/md] = [L T^{-2}]/[L] = [T^{-2}]$
Taking square root gives $[T^{-1}]$

But the problem only gives $e$ , $m$ , $E$ , and $B$ — no length scale. So without additional parameters, $eE/m$ alone cannot yield a frequency.

Watch out

A common mistake is to think $eE/m$ gives a frequency. It doesn't — it gives acceleration. You need a length or velocity to convert it into a frequency.

4. The mixed combination: $E/B$

What about $E/B$ ?

$[E/B] = [M L T^{-3} I^{-1}]/[M T^{-2} I^{-1}] = [L T^{-1}]$

This is a velocity — the drift velocity $v_d = E/B$ for crossed fields. If we then divide by a length, we get frequency. But again, no length is given.

5. What dimensionless quantities can we form?

A dimensionless quantity requires the product of powers of $e$ , $m$ , $E$ , $B$ to have dimension $[M^0 L^0 T^0 I^0]$ . Let's try:

$e^a m^b E^c B^d$ has dimension:

$[I^a T^a] [M^b] [M^c L^c T^{-3c} I^{-c}] [M^d T^{-2d} I^{-d}]$

Collecting: …

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Q.A charged particle of charge $e$ and mass $m$ is moving in an electric field $\vec{E}$ and magnetic field $\vec{B}$ . Construct dimensionless quantities and quantities of dimension $[T]^{-1}$ .

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 …

Why Magnetic Force Balance?

Step-by-step construction

Method: Constructing Quantities of a Target Dimension From a Given Set of Symbols

Steps

Q.A charged particle of charge eee and mass mmm is moving in an electric field E⃗\vec{E}E and magnetic field B⃗\vec{B}B. Construct dimensionless quantities and quantities of dimension [T]−1[T]^{-1}[T]−1.

Why Magnetic Force Balance?

Step-by-step construction

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Q.A charged particle of charge $e$ and mass $m$ is moving in an electric field $\vec{E}$ and magnetic field $\vec{B}$ . Construct dimensionless quantities and quantities of dimension $[T]^{-1}$ .

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