Show that a force that does no work must be a velocity dependent force.

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Q. Show that a force that does no work must be a velocity dependent force.

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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 that if a force does no work on a particle, its instantaneous power $\vec{F} \cdot \vec{v}$ must be zero at all times. This forces $\vec{F}$ to be perpendicular to $\vec{v}$ , which means $\vec{F}$ must depend on $\vec{v}$ — otherwise, you could always find a path where the force does work.

Why This Approach Works

The statement "a force that does no work must be velocity dependent" is a logical consequence of how work is defined. Work is the line integral of force along a path: $W = \int \vec{F} \cdot d\vec{r}$ . If this integral is zero for every possible path between two points, then the force must be perpendicular to the displacement at every instant. But displacement is along the velocity direction ( $d\vec{r} = \vec{v}\, dt$ ), so $\vec{F} \cdot \vec{v} = 0$ at all times.

Now, if $\vec{F}$ were independent of velocity — say, a function of position only, like gravity — then you could always choose a path where $\vec{F}$ has a component along the motion, and work would be done. The only way to guarantee $\vec{F} \cdot \vec{v} = 0$ for all possible motions is for $\vec{F}$ to depend on $\vec{v}$ itself, so that it always adjusts to be perpendicular to the instantaneous velocity.

Tip

The classic example is the magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ . It is always perpendicular to $\vec{v}$ , so it does no work. But notice: it depends explicitly on $\vec{v}$ — if you change the velocity, the force changes direction to stay perpendicular.

Step-by-Step Reasoning

Define "does no work" precisely. A force does no work if the work done by it is zero for every possible path the particle takes. Mathematically, for any path from point A to point B:

W_{AB} = \int_A^B \vec{F} \cdot d\vec{r} = 0

This must hold for all paths, not just one particular path.

Relate work to instantaneous power. Since $d\vec{r} = \vec{v}\, dt$ , the work integral becomes:

W = \int \vec{F} \cdot \vec{v} \, dt

For this to be zero for all possible motions, the integrand $\vec{F} \cdot \vec{v}$ must be zero at every instant. If it were nonzero at any time, you could choose a short path around that instant and get nonzero work. Therefore:

\vec{F} \cdot \vec{v} = 0 \quad \text{at all times}

Interpret the condition.

The dot product being zero means $\vec{F}$ is always perpendicular to $\vec{v}$ . So the force can never have a component along the direction of motion — it can only change the direction of velocity, not its magnitude.
Why must $\vec{F}$ depend on $\vec{v}$ ?

Suppose $\vec{F}$ did not depend on velocity — say, it was a function of position only: $\vec{F} = \vec{F}(\vec{r})$ . Then for a given position, $\vec{F}$ is fixed. But the particle's velocity $\vec{v}$ at that position can be in any direction (depending on initial conditions). For $\vec{F} \cdot \vec{v} = 0$ to hold for all possible velocities at that point, $\vec{F}$ would have to be zero — because a fixed vector cannot be perpendicular to every possible direction. …

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Q.Show that a force that does no work must be a velocity dependent force.

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 This Approach Works

Step-by-Step Reasoning

Method: Proving a Force Must Be Velocity-Dependent From "Does No Work"

Steps

Q.Show that a force that does no work must be a velocity dependent force.

Why This Approach Works

Step-by-Step Reasoning

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