Is the work-done by a magnetic field on a moving point charge zero?

Question

Blue vector is the magnetic field vector. Red vector is the velocity vector of the point charge (light grey) q. Green vector is the force vector (force on moving charge q due to the magnetic field). I have heard that workdone by a magnetic field on a moving charge is zero. Let's say that a charge q which was already moving with a velocity 'V' came under the influence of a magnetic field 'B'. It started experiencing a force 'F' due to the magnetic field given by F = q(V vector X B vector). Now due to this it will also start moving along the 'y' axis with some acceleration. So wouldn't we say that the work-done = F x displacement in y ?
Edit: What I am trying to say is that the displacement of the charge will have a 'y' component and the force here is also acting in the 'y' direction', so shouldn't the expression I wrote above be valid?

Answer 1

Work done is defined as the dot product between displacement and Force,

$$W=\mathbf{F} \cdot \mathbf{s}$$

where both Force $\mathbf{F}$ and displacement $\mathbf{s}$ are vectors and hence in bold. You may know that if two vectors are perpendicular, then their dot product is zero.

If we plug the expression of the Lorentz Force in the above equation, we get:

$$W=\mathbf{F}\cdot\mathbf{s}=q(\mathbf{v}\times\mathbf{B})\cdot\mathbf{s}$$

Now, note that Lorentz force is perpendicular to the velocity (it's a property of the cross product!!). Also note that the displacement is parallel to the velocity, this is because the displacement is in the direction where you are moving.

Hence, in the expression I just wrote, there is a dot product between a force which is perpendicular to the velocity and a displacement which is parallel to the velocity. Hence the expression gives $W=0$, since the $\mathbf{F}$ and $\mathbf{s}$ are perpendicular!

It's like going on a bicycle on a flat ground just with your inertia. You exert a force on the bicycle towards the ground, but you move forwards hence no work is done (if you are not using your legs of course).

Answer 2

The work done by magnetic field in this case is zero. Why? Because the magnetic field always acts perpendicular to the velocity, thus

$$W=\int_{\mathbf s_1}^{\mathbf s_2} \mathbf F\cdot \mathrm d\mathbf s=\int_{t_1}^{t_2} (\mathbf F\cdot \mathbf v) \mathrm d t=0 \tag{\(\because \mathbf F\perp \mathbf v\)}$$

Now, you might be wondering that how does this reconcile with the acceleration in the $y$ direction that you observed. Well, the reason why you ended up with wrong conclusions is because you only considered the work done by the component of the magnetic force in the $y$ direction. The correct version of your statement should be

\begin{align} W&=\int_{\mathbf s_1}^{\mathbf s_2} (\mathbf F_x +\mathbf F_y)\cdot (\mathrm d\mathbf s_x +\mathrm d \mathbf s_y)\\ &=\underbrace{\int_{s_{x1}}^{ s_{x2}} F_x\: \mathrm ds_x}_{\large{<\:0}}+\underbrace{\int_{ s_{y1}}^{ s_{y2}} F_y\: \mathrm ds_y}_{\large{>\:0}}=0 \end{align}

(Here $\mathbf F_x$, $\mathbf F_y$ and $\mathbf s_x$ and $\mathbf s_y$ are the forces and displacements, respectively, in the $x$ and $y$ directions, respectively)

As you can see that despite the vertical force of magnetic force doing some positive work, the horizontal component of the force would have done equal and opposite negative work, thus cancelling out each other to yield a zero net work done by the magnetic force.