Animated Solution for Physics - Magnetic Effects of Current: A particle of mass $M$ and charge $Q$ moving with velocity $\mathbf{v}$ describes a circular path of radius $R$ when subjected to a uniform transverse magnetic field of induction $B$. The work done by the field when the particle completes one full circle is

A Dance in the Magnetic Field

Imagine a charged particle, possessing mass $M$ and charge $Q$, entering a region where a uniform magnetic field $\mathbf{B}$ exists.

If the particle's velocity $\mathbf{v}$ is perfectly perpendicular to this magnetic field, something beautiful happens. The particle gets trapped in a continuous, perfect circular path of radius $R$.

But here is the million-dollar question: as the particle completes one full circle, how much work does the magnetic field actually do on it?

Lorentz Force

To answer this, we must look at the fundamental law governing moving charges in magnetic fields. The magnetic Lorentz force acting on the particle is given by the cross product:

By the very mathematical definition of a cross product, the resulting force vector $\mathbf{F}$ is always strictly perpendicular to both the velocity vector $\mathbf{v}$ and the magnetic field vector $\mathbf{B}$.

This means that at every single instant of the particle's journey, the angle between its velocity and the force acting upon it is exactly $90^{\circ }$.

Now, let's bring in the concept of power. In physics, the mechanical power $P$ delivered by a force is the dot product of the force and the velocity:

Since the angle $\theta$ between the force and velocity is always $90^{\circ }$, and we know that $\cos (90^{\circ })=0$, the power delivered by the magnetic field is exactly zero.

The Work Done

Work done is simply the total power delivered over a period of time. Mathematically, it is the integral of power:

Because the power is zero at every infinitesimal moment, the total work done over any time interval is also zero. It does not matter if the particle completes a quarter circle, a half circle, or one full circle. The work done by a static magnetic field on a moving charge is always, unequivocally, zero.

We can also understand this through the Work-Energy Theorem, which states that the net work done on an object equals its change in kinetic energy:

Since $W=0$, the change in kinetic energy $\Delta K$ must also be zero. This tells us that the kinetic energy, and therefore the speed of the particle, remains absolutely constant.

The magnetic field acts purely as a steering wheel, never as an accelerator or a brake. It changes the direction of the velocity vector, but never its magnitude!