Mathematical Proof That Magnetic Force Does No Work
In the study of electromagnetism, the behavior of a charged particle moving through electromagnetic fields is governed by the Lorentz Force. For a particle with charge $q$ moving with an instantaneous velocity $\mathbf{v}$ in the presence of an electric field $\mathbf{E}$ and a magnetic field $\mathbf{B}$, the total force $\mathbf{F}$ is expressed as:
$$\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})$$
This force is composed of two distinct components:
- The Electric Force ($\mathbf{F}_e = q\mathbf{E}$): This force acts along the direction of the electric field lines (for positive charges) and can either accelerate or decelerate the particle.
- The Magnetic Force ($\mathbf{F}_m = q(\mathbf{v} \times \mathbf{B})$): Also known as the Ampère force, its direction is determined by the right-hand rule and is always perpendicular to both the velocity of the particle and the magnetic field.
A fundamental property of the magnetic force is that it does no work on the particle. Below is the mathematical proof of this claim and an analysis of its physical implications.
To demonstrate that the magnetic force does no work, we can approach the problem from two perspectives: instantaneous power and the integral of work.
1. Proof via Instantaneous Power
In classical mechanics, the instantaneous power $P$ delivered by a force $\mathbf{F}$ to an object moving with velocity $\mathbf{v}$ is defined as the dot product (scalar product) of the force and the velocity:
$$P = \mathbf{F} \cdot \mathbf{v}$$
Substituting the expression for the magnetic force $\mathbf{F}_m = q(\mathbf{v} \times \mathbf{B})$ into the power formula, we get:
$$P_m = [q(\mathbf{v} \times \mathbf{B})] \cdot \mathbf{v}$$
According to the properties of the scalar triple product in vector calculus, the vector resulting from a cross product ($\mathbf{v} \times \mathbf{B}$) is, by definition, orthogonal (perpendicular) to both original vectors, $\mathbf{v}$ and $\mathbf{B}$.
Since the vector $(\mathbf{v} \times \mathbf{B})$ is always perpendicular to $\mathbf{v}$, the angle $\theta$ between them is $90^\circ$. Given that the dot product $\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}||\mathbf{B}|\cos\theta$, and $\cos(90^\circ) = 0$, it follows that:
$$(\mathbf{v} \times \mathbf{B}) \cdot \mathbf{v} = 0 \implies P_m = 0$$
2. Proof via the Work Integral
Work $W$ is defined as the integral of the force along the path of displacement. Mathematically, this is expressed as:
$$W = \int_{t_1}^{t_2} \mathbf{F}_m \cdot \mathbf{v} , dt$$
Substituting the magnetic force expression:
$$W = \int_{t_1}^{t_2} q(\mathbf{v} \times \mathbf{B}) \cdot \mathbf{v} , dt$$
As established in the power proof, the term $(\mathbf{v} \times \mathbf{B}) \cdot \mathbf{v}$ is identically zero at every single instant $t$ during the particle's motion. Therefore, the integral of zero over any time interval is zero:
$$W = \int_{t_1}^{t_2} 0 , dt = 0$$
Conclusion: The work done by the magnetic force on a moving charge is always zero.
Physical Intuition and Interpretation
While the mathematics are definitive, the physical "picture" provides a deeper understanding of how particles behave in magnetic fields.
The "Steering" Force
The magnetic force acts as a deflecting force rather than an accelerating force. Because $\mathbf{F}_m$ is always perpendicular to the direction of motion, it cannot change the magnitude of the velocity (the speed); it can only change the direction of the velocity vector. In essence, the magnetic field "steers" the particle without adding or removing energy from it.
Perspective from the Work-Energy Theorem
The Work-Energy Theorem states that the change in kinetic energy $\Delta K$ of a particle is equal to the net work done on it:
$$\Delta K = W_{net}$$
In a region where only a magnetic field is present ($W_{net} = W_m = 0$), the change in kinetic energy is zero:
$$\Delta K = 0 \implies \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = 0 \implies v_f = v_i$$
This confirms that a particle moving in a pure magnetic field maintains a constant speed.
Comparison: Magnetic vs. Electric Forces
To highlight the uniqueness of the magnetic force, it is helpful to contrast it with the electric force:
- Electric Force ($\mathbf{F}_e = q\mathbf{E}$): The force is independent of the particle's velocity. It can act parallel to the motion, meaning it can do work, change the particle's speed, and alter its kinetic energy.
- Magnetic Force ($\mathbf{F}_m = q(\mathbf{v} \times \mathbf{B})$): The force is velocity-dependent and strictly perpendicular to the motion. It acts like a centripetal force, altering the trajectory without affecting the energy.
Practical Example: Motion in a Uniform Magnetic Field
Consider a particle of charge $q$ and mass $m$ entering a uniform magnetic field $\mathbf{B}$ with an initial velocity $v_0$ perpendicular to the field lines.
- Trajectory: The force $\mathbf{F}_m = q(\mathbf{v} \times \mathbf{B})$ acts as a centripetal force, pulling the particle into a circular path.
- Mathematical Balance: The magnetic force provides the necessary centripetal acceleration:
$$qv_0B = \frac{mv_0^2}{r} \implies r = \frac{mv_0}{qB}$$ - Energy Analysis: Throughout this circular motion, the force $\mathbf{F}_m$ is always tangent to the circle's radius and perpendicular to the velocity vector. Consequently, the speed $v_0$ remains constant, and the kinetic energy $K = \frac{1}{2}mv_0^2$ remains unchanged.
This scenario perfectly illustrates the principle: the particle is under the influence of a constant force (causing a change in direction), yet the energy of the particle remains perfectly conserved.
Summary
The fact that magnetic forces do no work is a cornerstone of electrodynamics. By proving that $\mathbf{F}_m \cdot \mathbf{v} = 0$, we establish that magnetic fields cannot change the kinetic energy of a charged particle. This property is exploited in various scientific instruments—such as mass spectrometers, cyclotrons, and particle accelerators—where magnetic fields are used to precisely bend and focus particle beams without altering their energy levels.