# Why is the magnetic field perpendicular to the magnetic force?

In the case of the electric field, we define the field as the force at each point in space exerted on a unit charge. This is intuitive, as the field will give me a good idea of how a charge will move in space under the influence of the field. But in the case of the magnetic field, what is the significance in defining the magnetic field in a way such that we must take the cross product to find the magnetic force: $$F_B = qv \times B$$ Couldn't we define it in a way similar to the electric field, such that the magnetic force points in the same direction as the field? I understand that a charged particle follows a circular pattern under the influence of a steady magnetostatic field, so it may lead to simpler equations for B this way, but is this the only reason?

The equivalent for magnetic force would be $F_M = q_mB$ where $q_m$ is the magnetic charge, and the Lorentz force due to the interaction of an electric field and magnetic charge would involve the cross product - so it has a nice symmetry if magnetic monopoles exist.

The most succinct answer is relativistic covariance. The scalar electric potential and the vector magnetic potential are the components of a four-vector - the four-potential.

The electric and magnetic fields are then components of a four-tensor - the Faraday tensor.

This is so that the electric and magnetic fields transform properly under a Lorentz transformation.

Interestingly, when thinking in terms of spacetime and four-vectors, the electromagnetic four-force on a particle is (Minkowski) orthogonal to the particle's four-velocity.

Simply put, a particle's four-velocity has constant length and thus, the four-acceleration must by (Minkowski) orthogonal, i.e., acceleration can only change the direction of the four-velocity, not the length.

The Lorentz force, expressed in four-vector notation is

$$\frac{dp_{\alpha}}{d\tau} = qF_{\alpha \beta}u^{\beta}$$

The left hand side is the four-force, the right hand side is the product of the charge of the particle with the contraction of the Faraday tensor and the four-velocity.

For a particle at rest, the four-velocity points in the time direction and it's straight forward to show that the four-force is due to the electric field and points in a space direction, i.e., the four-force due to the electric field is orthogonal to the particle's four-velocity.

Thus, we see that, in the relativistic four-vector context, both the electric and magnetic forces are orthogonal to the particle's four-velocity.

To summarize, starting with the manifestly covariant four-potential and its exterior derivative, the Faraday tensor, and the Lorentz force law, we find that the 3D + 1 expression for the Lorentz force is

$$\frac{d\vec p}{dt} = q(\vec E + \vec v \times \vec B)$$

$$\frac{dE}{dt} = q\vec v \cdot \vec E$$

where $E$ is the particle's energy.

While not sure, If I completely understand the question.However, I see where this sort of relation comes from. If you look at the Maxwell's equations, specifically the Faraday's law of induction and Ampère's circuital law. It becomes clear that Electric field is proportional to the curl of magnetic field and vice versa. Therefore, it would be unreasonable to expect similar relation between electric and magnetic forces and their respective fields.