## Is magnetic potential vector quantity?

Magnetic fields can be written as the gradient of magnetic scalar potential. It is a vector field. So electric fields and magnetic fields are vector quantities.

## Is magnetic vector potential continuous?

potential are continuous. The discontinuity in the tangential component of the magnetic field, however, translates to a discontinuity in the normal derivative of the vector potential.

**Is electric flux is a vector quantity?**

Is Electric flux a scalar or a vector quantity? Answer: Electric flux is a scalar quantity. It is a scalar because it is the dot product of two vector quantities, electric field and the perpendicular differential area.

**Is electric field a vector quantity?**

Electric field strength is a vector quantity; it has both magnitude and direction. The magnitude of the electric field strength is defined in terms of how it is measured.

### Is vector potential conservative?

The vector field F is indeed conservative. Since F is conservative, we know there exists some potential function f so that ∇f=F.

### Is the vector potential admitted by a solenoidal field unique?

The vector potential admitted by a solenoidal field is not unique. If A is a vector potential for v, then so is where f is any continuously differentiable scalar function.

**How is a vector potential related to a vector field?**

In vector calculus, a vector potential is a vector field whose curl is a given vector field. This is analogous to a scalar potential, which is a scalar field whose gradient is a given vector field. v = ∇ × A . {displaystyle mathbf {v} = abla imes mathbf {A} .}

**Which is vector potential for V is nonunique?**

If A is a vector potential for v, then so is where m is any continuously differentiable scalar function. This follows from the fact that the curl of the gradient is zero. This nonuniqueness leads to a degree of freedom in the formulation of electrodynamics, or gauge freedom, and requires choosing a gauge .

#### Why is the force on a small current loop proportional to the derivative?

Only now we see why it is that the force on a small current loop is proportional to the derivative of the magnetic field, as we would expect from FxΔx = − ΔUmech = − Δ( − μ ⋅ B).