Exercise 3

The following vector potentials are given: \[ \vec{A} = \left(\begin{matrix} 0 \\ bx \\ 0 \end{matrix} \right) \]

and \[ \vec{A'} = \left(\begin{matrix} -by \\ 0 \\ 0 \end{matrix} \right) \]

The curl of \(\vec{A}\) is calculated as \[ \vec{B} = \vec{\nabla} \times \vec{A} = \left(\begin{matrix} 0 \\ 0 \\ b \end{matrix} \right) \]

since the only component which doesn’t vanish is \(\frac{\partial A_y}{\partial x} = b\).

The curl of \(\vec{A'}\) is also calculated as \[ \vec{B} = \vec{\nabla} \times \vec{A'} = \left(\begin{matrix} 0 \\ 0 \\ b \end{matrix} \right) \]

since the only component which doesn’t vanish is \(-\frac{\partial A'_x}{\partial y} = b\).

Both vector potentials are describing the same magnetic Field. This means, the vector potentials differ by the gradient of a scalar field \(s\):

\[ \vec{A'} = \vec{A} + \vec{\nabla} s \]

Or in component notation: \[ -by = 0 + \frac{\partial s}{\partial x} \Rightarrow s = -bxy + C \] \[ 0 = bx + \frac{\partial s}{\partial y} \Rightarrow s = -bxy + C \] \[ 0 = 0 + \frac{\partial s}{\partial z} \Rightarrow s = s(x, y)\]

The scalar field \[ s(x, y, z) = -bxy + C \] satisfies all three conditions.

The gradient of \(s\) is \[ \vec{\nabla}s = \left(\begin{matrix} -by \\ -bx \\ 0 \end{matrix} \right) \]

and the curl of this gradient is equal to zero: \[ \vec{\nabla} \times \vec{\nabla}s = 0 \]