Question

The vector field \(\mathbf{F}\) for a region is described by \(\mathbf{F} = a r^n \hat{r}\) \((r\neq 0)\), where \(a\) is a non-zero constant and \(r\) is the radial distance from the source. For what value(s) of \(n\), \(\mathbf{F}\) becomes both solenoidal and irrotational?

• A. 2
• B. -2
• C. 1
• D. -1

Hint:

The curl of any purely radial field is automatically zero for any n; the real constraint comes from setting the divergence to zero.

Answer 1

The Correct Option is B

An alternative route uses the scalar potential. Any irrotational field can be written as \(\mathbf{F}=-\nabla\phi\) for some potential \(\phi(r)\). Matching \(-d\phi/dr = ar^n\) for the purely radial field gives \(\phi(r) = -\dfrac{ar^{n+1}}{n+1}\) (for \(n\neq -1\)) - a valid scalar potential exists for every such \(n\), confirming \(\mathbf{F}\) is a gradient (hence irrotational) field for any \(n\), consistent with the direct curl argument.

Now impose the additional solenoidal requirement \(\nabla\cdot\mathbf{F}=0\), i.e. \(\nabla^2\phi=0\) (Laplace's equation) away from the source. Using the radial Laplacian and \(d\phi/dr=-ar^n\) reproduces the same condition as before:

\[\frac{d}{dr}\left(r^{n+2}\right)=0 \implies n=-2\]

This is the geophysically familiar potential of a point mass or charge, \(\phi\propto 1/r\), whose gradient field \(\propto r^{-2}\hat{r}\) is both irrotational (any gradient field is) and solenoidal away from the source - exactly the behaviour of the gravitational and Coulomb fields used throughout potential-field geophysics.

So \(n=-2\), matching option (B).