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

We are given the radial vector field \(\mathbf{F} = a r^n \hat{r}\), \(r\neq 0\).

Step 1 - Irrotational check: For any purely radial vector field that depends only on \(r\) (no angular dependence), the curl vanishes identically: \(\nabla\times(f(r)\hat{r}) = 0\) for any function \(f(r)\), because none of the angular derivatives that would produce a non-zero curl component are present. So \(\mathbf{F}\) is irrotational for EVERY value of \(n\) - this condition alone places no restriction on \(n\).

Step 2 - Solenoidal check: The divergence of a radial field in spherical coordinates is

\[\nabla\cdot\mathbf{F} = \frac{1}{r^2}\frac{d}{dr}\left(r^2\cdot a r^n\right) = \frac{a}{r^2}\frac{d}{dr}\left(r^{n+2}\right) = a(n+2)r^{n-1}\]

For \(\mathbf{F}\) to be solenoidal, \(\nabla\cdot\mathbf{F}=0\) for all \(r\neq 0\), which (since \(a\neq 0\)) requires

\[n+2=0 \implies n=-2\]

Step 3: Since the irrotational condition holds for every \(n\), the binding constraint is the solenoidal one, giving \(n=-2\). This is exactly the exponent of the familiar inverse-square-law fields (gravitational field of a point mass, Coulomb field of a point charge), which are known to be divergence-free away from their source.

\(\boxed{n=-2}\), option (B).