GATE EC 2012 Electromagnetics Solution Q 2

GATE EC 2012 Electromagnetics Solution Q 2

The direction of vector A is radially outward from the origin, with \left|A\right|=Kr^n where r^2=x^2+y^2+z^2 and K is a constant. The value of n for which \nabla\cdot A=0 is

a) -2

b) 2

c) 1

d) 0

Ans- a

Explanation

\left|A\right|=Kr^n \overrightarrow A=Kr^n{\widehat{\;a}}_r\left(\sin ce\;it\;is\;radially\;outward\right) \nabla\cdot\overrightarrow A\;in\;spherical\;coordinate\;is \nabla\cdot\overrightarrow A=\frac1{r^2}\frac\partial{\partial r}\left(r^2A_r\right)+\frac1{r\;\sin\theta}\frac\partial{\partial r}\left(A_\theta\sin_\theta\right)+\frac1{r\;\sin\theta}\frac\partial{\partial\phi}A_\phi \nabla\cdot\overrightarrow A=\frac1{r^2}\frac\partial{\partial r}\left(r^2Kr^2\right)+0+0 \nabla\cdot\overrightarrow A=\frac K{r^2}\frac\partial{\partial r}\left(r^{n+2}\right)

For \nabla\cdot\overline A to be zero, derivative must be zero in other words (rn+2) must be constant.

This could be possible only if n = -2.

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