GATE EC 2011 Electromagnetics Solution Q 1

GATE EC 2011 Electromagnetics Solution Q 1

Consider a closed surface S surrounding a volume V. If \overrightarrow r is the position vector of a point inside s, with \widehat n the unit normal on S,the value of the integral \oint\limits_S5\overrightarrow r\cdot\widehat nds

a) 3 V

b) 5 V

c) 10 V

d) 10 V

Ans- d

Explanation

For close surface, we can use Divergence Theorem.

\oint_s\overrightarrow A\cdot\overrightarrow{ds}=\int_v\left(\nabla\cdot\overrightarrow A\right)dv \oint\limits_S5\overrightarrow r\cdot\widehat nds=\underset V{\int\int\int}\nabla\cdot5\overrightarrow rdv \begin{array}{l}\nabla\cdot r=\frac1{r^2}\frac\partial{\partial r}\left(r^2r\right)\\\\\;\;\;\;\;\;=\frac3{r^2}\cdot r^2=3\end{array} 5\underset V{\int\int\int}\nabla\cdot\overrightarrow rdv=5\times3V=15V \oint_s\overrightarrow A\cdot\overrightarrow{ds}=\int_v\left(\nabla\cdot\overrightarrow A\right)dv \oint\limits_S5\overrightarrow r\cdot\widehat nds=\underset V{\int\int\int}\nabla\cdot5\overrightarrow rdv \oint\limits_S5\overrightarrow r\cdot\widehat nds=\underset V{\int\int\int}\nabla\cdot5\overrightarrow rdv \oint\limits_S5\overrightarrow r\cdot\widehat nds=\underset V{\int\int\int}\nabla\cdot5\overrightarrow rdv \oint_s\overrightarrow A\cdot\overrightarrow{ds}=\int_v\left(\nabla\cdot\overrightarrow A\right)dv \oint_s\overrightarrow A\cdot\overrightarrow{ds}=\int_v\left(\nabla\cdot\overrightarrow A\right)dv \oint_s\overrightarrow A\cdot\overrightarrow{ds}=\int_v\left(\nabla\cdot\overrightarrow A\right)dv \oint_s\overrightarrow A\cdot\overrightarrow{ds}=\int_v\left(\nabla\cdot\overrightarrow A\right)dv

We know that the \oint_s\overrightarrow A\cdot\overrightarrow{ds}=\int_v\left(\nabla\cdot\overrightarrow A\right)dv

and \oint_s\overrightarrow A\cdot\overrightarrow{ds}=\int_v\left(\nabla\cdot\overrightarrow A\right)dv

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