Basics Of Electromagnetics – GATE ECE Electromagnetics Practice Questions & PYQs | BattleExam | BattleExam

Q8GATE 2021NAT

For a vector field $D=\rho\cos^{2}\:\varphi \:a_{\rho }+z^{2}\sin^{2}\:\varphi \:a_{\varphi }$ in a cylindrical coordinate system $\left ( \rho ,\varphi ,z \right )$ with unit vectors $a_{\rho },a_{\varphi }$ and $a_{z}$ , the net flux of D leaving the closed surface of the cylinder $\left ( \rho =3, 0\leq z\leq 2 \right )$ (rounded off to two decimal places) is ________________

🚀 Solve in Practice Mode

📖 Explanation

Method 1: $\vec{D}=\rho \cos ^{2} \phi \hat{a}_{\rho}+z^{2} \sin ^{2} \phi \hat{a}_{\phi}$ Electric flux crossing the closed surface is

\begin{aligned} \psi &=\oiint \vec{D} \cdot \overrightarrow{d S}=\iiint(\vec{\nabla} \cdot \vec{D}) d v \\ \vec{\nabla} \cdot \vec{D} &=\frac{1}{\rho} \frac{\partial}{\partial p}\left(\rho D_{\rho}\right)+\frac{1}{\rho} \frac{\partial D_{\phi}}{\partial \phi}+\frac{\partial D_{z}}{\partial z} \\ &=\frac{1}{\rho} \frac{\partial}{\partial p}\left(\rho \rho \cos ^{2} \phi\right)+\frac{1}{\rho} \frac{\partial}{\partial \phi}\left(z^{2} \sin ^{2} \phi\right)+0 \\ &=\frac{1}{\rho}(2 \rho) \cos ^{2} \phi+\frac{1}{\rho} z^{2} 2 \sin \phi \cos \phi=2 \cos ^{2} \phi+\frac{z^{2}}{\rho} \sin 2 \phi\\ \iiint(\vec{\nabla} \cdot \vec{D}) d v &=\iiint 2 \cos ^{2} \phi(\rho d \rho d \phi d z)+\iiint\left(\frac{z^{2}}{\rho} \sin 2 \phi\right) \rho d \rho d \phi d z \\ &=2 \int_{\rho=0}^{3} \rho d \rho \int_{\phi=0}^{2 \pi}\left(\frac{1+\cos 2 \phi}{2}\right) d \phi \int_{z=0}^{2} d z+\int_{\rho=0}^{2} d \rho \int_{\phi=0}^{2 \pi} \sin 2 \phi d \phi \int_{z=0}^{2} z^{2} d z \\ &=2\left(\frac{\rho^{2}}{-2}\right)_{\rho=0}^{3} \frac{1}{2}(2 \pi)(z)_{z=0}^{2}+0 \\ &=2\left(\frac{3^{2}}{2}\right) \pi(2)=18 \pi(\text { Coulomb })=56.55(\text { Coulomb }) \end{aligned}

Method 2: Electric flux crossing the closed surface is $\psi=\oiint \vec{D} \cdot \overrightarrow{d S}$ Electric flux crossing \rho=3 cylindrical surface is

\begin{aligned} \left.\psi\right|_{\rho=3} &=\oiint\left(\rho \cos ^{2} \phi \hat{a}_{p}\right) \cdot(\rho d \phi d z) \hat{a}_{p} \\ &=3^{2} \int_{\phi=0}^{2 \pi} \cos ^{2} \phi d \phi \int_{z=0}^{2} d z \\ &=9 \frac{1}{2}(2 \pi)(2)=18 \pi(\text { coulomb })=56.55(\text { coulomb }) \end{aligned}

Q17GATE 2016MCQ

The current density in a medium is given by $\vec{j}=\frac{400 sin\theta }{2\pi (r^{2})+4}\hat{a}_{r}Am^{-2}$ The total current and the average current density flowing through the portion of a spherical surface r = 0.8 m, $\frac{\pi }{12}\leq \theta \leq \frac{\pi }{4},0\leq \phi \leq 2\pi$ are given, respectively, by

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} I &=\int \overrightarrow{J} \cdot \overrightarrow{d s} \\ &=\int_{\theta=\frac{\pi}{12}}^{\pi / 4} \int_{\phi=0}^{2 \pi} \frac{400 \sin \theta}{2 \pi\left(r^{2}+4\right)} r^{2} \sin \theta d \theta d \phi \\ &=\left.\frac{400}{2 \pi\left(r^{2}+4\right)} \cdot r^{2} \cdot \phi\right|_{0} ^{2 \pi} \int_{\pi / 12}^{\pi / 4} \sin ^{2} \theta d \theta \\ &=\frac{400 r^{2}}{\left(r^{2}+4\right)} \int_{\pi / 12}^{\pi / 4}\left(\frac{1-\cos 2 \theta}{2}\right) d \theta \\ &=\frac{400 \cdot r^{2}}{\left(r^{2}+4\right)}\left(\frac{\left(\frac{\pi}{4}-\frac{\pi}{12}\right)}{2}-\left(\frac{\sin 2 \theta}{4}\right)_{\pi / 12}^{\pi / 4}\right) \\ &=\left.\frac{400 \cdot r^{2}}{\left(r^{2}+4\right)}\left(\frac{\pi}{12}-\left(\frac{1-1 / 2}{4}\right)\right)\right|_{r=0.8} \\ &=\frac{400 \times 0.8 \times 0.8}{4.64} \times 0.13=7.17 \mathrm{Amp} \end{aligned}

\begin{array}{l} \text { Total area }=\int d s=\iint r^{2} \sin \theta d \theta d \phi \\ =r^{2} \int_{0=\frac{\pi}{12}}^{\pi / 4} \sin \theta d \theta \cdot 2 \pi \\ =\left.r^{2} \cdot 2 \pi \cdot 0.259\right|_{r=0.8} \\ =0.8^{2} \times 0.5 \times 2 \pi \times \frac{1}{4} \\ =1.041 \mathrm{m}^{2} \\ \text { Average current }=\frac{7.17}{1.041}=6.88 \mathrm{A} / \mathrm{m}^{2} \end{array}

Note : GATE key mentioned (MTA - marks to all)