Question 1 |
An infinite surface of linear current density K=5 \hat{a}_{x} A / m exists on the x-y plane, as shown in the figure. The magnitude of the magnetic field intensity (H) at a point (1,1,1) due to the surface current in Ampere/meter is ___ (Round off to 2 decimal places).
1.45 | |
6.36 | |
2.5 | |
3.54 |
Question 1 Explanation:
\begin{aligned}
\overrightarrow{\mathrm{a}}_{\mathrm{N}}&=\overrightarrow{\mathrm{a}}_{\mathrm{Z}}\\
\text{Now, }\quad \overrightarrow{\mathrm{H}}&=\frac{1}{2}\left(\overrightarrow{\mathrm{K}} \times \overrightarrow{\mathrm{a}}_{\mathrm{N}}\right) \\
& =\frac{1}{2}\left(5 \hat{a}_{x} \times \hat{a}_{z}\right) \\
& =-2.5 \hat{a}_{y} A / m
\end{aligned}
\therefore Magnitude, |\overrightarrow{\mathrm{H}}|=2.5 \mathrm{~A} / \mathrm{m}
\therefore Magnitude, |\overrightarrow{\mathrm{H}}|=2.5 \mathrm{~A} / \mathrm{m}
Question 2 |
If the magnetic field intensity (H) in a conducting region is given by the expression, H=x^2\hat{i}+x^2y^2\hat{j}+x^2y^2z^2\hat{k}\;A/m. The magnitude of the current density, in A/m^2
, at x=1m, y=2m and z=1m , is
8 | |
12 | |
16 | |
20 |
Question 2 Explanation:
We have,
\begin{aligned} \vec{J}&=\triangledown \times \vec{H}\\ \vec{J}&=\begin{vmatrix} \hat{a}_x &\hat{a}_y & \hat{a}_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ x^2& x^2y^2 & x^2y^2z^2 \end{vmatrix} \\ &=\hat{a}_x(2x^2yz^2-0)-\hat{a}_y(2xy^2z^2-0)+\hat{a}_z(2xy^2-0)\\ &=2x^2yz^2\hat{a}_x-2xy^2z^2\hat{a}_y+2xy^2\hat{a}_z\\ &\text{At point (1,2,1)}\\ \vec{J}&=4\hat{a}_x-8\hat{a}_y+8\hat{a}_z\\ |\vec{J}|&=\sqrt{4^2+8^2+8^2}=12A/m^2 \end{aligned}
\begin{aligned} \vec{J}&=\triangledown \times \vec{H}\\ \vec{J}&=\begin{vmatrix} \hat{a}_x &\hat{a}_y & \hat{a}_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ x^2& x^2y^2 & x^2y^2z^2 \end{vmatrix} \\ &=\hat{a}_x(2x^2yz^2-0)-\hat{a}_y(2xy^2z^2-0)+\hat{a}_z(2xy^2-0)\\ &=2x^2yz^2\hat{a}_x-2xy^2z^2\hat{a}_y+2xy^2\hat{a}_z\\ &\text{At point (1,2,1)}\\ \vec{J}&=4\hat{a}_x-8\hat{a}_y+8\hat{a}_z\\ |\vec{J}|&=\sqrt{4^2+8^2+8^2}=12A/m^2 \end{aligned}
Question 3 |
A long conducting cylinder having a radius 'b' is placed along the z axis. The current
density is J=J_ar^3\hat{z} for the region r \lt b where r is the distance in the radial direction.
The magnetic field intensity (H) for the region inside the conductor (i.e. for r \lt b) is
\frac{J_a}{4}r^4 | |
\frac{J_a}{3}r^3 | |
\frac{J_a}{5}r^4 | |
J_a r^3 |
Question 3 Explanation:
Using Ampere?s law,
\begin{aligned} \oint \vec{H}\cdot d\vec{l}&=I_{enc}\\ \text{where, }I_{enc}&=\int \int \vec{J}\cdot d\vec{s}\\ d\vec{s}&=rdrd\phi \hat{a}_z\\ \therefore \oint \vec{H}\cdot d\vec{l}&=\int \int \vec{J}\cdot r^3\hat{a}_z\cdot rdrd\phi \hat{a}_z\\ \vec{H}\cdot 2 \pi r&=\frac{J_ar^5}{5}(2 \pi)\\ \Rightarrow \vec{H}&=\frac{J_ar^4}{5}A/m \end{aligned}
\begin{aligned} \oint \vec{H}\cdot d\vec{l}&=I_{enc}\\ \text{where, }I_{enc}&=\int \int \vec{J}\cdot d\vec{s}\\ d\vec{s}&=rdrd\phi \hat{a}_z\\ \therefore \oint \vec{H}\cdot d\vec{l}&=\int \int \vec{J}\cdot r^3\hat{a}_z\cdot rdrd\phi \hat{a}_z\\ \vec{H}\cdot 2 \pi r&=\frac{J_ar^5}{5}(2 \pi)\\ \Rightarrow \vec{H}&=\frac{J_ar^4}{5}A/m \end{aligned}
Question 4 |
One coulomb of point charge moving with a uniform velocity 10\:\hat{X} \text{m/s} enters the region x\geq 0 having a magnetic flux density \overrightarrow{B}=\left ( 10y\:\hat{X} + 10x\:\hat{Y}+10\:\hat{Z}\right )T. The magnitude of force on the charge at x=0^{+} is _____________ N.
(\hat{X}, \hat{Y}, and \hat{Z} are unit vectors along x-axis, y-axis and z-axis, respectively.)
(\hat{X}, \hat{Y}, and \hat{Z} are unit vectors along x-axis, y-axis and z-axis, respectively.)
80 | |
125 | |
112 | |
100 |
Question 4 Explanation:
Force on a charge moving with \vec{v} velocity due to magnetic field is
\begin{aligned} \vec{F} &=Q(\vec{v} \times \vec{B}) \\ &=1[10 \hat{x} \times(10 y \hat{x}+10 x \hat{y}+10 \hat{z}]\\ &=10(10 x) \hat{z}+10(10)(-\hat{y}) \\ &=100 x \hat{z}-100 \hat{y} \\ \left.\vec{F}\right|_{x=0^{+}} &=-100 \hat{y} \\ |\vec{F}| &=\sqrt{(-100)^{2}}=100 \text { Newton } \end{aligned}
\begin{aligned} \vec{F} &=Q(\vec{v} \times \vec{B}) \\ &=1[10 \hat{x} \times(10 y \hat{x}+10 x \hat{y}+10 \hat{z}]\\ &=10(10 x) \hat{z}+10(10)(-\hat{y}) \\ &=100 x \hat{z}-100 \hat{y} \\ \left.\vec{F}\right|_{x=0^{+}} &=-100 \hat{y} \\ |\vec{F}| &=\sqrt{(-100)^{2}}=100 \text { Newton } \end{aligned}
Question 5 |
Which one of the following vector functions represents a magnetic field \overrightarrow{B}?
(\hat{X}, \hat{Y} and \hat{Z} are unit vectors along x-axis, y-axis, and z-axis, respectively)
(\hat{X}, \hat{Y} and \hat{Z} are unit vectors along x-axis, y-axis, and z-axis, respectively)
10x\hat{X}+20y\hat{Y}-30z\hat{Z} | |
10y\hat{X}+20x\hat{Y}-10z\hat{Z} | |
10z\hat{X}+20y\hat{Y}-30x\hat{Z} | |
10x\hat{X}-30z\hat{Y}+20y\hat{Z} |
Question 5 Explanation:
If \vec{B} is magnetic flux density then \vec{\nabla} \cdot \vec{B}=0
\begin{aligned} &\vec{\nabla} \cdot \vec{B}=\frac{\partial B x}{\partial x}+\frac{\partial B y}{\partial y}+\frac{\partial B z}{\partial z}\\ &\frac{\partial}{\partial x}(10 x)+\frac{\partial}{\partial y}(20 y)+\frac{\partial}{\partial z}(-30 z)=\vec{\nabla} \cdot \vec{B}\\ &\qquad \qquad \vec{\nabla} \cdot \vec{B}=10+20-30=0 \end{aligned}
\begin{aligned} &\vec{\nabla} \cdot \vec{B}=\frac{\partial B x}{\partial x}+\frac{\partial B y}{\partial y}+\frac{\partial B z}{\partial z}\\ &\frac{\partial}{\partial x}(10 x)+\frac{\partial}{\partial y}(20 y)+\frac{\partial}{\partial z}(-30 z)=\vec{\nabla} \cdot \vec{B}\\ &\qquad \qquad \vec{\nabla} \cdot \vec{B}=10+20-30=0 \end{aligned}
There are 5 questions to complete.