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 |
In the figure, the electric field E and the magnetic field B point to \mathrm{x} and \mathrm{z} directions, respectively, and have constant magnitudes. A positive charge 'q' is released from rest at the origin. Which of the following statement(s) is/ are true.
The charge will move in the direction of z with constant velocity. | |
The charge will al ways move on the y-z plane only. | |
The trajectory of the charge will be a cycle. | |
The charge will progress in the direction of y. |
Question 2 Explanation:
As per Answer key of IIT Official : MTA (Marks to All)
Given :
Given :
Question 3 |
As shown in the figure below, two concentric conducting spherical shells, centered
at r=0 and having radii r=c and r=d are maintained at potentials such that
the potential V(r) at r=c is V_1 and V(r) at r=d is V_2. Assume that V(r) depends
only on r , where r is the radial distance. The expression for V(r) in the region
between r=c and r=d is
V(r)=\frac{cd(V_2-V_1)}{(d-c)r}-\frac{V_1c+V_2d-2V_1d}{d-c} | |
V(r)=\frac{cd(V_1-V_2)}{(d-c)r}-\frac{V_2d-V_1c}{d-c} | |
V(r)=\frac{cd(V_1-V_2)}{(d-c)r}-\frac{V_1c-V_2c}{d-c} | |
V(r)=\frac{cd(V_2-V_1)}{(d-c)r}-\frac{V_2c-V_1c}{d-c} |
Question 3 Explanation:
We have, Laplace equation
\triangledown ^2V=0 ...(1)
where, \triangledown ^2V=\frac{1}{r^2}\frac{\partial }{\partial r}\left ( r^2\frac{\partial V}{\partial r} \right )+\frac{1}{r^2 \sin \theta }\frac{\partial }{\partial \theta }\left ( \sin \theta \frac{\partial V}{\partial \theta } \right )+ \frac{1}{r^2 \sin ^2\theta } \frac{\partial^2 V}{\partial \phi ^2}
V will have only radial component.
From equation (1)
\frac{1}{r^2}\frac{\partial }{\partial r}\left ( r^2\frac{\partial V}{\partial r} \right )=0
Integrate both side r^2\frac{\partial V}{\partial r} =P
Again integrate, V=-\frac{P}{r}+Q
Here, P and Q are constant.
Given: at r=c, V=V_1
V_1=-\frac{P}{r}+Q ...(2)
at r=d, V=V_2
V_2=-\frac{P}{d}+Q ...(3)
From eq. (2) & (3)
P=\frac{(V_2-V_1)cd}{d-c}
Q=V_1+\frac{d(V_2-V_1)}{d-c}
From eq. (1),
V=-\frac{(V_2-V_1)cd}{r(d-c)}+V_1+\frac{(V_2-V_1)cd}{(d-c)}
V=\frac{(V_1-V_2)cd}{r(d-c)}+\frac{(V_2d-V_1c)}{(d-c)}
\triangledown ^2V=0 ...(1)
where, \triangledown ^2V=\frac{1}{r^2}\frac{\partial }{\partial r}\left ( r^2\frac{\partial V}{\partial r} \right )+\frac{1}{r^2 \sin \theta }\frac{\partial }{\partial \theta }\left ( \sin \theta \frac{\partial V}{\partial \theta } \right )+ \frac{1}{r^2 \sin ^2\theta } \frac{\partial^2 V}{\partial \phi ^2}
V will have only radial component.
From equation (1)
\frac{1}{r^2}\frac{\partial }{\partial r}\left ( r^2\frac{\partial V}{\partial r} \right )=0
Integrate both side r^2\frac{\partial V}{\partial r} =P
Again integrate, V=-\frac{P}{r}+Q
Here, P and Q are constant.
Given: at r=c, V=V_1
V_1=-\frac{P}{r}+Q ...(2)
at r=d, V=V_2
V_2=-\frac{P}{d}+Q ...(3)
From eq. (2) & (3)
P=\frac{(V_2-V_1)cd}{d-c}
Q=V_1+\frac{d(V_2-V_1)}{d-c}
From eq. (1),
V=-\frac{(V_2-V_1)cd}{r(d-c)}+V_1+\frac{(V_2-V_1)cd}{(d-c)}
V=\frac{(V_1-V_2)cd}{r(d-c)}+\frac{(V_2d-V_1c)}{(d-c)}
Question 4 |
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 4 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 5 |
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 5 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}
There are 5 questions to complete.