Magnetostatic Fields

Question 1
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
A
8
B
12
C
16
D
20
GATE EE 2022   Electromagnetic Theory
Question 1 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}
Question 2
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
A
\frac{J_a}{4}r^4
B
\frac{J_a}{3}r^3
C
\frac{J_a}{5}r^4
D
J_a r^3
GATE EE 2022   Electromagnetic Theory
Question 2 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}
Question 3
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.)
A
80
B
125
C
112
D
100
GATE EE 2021   Electromagnetic Theory
Question 3 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}

Question 4
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)
A
10x\hat{X}+20y\hat{Y}-30z\hat{Z}
B
10y\hat{X}+20x\hat{Y}-10z\hat{Z}
C
10z\hat{X}+20y\hat{Y}-30x\hat{Z}
D
10x\hat{X}-30z\hat{Y}+20y\hat{Z}
GATE EE 2021   Electromagnetic Theory
Question 4 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}
Question 5
The magnitude of magnetic flux density (B) in micro Teslas (\muT) at the center of a loop of wire wound as a regular hexagon of side length 1m carrying a current (I=1A), and placed in vacuum as shown in the figure is __________.
A
1.4
B
2.8
C
0.7
D
0.2
GATE EE 2017-SET-1   Electromagnetic Theory
Question 5 Explanation: 
i= 1 A

Here, B at point P is
B=\frac{\mu _0I}{4 \pi d}(\sin \theta _1 +\sin \theta _2)
We know for each segment of hexagon

magnetic field intensity due to element length dx,
dH=\frac{Idx}{4 \pi r^2}i \times u_r
where,
i \times u_r=(\sin \theta)\hat{k}....perpendicular to plane of paper
\begin{aligned} H &=\int dH=\int \frac{I \sin \theta}{4 \pi r^2}(dx)\hat{k} \\ B&=\mu _0 H \\ B_p&=\frac{\mu _0I}{4 \pi d}(\cos \alpha _1+\cos \alpha _2)\hat{k} \\ d&=\frac{\sqrt{3}}{2} l\;\;\;...\text{where}\; l=1m\\ \alpha _1&=\alpha _2 =60^{\circ}\\ |B_p| &=\frac{\mu _0 I}{4 \pi \frac{\sqrt{3}}{2} }(\cos 60^{\circ} +\cos 60^{\circ}) \\ &\text{for all six segment} \\ 6 \times |B_p| &=\frac{4 \pi \times 10^{-7} \times 6 \times 1}{4 \pi \times \frac{\sqrt{3}}{2} }(1) \\ &=\frac{12}{\sqrt{3}}\times 10^{-7}=\frac{1.20}{\sqrt{3}}\times 10^{-6}T \\ B&=0.69 \mu T \end{aligned}
Question 6
A solid iron cylinder is placed in a region containing a uniform magnetic field such that the cylinder axis is parallel to the magnetic field direction. The magnetic field lines inside the cylinder will
A
bend closer to the cylinder axis
B
bend farther away from the axis
C
remain uniform as before
D
cease to exist inside the cylinder
GATE EE 2017-SET-1   Electromagnetic Theory
Question 6 Explanation: 
Iron being a ferromagnetic material, magnetic lines of force bend closer to cylindrical axis.
Question 7
A soft-iron toroid is concentric with a long straight conductor carrying a direct current I. If the relative permeability \mu _r of soft-iron is 100, the ratio of the magnetic flux densities at two adjacent points located just inside and just outside the toroid, is _______.
A
100
B
200
C
300
D
350
GATE EE 2016-SET-1   Electromagnetic Theory
Question 7 Explanation: 
Toroid has field,
\begin{aligned} B &\propto \mu \\ As\;\; \mu &=100 \;(\text{inside field}) \end{aligned}
Magnetic field density B at any point at a distance at r is
\begin{aligned} B &=\frac{\mu I}{2 \pi r} \\ Now\;\;B_{at\; r^-}&=\frac{\mu_0 \mu_r I}{2 \pi r^-}\;(\text{just inside toroid})\\ Now\;\;B_{at\; r^+}&=\frac{\mu_0 I}{2 \pi r^+}\;(\text{just outside toroid})\\ \frac{B_{at\; r^-}}{B_{at\; r^+}}&=\mu _r=100 \end{aligned}
Question 8
A steady current I is flowing in the -x direction through each of two infinitely long wires at y=\pm \frac{L}{2} as shown in the figure. The permeability of the medium is \mu _{0}. The \vec{B}-field at (0,L,0) is
A
-\frac{4\mu _0I}{3 \pi L}\hat{Z}
B
+\frac{4\mu _0I}{3 \pi L}\hat{Z}
C
0
D
-\frac{3\mu _0I}{4 \pi L}\hat{Z}
GATE EE 2015-SET-1   Electromagnetic Theory
Question 8 Explanation: 
\begin{aligned} \vec{B}&=\frac{\mu _0 I}{2\pi \left ( \frac{L}{2} \right )}(-\hat{z})+\frac{\mu _0 I}{2\pi \left ( \frac{3L}{2} \right )}(-\hat{z}) \\ &=\frac{\mu _0 I(-\hat{z})}{2\pi L}\left ( 2+\frac{2}{3} \right )\\ &=-\frac{4}{3}\frac{\mu _0 I}{\pi L}\hat{z} \end{aligned}
Question 9
The magnitude of magnetic flux density \vec{B} at a point having normal distance d meters from an infinitely extended wire carrying current of I A is \frac{\mu _{0}I}{2\pi d} (in SI units). An infinitely extended wire is laid along the x -axis and is carrying current of 4 A in the +ve x direction. Another infinitely extended wire is laid along the y axis and is carrying 2 A current in the +ve y direction. \mu _0 is permeability of free space. Assume \hat{i},\hat{j},\hat{k} to be unit vectors along x,y and z axes respectively.

Assuming right handed coordinate system, magnetic field intensity, \hat{H} at coordinate (2,1,0) will be
A
\frac{3}{2\pi }\hat{k} A/m^{2}
B
\frac{4}{3\pi }\hat{i} A/m
C
\frac{3}{2\pi }\hat{k} A/m
D
0 A/m
GATE EE 2014-SET-2   Electromagnetic Theory
Question 10
The following four vector fields are given in Cartesian co-ordinate system. The vector field which does not satisfy the property of magnetic flux density is
A
y^{2}a_{x}+z^{2}a_{y}+x^{2}a_{z}
B
z^{2}a_{x}+x^{2}a_{y}+y^{2}a_{z}
C
x^{2}a_{x}+y^{2}a_{y}+z^{2}a_{z}
D
y^{2}z^{2}a_{x}+x^{2}z^{2}a_{y}+x^{2}y^{2}a_{z}
GATE EE 2014-SET-1   Electromagnetic Theory
Question 10 Explanation: 
The property of a solid magnetic field is
\bigtriangledown \cdot B=0
(i.e. solenoidal property)
when,
\begin{aligned} B&=x^2\hat{a_x}+y^2\hat{a_y}+z^2\hat{a_z} \\ \bigtriangledown \cdot B&=2x+2y+2z\neq 0 \end{aligned}
There are 10 questions to complete.