Magnetostatic Fields

Question 1
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 1 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 2
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 2 Explanation: 
Iron being a ferromagnetic material, magnetic lines of force bend closer to cylindrical axis.
Question 3
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 3 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 4
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 4 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 5
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 6
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 6 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}
Question 7
The flux density at a point in space is given by B=4xa_{x}+2kya_{y}+8a_{z} Wb/m^{2}. The value of constant k must be equal to
A
-2
B
-0.5
C
0.5
D
2
GATE EE 2013   Electromagnetic Theory
Question 7 Explanation: 
\begin{aligned} \bigtriangledown \cdot B &=0 \\ \left ( \frac{\partial }{\partial x}a_x +\frac{\partial }{\partial y}a_y + \frac{\partial }{\partial z}a_z\right )&(4xa_x+2kya_y+8a_z)=0 \\ 4+2k&=0\\ k&=-2 \end{aligned}
Question 8
A coil of 300 turns is wound on a non-magnetic core having a mean circumference of 300 mm and a cross-sectional area of 300 mm^{2} . The inductance of the coil corresponding to a magnetizing current of 3 A will be (Given that \mu _{0}=4\pi \times 10^{-7} H/m)
A
37.68 \muH
B
113.04 \muH
C
37.68 \muH
D
113.04 \muH
GATE EE 2008   Electromagnetic Theory
Question 8 Explanation: 
\begin{aligned} n &=300 \\ (I)\; \text{circumference}&=2 \pi r=300\; mm \\ A &=30\; mm^2=300 \times 10^{-6} \; m^2 \\ &\therefore \; \text{Inductance of coil} \\ L &=\frac{\mu _0 N^2 A}{I} \\ &=\frac{4 \pi \times 10^{-7} \times 300 ^3 \times 10^{-6}}{300 \times 10^{-3}} \\ &=113.04\; \mu H \end{aligned}
Question 9
An inductor designed with 400 turns coil wound on an iron core of 16 cm^2 cross sectional area and with a cut of an air gap length of 1 mm. The coil is connected to a 230 V, 50 Hz ac supply. Neglect coil resistance, core loss, iron reluctance and leakage inductance, \mu _0=4 \pi \times 10^{-7} H/M

The average force on the core to reduce the air gap will be
A
832.29N
B
1666.22N
C
3332.47N
D
6664.84N
GATE EE 2007   Electromagnetic Theory
Question 9 Explanation: 
\begin{aligned} \text{Energy stored}&=\frac{1}{2} \times L \times I^2 \\ &= \frac{1}{2} \times 0.3215 \times 2.28^2\\ &=0.8356 \; J \end{aligned}
Force to reduce 1 mm air gap =\frac{E}{d}=\frac{0.8356}{1 \times 10^{-3}}=835.6\; N
Question 10
An inductor designed with 400 turns coil wound on an iron core of 16 cm^2 cross sectional area and with a cut of an air gap length of 1 mm. The coil is connected to a 230 V, 50 Hz ac supply. Neglect coil resistance, core loss, iron reluctance and leakage inductance, \mu _0=4 \pi \times 10^{-7} H/M

The current in the inductor is
A
18.08A
B
9.04A
C
4.56A
D
2.28A
GATE EE 2007   Electromagnetic Theory
Question 10 Explanation: 
\begin{aligned} L&=\frac{\mu _0 N^2 A}{l} \\ &= \frac{4 \pi \times 10^{-7} \times (400)^2 \times 16 \times 10^{-4}}{1 \times 10^{-3}}\\ &= 0.3215\\ X_L&= 2 \pi fL\\ \therefore \;\; I&=\frac{V}{X_L}=\frac{V}{2 \pi f L} \\ &= \frac{230}{2 \pi \times 50 \times 0.3215}=2.28\;A \end{aligned}
There are 10 questions to complete.
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