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GATE EE · Electromagnetic Theory

Magnetostatic Fields Practice Questions

Generate GATE-level questions on Magnetostatics. Focus on: 1. Biot-Savart Law and Ampere's Circuital Law. 2. Magnetic Flux Density and Magnetic Scalar/Vector Potentials. 3. Magnetic forces, materials, and Inductance.

19 questions · 19 PYQs · 0 AI practice · GATE EE 2027

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Q1GATE 2024NAT

The given equation represents a magnetic field strength $\bar{H}(r, \theta, \phi)$ in the spherical coordinate system, in free space. Here, $\hat{r}$ and $\hat{\theta}$ represent the unit vectors along $r$ and $\theta$ , respectively. The value of $P$ in the equation should be _____ (rounded off to the nearest integer). $\bar{H}(r, \theta, \phi)=\frac{1}{r^{3}}(\hat{r} P \cos \theta+\hat{\theta} \sin \theta)$

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📖 Explanation

\begin{aligned} & \nabla \cdot \bar{B}=0 \\ & \nabla \cdot \bar{H}=0 \end{aligned}

Spherical coordinate system, On substituting $P=2$ .

Q2GATE 2023NAT

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).

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📖 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, $|\overrightarrow{\mathrm{H}}|=2.5 \mathrm{~A} / \mathrm{m}$

Q3GATE 2022MCQ

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

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📖 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}

Q4GATE 2022MCQ

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

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📖 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}

Q5GATE 2021NAT

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.)

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📖 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}

Explanation Figure 1

Q6GATE 2021MCQ

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)

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📖 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}

Q7GATE 2017NAT

The magnitude of magnetic flux density (B) in micro Teslas ( $\mu$ T) 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 __________.

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📖 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}

Explanation Figure 2

Explanation Figure 3

Q8GATE 2017MCQ

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

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📖 Explanation

Iron being a ferromagnetic material, magnetic lines of force bend closer to cylindrical axis.

Q9GATE 2016NAT

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 _______.

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📖 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}

Q10GATE 2015MCQ

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

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📖 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}

Q11GATE 2014MCQ

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

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📖 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}

Q12GATE 2014MCQ

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

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📖 Explanation

Currently no explanation available

Q13GATE 2013MCQ

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

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📖 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}

Q14GATE 2008MCQ

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 $\mu _{0}=4\pi \times 10^{-7} H/m$ )

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📖 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}

Q15GATE 2007MCQ

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

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📖 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 $=\frac{E}{d}=\frac{0.8356}{1 \times 10^{-3}}=835.6\; N$

Q16GATE 2006MCQ

Which of the following statement holds for the divergence of electric and magnetic flux densities ?

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📖 Explanation

The divergence of magnetic field is always zzero because magnetic flux makes always a closed path. So, $\bigtriangledown \cdot B=0$ (Maxweel's equation) while divergence of electric field, $\bigtriangledown \cdot \vec{E}=\frac{\rho _v}{\in}$

Q17GATE 2004MCQ

The inductance of a long solenoid of length 1000 mm wound uniformly with 3000 turns on a cylindrical paper tube of 60 mm diameter is

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📖 Explanation

Let, $B=\mu _0nI$ Then, $\phi =B.S=\mu _0nSI$ Where 'S' is the cross sectional area of solenoid flux linkage, $a'=n\phi =\mu _0n^2SI$ Hence, Inductance length $=\mu _0n^2S$ For, $l=1000mm=1m$ $\therefore \;\;L=4 \pi \times 10^{-7} \times 3000^2 \times \pi \times (30 \times 10^{-3})^2$ $\;\;\;=32mH$

Q18GATE 2003MCQ

Two conductors are carrying forward and return current of +I and -I as shown in figure. The magnetic field intensity H at point P is

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📖 Explanation

$\bar{H}=\bar{H_1}+\bar{H_2}=\frac{I}{2 \pi d}\vec{y}+\frac{I}{2 \pi d}\vec{y}=\frac{I}{ \pi d}\vec{y}$

Explanation Figure 2

Q19GATE 2003MCQ

Two infinite strips of width w m in x -direction as shown in figure, are carrying forward and return currents of +I and -I in the z - direction. The strips are separated by distance of x m. The inductance per unit length of the configuration is measured to be L H/m. If the distance of separation between the strips in snow reduced to x/2 m, the inductance per unit length of the configuration is

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📖 Explanation

Inductance is proportional to separation between the strips so when separation is reduced to x/2 then inductance will become L/2 H/m.

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