GATE 2026 Electromagnetic Fields Questions and Solutions

1 mark, numerical answer. A solid sphere of radius \(a = 2\,\mathrm{cm}\) carries a uniform volume charge density \(\rho_v = 5\,\mu\mathrm{C/m^3}\) and is surrounded by free space (\(\varepsilon_0 = 8.854\times10^{-12}\,\mathrm{F/m}\)). The magnitude of the electric field intensity at a radial distance \(r = 5\,\mathrm{cm}\) from the centre, in V/m rounded off to one decimal place, is _____.

Solution

The charge distribution is spherically symmetric, so a concentric spherical Gaussian surface of radius \(r > a\) carries a uniform, purely radial \(\mathbf{D}\). Gauss's law gives

Substituting \(\rho_v = 5\times10^{-6}\,\mathrm{C/m^3}\), \(a = 0.02\,\mathrm{m}\) and \(r = 0.05\,\mathrm{m}\):

Background: Chapter 5.

1 mark, multiple choice. In a charge-free region the electric scalar potential is \(V = 2x^{2}y - 5z\) volts, with \(x, y, z\) in metres. The electric field intensity at the point \((1, 2, 3)\) is

1. \(8\hat{a}_x + 2\hat{a}_y - 5\hat{a}_z\) V/m
2. \(-8\hat{a}_x - 2\hat{a}_y + 5\hat{a}_z\) V/m
3. \(-4\hat{a}_x - 2\hat{a}_y + 5\hat{a}_z\) V/m
4. \(-8\hat{a}_x - 4\hat{a}_y - 5\hat{a}_z\) V/m

Solution

The field is the negative gradient of the potential:

At \((1,2,3)\): \(-4(1)(2) = -8\) and \(-2(1)^2 = -2\), the \(z\) component being independent of position, so

Background: Chapter 6.

2 marks, numerical answer. A parallel-plate capacitor has plates of area \(A = 100\,\mathrm{cm^2}\). The gap between the plates is completely filled by two lossless dielectric slabs stacked one above the other, both spanning the full plate area: slab 1 of thickness \(1\,\mathrm{mm}\) with \(\varepsilon_{r1} = 2\), and slab 2 of thickness \(2\,\mathrm{mm}\) with \(\varepsilon_{r2} = 4\). Fringing is neglected. The capacitance, in pF rounded off to two decimal places, is _____.

Solution

The interface between the slabs is an equipotential surface perpendicular to \(\mathbf{E}\), so the arrangement is two capacitors in series:

With \(A = 0.01\,\mathrm{m^2}\), \(d_1 = 10^{-3}\,\mathrm{m}\) and \(d_2 = 2\times10^{-3}\,\mathrm{m}\), the effective electrical thickness is

Background: Chapter 8.

2 marks, numerical answer. A parallel-plate capacitor with plate area \(A = 0.02\,\mathrm{m^2}\) and plate separation \(d = 0.5\,\mathrm{mm}\) is filled with a lossless dielectric of relative permittivity \(\varepsilon_r = 2\). A voltage \(v(t) = 50\sin(2\pi \times 10^{6} t)\) volts is applied across the plates. Fringing and conduction current in the dielectric are neglected. The peak value of the displacement current flowing between the plates, in mA rounded off to two decimal places, is _____.

Solution

Maxwell's correction to Ampere's law introduces the displacement current density \(\mathbf{J}_d = \partial \mathbf{D}/\partial t\). Inside the capacitor the field is uniform, \(E = v/d\), so

which is exactly the terminal current of the capacitor, as continuity demands. The capacitance is

With \(v = V_m\sin\omega t\), the peak of \(dv/dt\) is \(V_m\omega\):

Background: Chapter 16.

1 mark, numerical answer. The solid inner conductor of a long coaxial cable has radius \(a = 1\,\mathrm{mm}\) and carries a total current of 5 A distributed uniformly over its cross-section. The outer conductor carries the same current in the opposite direction. The magnitude of the magnetic field intensity at a radial distance \(\rho = 0.5\,\mathrm{mm}\) from the axis, in A/m rounded off to one decimal place, is _____.

Solution

Symmetry makes \(\mathbf{H}\) purely azimuthal and constant on a circle of radius \(\rho\). Ampere's circuital law on that circle gives

Inside the inner conductor only the fraction of the current passing through the area \(\pi\rho^{2}\) is enclosed:

Background: Chapter 11.

2 marks, multiple choice. A circular loop of radius \(5\,\mathrm{cm}\) lies in the \(z = 0\) plane with its centre at the origin and carries a steady current of 10 A in free space. The magnitude of the magnetic flux density on the axis of the loop at the point \((0,0,12\,\mathrm{cm})\) is

1. \(1.26\,\mu\mathrm{T}\)
2. \(7.15\,\mu\mathrm{T}\)
3. \(48.3\,\mu\mathrm{T}\)
4. \(126\,\mu\mathrm{T}\)

Solution

For a current element \(I\,d\mathbf{l}\) on the loop, the Biot-Savart contribution at an axial point a distance \(R = \sqrt{a^{2}+z^{2}}\) away has magnitude \(\mu_0 I\,dl/(4\pi R^{2})\), since \(d\mathbf{l}\) is perpendicular to \(\mathbf{R}\). Going round the loop, the radial components cancel by symmetry and only the axial components survive, each scaled by \(\cos\alpha = a/R\):

With \(a = 0.05\,\mathrm{m}\) and \(z = 0.12\,\mathrm{m}\), \(a^{2}+z^{2} = 0.0169\,\mathrm{m^2}\) and \(\sqrt{0.0169} = 0.13\), so \((a^{2}+z^{2})^{3/2} = 0.0169 \times 0.13 = 2.197\times10^{-3}\).

Background: Chapter 12.

2 marks, numerical answer. In the region outside a long straight filamentary conductor lying along the \(z\) axis and carrying a current \(I = 10\,\mathrm{A}\) in free space, the magnetic vector potential is

where \(\rho\) is the cylindrical radial coordinate and \(b\) is a constant reference radius. The magnitude of the magnetic flux density at \(\rho = 2\,\mathrm{cm}\), in \(\mu\mathrm{T}\), is _____.

Solution

The flux density follows from \(\mathbf{B} = \nabla \times \mathbf{A}\). For a vector potential with only a \(z\) component that depends on \(\rho\) alone, the curl in cylindrical coordinates reduces to

which is the familiar field of an infinite straight wire, confirming the vector potential. Numerically, at \(\rho = 0.02\,\mathrm{m}\):

Background: Chapter 12.

2 marks, numerical answer. A toroidal coil of 200 turns is wound uniformly on a non-magnetic core (\(\mu_r = 1\)) of mean radius \(R = 10\,\mathrm{cm}\) and cross-sectional area \(A = 4\,\mathrm{cm^2}\). The cross-section is small compared with the mean radius, so the flux density may be taken as uniform over it. The self-inductance of the coil, in \(\mu\mathrm{H}\), is _____.

Solution

Ampere's law applied to a circular path of radius \(R\) inside the core encloses the current \(NI\):

The flux through one turn is \(\Phi = BA\), and each of the \(N\) turns links it, so the flux linkage is \(\lambda = N\Phi\) and

Background: Chapter 14.

2 marks, numerical answer. A uniform plane wave at 300 MHz propagates in a lossless non-magnetic dielectric with \(\varepsilon_r = 4\) and \(\mu_r = 1\). The electric field amplitude of the wave is \(10\,\mathrm{V/m}\) (peak, not RMS). Taking the intrinsic impedance of free space as \(120\pi\,\Omega\), the time-average power density carried by the wave, in \(\mathrm{mW/m^2}\) rounded off to one decimal place, is _____.

Solution

For a lossless medium the intrinsic impedance is

The magnetic field amplitude is \(H_0 = E_0/\eta\), and \(\mathbf{E}\), \(\mathbf{H}\) are in phase and mutually perpendicular, so the time-average Poynting vector has magnitude

The frequency does not enter the power calculation; it only fixes the wavelength, \(\lambda = c/(f\sqrt{\varepsilon_r}) = 3\times10^{8}/(6\times10^{8}) = 0.5\,\mathrm{m}\). Background: Chapter 20.

2 marks, multiple choice. A lossless transmission line of characteristic impedance \(75\,\Omega\) and electrical length exactly one quarter of a wavelength at the operating frequency is terminated in a purely resistive load of \(300\,\Omega\). The impedance seen at the input terminals of the line, and the standing wave ratio on the line, are respectively

1. \(18.75\,\Omega\) and 4
2. \(300\,\Omega\) and 1
3. \(18.75\,\Omega\) and 1
4. \(1200\,\Omega\) and 4

Solution

For a lossless line of length \(l\) the input impedance is

A quarter-wave line has \(\beta l = (2\pi/\lambda)(\lambda/4) = \pi/2\), so \(\tan\beta l \to \infty\). Dividing numerator and denominator by \(\tan\beta l\) and letting it grow without bound leaves only the \(j\) terms:

The standing wave ratio is set by the load mismatch and is the same everywhere on a lossless line:

Background: Chapter 24.