GATE 2024 Electromagnetic Fields Questions and Solutions

Question 01

Question 1

In the \((x, y, z)\) coordinate system, three point-charges \(Q, Q\) and \(\alpha Q\) are located in free space at \((-1, 0, 0)\), \((1, 0, 0)\) and \((0, -1, 0)\) respectively. The value of \(\alpha\) for the electric field to be zero at \((0, 0.5, 0)\) is ________ (rounded off to 1 decimal place).

Solution

Vector diagram showing field cancellation at \(y=0.5\)

From the figure, for the net Electric Field to be zero:

Equation

\[\vec{E}_{\text{net}} = \vec{E}_{1y} + \vec{E}_{2y} + \vec{E}|_{\alpha Q} = 0 \quad \text{[as per question]}\]

Due to symmetry of charges at \(x=-1\) and \(x=1\), their horizontal components cancel, and their vertical components add up. Let:

Equation

\[\vec{E}_{1y} = \vec{E}_{2y} = \vec{E}|_{Q}\]

Equation

\[\therefore \quad \vec{E}_{\text{net}} = 2\vec{E}|_{Q} + \vec{E}|_{\alpha Q} = 0\]

Calculating the field due to one charge \(Q\) at \((1,0,0)\) acting on point \((0, 0.5, 0)\):

Equation

\[\begin{aligned} \text{Distance } r &= \sqrt{1^2 + 0.5^2} = \sqrt{1.25} \\ \sin\theta &= \frac{0.5}{\sqrt{1.25}} \\ \vec{E}|_{Q} &= \frac{kQ}{r^2} \sin\theta \, \hat{a}_y \quad \text{where } k = \frac{1}{4\pi \epsilon_0} \end{aligned}\]

Equation

\[\begin{aligned} \vec{E}|_{Q} &= \frac{kQ}{1.25} \times \frac{0.5}{\sqrt{1.25}} \, \hat{a}_y \\ &= \frac{0.5}{(1.25)^{1.5}} kQ \, \hat{a}_y \\ &\approx 0.357 \, kQ \, \hat{a}_y \end{aligned}\]

Calculating the field due to charge \(\alpha Q\) at \((0, -1, 0)\) acting on \((0, 0.5, 0)\):

Equation

\[\text{Distance } d = 0.5 - (-1) = 1.5\]

Equation

\[\vec{E}|_{\alpha Q} = \frac{k(\alpha Q)}{(1.5)^2} \, \hat{a}_y = \frac{1}{2.25} k(\alpha Q) \, \hat{a}_y = 0.444 \, k(\alpha Q) \, \hat{a}_y\]

Equating the net force to zero:

Equation

\[\begin{aligned} \vec{E}_{\text{net}} = 2(0.357 \, kQ) \hat{a}_y + 0.444 \, k(\alpha Q) \hat{a}_y &= 0 \\ 0.715 \, kQ + 0.444 \, k\alpha Q &= 0 \\ 0.715 &= -0.444 \alpha \\ \alpha &= \frac{-0.715}{0.444} \\ \alpha &\approx -1.61 \end{aligned}\]

✓

Final Answer

Answer: -1.6 (-1.7 to -1.5).

Question 02

Question 2

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).

Equation

\[\bar{H}(r, \theta, \phi) = \frac{1}{r^3} (\hat{r} P \cos\theta + \hat{\theta} \sin\theta)\]

Solution

Maxwell's equation for magnetic fields in free space states that the divergence of the magnetic flux density is zero:

Equation

\[\nabla \cdot \bar{B} = 0 \implies \nabla \cdot \bar{H} = 0\]

The formula for divergence in the spherical coordinate system is:

Equation

\[\nabla \cdot \bar{H} = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 H_r) + \frac{1}{r \sin\theta} \frac{\partial}{\partial \theta} (H_\theta \sin\theta) + \frac{1}{r \sin\theta} \frac{\partial H_\phi}{\partial \phi}\]

Identifying the components from the given \(\bar{H}\) field:

Equation

\[H_r = \frac{P \cos\theta}{r^3}, \quad H_\theta = \frac{\sin\theta}{r^3}, \quad H_\phi = 0\]

Substituting these into the divergence equation:

Equation

\[\begin{aligned} \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \cdot \frac{P \cos\theta}{r^3} \right) + \frac{1}{r \sin\theta} \frac{\partial}{\partial \theta} \left( \frac{\sin\theta}{r^3} \cdot \sin\theta \right) &= 0 \\[10pt] \frac{1}{r^2} (P \cos\theta) \frac{\partial}{\partial r} (r^{-1}) + \frac{1}{r^4 \sin\theta} \frac{\partial}{\partial \theta} (\sin^2\theta) &= 0 \\[10pt] \frac{1}{r^2} (P \cos\theta) (-r^{-2}) + \frac{1}{r^4 \sin\theta} (2\sin\theta \cos\theta) &= 0 \\[10pt] -\frac{P \cos\theta}{r^4} + \frac{2 \cos\theta}{r^4} &= 0 \\[10pt] \frac{\cos\theta}{r^4} (-P + 2) &= 0 \end{aligned}\]

For this to be zero for all \(r\) and \(\theta\):

Equation

\[-P + 2 = 0 \implies P = 2\]

✓

Final Answer

Answer: 2.

Question 03

Question 3

For the circuit shown in the figure, the source frequency is \(5000\) rad/sec. The mutual inductance between the magnetically coupled inductors is \(5\) mH with their self inductances being \(125\) mH and \(1\) mH. The Thevenin's impedance, \(Z_{\text{th}}\), between the terminals \(P\) and \(Q\) in \(\Omega\) is ________ (rounded off to 2 decimal places).

Solution

Given: \(\omega = 5000\) rad/sec.

First, we analyze the input impedance of the transformer section.

We calculate the reactances:

Equation

\[\begin{aligned} X_{L1} &= \omega L_1 = 5000 \times 125 \times 10^{-3} = 625 \, \Omega \\ X_{L2} &= \omega L_2 = 5000 \times 1 \times 10^{-3} = 5 \, \Omega \\ X_{M} &= \omega M = 5000 \times 5 \times 10^{-3} = 25 \, \Omega \\ X_{C} &= \frac{1}{\omega C} = \frac{1}{5000 \times 50 \times 10^{-6}} = 4 \, \Omega \end{aligned}\]

The load impedance on the secondary side (\(Z_L\)) consists of the capacitor and the inductor \(L_2\):

Equation

\[Z_{\text{secondary loop}} = jX_{L2} - jX_C = j5 - j4 = j1 \, \Omega\]

The input impedance looking into the primary of the coupled coil is:

Equation

\[Z_{\text{in(primary)}} = jX_{L1} + \frac{(\omega M)^2}{Z_{\text{secondary loop}}}\]

Substituting the values:

Equation

\[\begin{aligned} Z_{\text{in(primary)}} &= j625 + \frac{(25)^2}{j1} \\ &= j625 + \frac{625}{j} \\ &= j625 - j625 \\ &= 0 \, \Omega \end{aligned}\]

Since the input impedance of the coupled section is \(0 \, \Omega\), it acts as a short circuit. We can replace the transformer section with a wire. The circuit simplifies to a resistive network to find \(Z_{\text{th}}\) across terminals P and Q.

Looking into terminals P-Q, we have the \(4\Omega\) resistor in parallel with the \(2\Omega\) resistor, and that combination is in series with the other \(4\Omega\) resistor.

Equation

\[\begin{aligned} Z_{\text{th}} &= (4 \, \Omega \parallel 2 \, \Omega) + 4 \, \Omega \\ &= \left( \frac{4 \times 2}{4 + 2} \right) + 4 \\ &= \frac{8}{6} + 4 \\ &= 1.33 + 4 \\ &= 5.33 \, \Omega \end{aligned}\]

✓

Final Answer

Answer: 5.33 (5.30 to 5.40).