GATE 2021 Electromagnetic Fields Questions and Solutions

Question 01

Question 1

One coulomb of point charge moving with a uniform velocity \(10\hat{x}\) m/s enters the region \(x \ge 0\) having a magnetic flux density \(\vec{B} = (10y\hat{x} + 10x\hat{y} + 10\hat{z})\) 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).

Solution

The force on a charge moving with velocity \(\vec{v}\) due to a magnetic field is given by the Lorentz force law:

Equation

\[\vec{F} = Q(\vec{v} \times \vec{B})\]

GATE 2021 Electromagnetic Fields Q1 figure

Substituting the given values (\(Q=1\) C, \(\vec{v} = 10\hat{x}\)):

Equation

\[\begin{aligned} \vec{F} &= 1 [10\hat{x} \times (10y\hat{x} + 10x\hat{y} + 10\hat{z})] \\ &= 100y(\hat{x} \times \hat{x}) + 100x(\hat{x} \times \hat{y}) + 100(\hat{x} \times \hat{z}) \\ &= 0 + 100x(\hat{z}) + 100(-\hat{y}) \\ &= 100x\hat{z} - 100\hat{y} \end{aligned}\]

At \(x = 0^+\):

Equation

\[\vec{F}|_{x=0^+} = -100\hat{y}\]

The magnitude is:

Equation

\[|\vec{F}| = \sqrt{0^2 + (-100)^2 + 0^2} = 100 \text{ N}\]

✓

Final Answer

Answer: 100.

Question 02

Question 2

Consider a large parallel plate capacitor. The gap '\(d\)' between the two plates is filled entirely with a dielectric slab of relative permittivity 5. The plates are initially charged to a potential difference of \(V\) volts and then disconnected from the source. If the dielectric slab is pulled out completely, then the ratio of the new electric field \(E_2\) in the gap to the original electric field \(E_1\) is ________ .

GATE 2021 Electromagnetic Fields Q2 figure

Solution

Since the voltage source is disconnected, the charge \(Q\) on the plates remains constant (\(Q_1 = Q_2 = Q\)).

Case-1 (With Dielectric): \(\epsilon_r = 5, V_1 = V\)

Equation

\[\begin{aligned} Q_1 &= C_1 V_1 \\ Q &= \frac{\epsilon_0 (5) A}{d} V_1 \\ Q &= 5 \left( \frac{\epsilon_0 A}{d} \right) V_1 \quad \dots(\text{i}) \end{aligned}\]

Case-2 (Air/Vacuum): \(\epsilon_r = 1, V_2 = ?\)

Equation

\[\begin{aligned} Q_2 &= C_2 V_2 \\ Q &= \frac{\epsilon_0 (1) A}{d} V_2 \\ Q &= \left( \frac{\epsilon_0 A}{d} \right) V_2 \quad \dots(\text{ii}) \end{aligned}\]

Equating (i) and (ii):

Equation

\[\begin{aligned} 5 \left( \frac{\epsilon_0 A}{d} \right) V_1 &= \left( \frac{\epsilon_0 A}{d} \right) V_2 \\ 5 V_1 &= V_2 \\ \frac{V_2}{V_1} &= 5 \end{aligned}\]

The electric field is given by \(E = V/d\). Thus:

Equation

\[\frac{E_2}{E_1} = \frac{V_2 / d}{V_1 / d} = \frac{V_2}{V_1} = 5\]

✓

Final Answer

Answer: 5.

Question 03

Question 3

Which one of the following vector functions represents a magnetic field \(\vec{B}\)? (\(\hat{x}\), \(\hat{y}\) and \(\hat{z}\) are unit vectors along \(x\)-axis, \(y\)-axis and \(z\)-axis, respectively).

(Note: Capital \(\hat{X}, \hat{Y}, \hat{Z}\) in options represent unit vectors)

\(10x\hat{X} + 20y\hat{Y} - 30z\hat{Z}\)
\(10x\hat{X} - 30z\hat{Y} + 20y\hat{Z}\)
\(10z\hat{X} + 20y\hat{Y} - 30x\hat{Z}\)
\(10y\hat{X} + 20x\hat{Y} - 10z\hat{Z}\)

Solution

For a vector field to represent a valid magnetic flux density \(\vec{B}\), it must satisfy Maxwell's equation for the non-existence of magnetic monopoles (Solenoidal nature):

Equation

\[\nabla \cdot \vec{B} = 0\]

The divergence in Cartesian coordinates is:

Equation

\[\nabla \cdot \vec{B} = \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z}\]

Checking option (a): \(\vec{B} = 10x\hat{x} + 20y\hat{y} - 30z\hat{z}\)

Equation

\[\begin{aligned} \nabla \cdot \vec{B} &= \frac{\partial}{\partial x}(10x) + \frac{\partial}{\partial y}(20y) + \frac{\partial}{\partial z}(-30z) \\ &= 10 + 20 - 30 \\ &= 0 \end{aligned}\]

Since the divergence is zero, option (a) represents a valid magnetic field.

Final Answer

Correct answer: A.