The flux density at a point in space is given by \(\vec{B} = 4x\hat{a}_x + 2ky\hat{a}_y + 8\hat{a}_z \text{ Wb/m}^2\). The value of constant \(k\) must be equal to:
- -2
- -0.5
- +0.5
- +2
Solution
The given field is \(\vec{B} = 4x\hat{a}_x + 2ky\hat{a}_y + 8\hat{a}_z \text{ Wb/m}^2\).
For a valid magnetic field, the divergence must be zero (Maxwell's equation for magnetic flux continuity: \(\nabla \cdot \vec{B} = 0\)).
Equation
\[\nabla \cdot \vec{B} = \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z} = 0\]
Substituting the components:
Equation
\[\frac{\partial (4x)}{\partial x} + \frac{\partial (2ky)}{\partial y} + \frac{\partial (8)}{\partial z} = 0\]
Equation
\[4 + 2k + 0 = 0\]
Equation
\[2k = -4 \implies k = -2\]
Therefore, the value of the constant \(k\) must be \(-2\).
A
Final Answer
Correct answer: A.
A dielectric slab with \(500 \text{ mm} \times 500 \text{ mm}\) cross-section is \(0.4 \text{ m}\) long. The slab is subjected to a uniform electric field of \(\vec{E} = 6\hat{a}_x + 8\hat{a}_y \text{ kV/mm}\). The relative permittivity of the dielectric material is equal to 2. The value of constant \(\epsilon_0\) is \(8.85 \times 10^{-12} \text{ F/m}\). The energy stored in the dielectric in Joules is:
- \(8.85 \times 10^{-11}\)
- \(8.85 \times 10^{-5}\)
- \(88.5\)
- \(885\)
Solution
1. Calculate Volume (\(V\))
The slab has a cross-section of \(500 \text{ mm} \times 500 \text{ mm}\) and a length of \(0.4 \text{ m}\).
Convert mm to meters: \(500 \text{ mm} = 0.5 \text{ m}\).
Equation
\[V = (\text{Area}) \times (\text{Length}) = (0.5 \text{ m} \times 0.5 \text{ m}) \times 0.4 \text{ m} = 0.1 \text{ m}^3\]
2. Calculate Magnitude of Electric Field (\(E\))
The field is given as \(\vec{E} = 6\hat{a}_x + 8\hat{a}_y \text{ kV/mm}\).
Equation
\[E = |\vec{E}| = \sqrt{6^2 + 8^2} \text{ kV/mm} = \sqrt{36 + 64} \text{ kV/mm} = 10 \text{ kV/mm}\]
Convert to V/m:
Equation
\[10 \frac{\text{kV}}{\text{mm}} = 10 \times \frac{10^3 \text{ V}}{10^{-3} \text{ m}} = 10^7 \text{ V/m}\]
3. Calculate Stored Energy (\(W_E\))
The total energy stored is \(W_E = (\text{Energy Density}) \times (\text{Volume})\).
Equation
\[W_E = \left( \frac{1}{2} \epsilon E^2 \right) \cdot V\]
Where \(\epsilon = \epsilon_0 \epsilon_r = (8.85 \times 10^{-12}) \times 2 = 17.7 \times 10^{-12} \text{ F/m}\).
Equation
\[W_E = \frac{1}{2} \cdot (17.7 \times 10^{-12}) \cdot (10^7)^2 \cdot (0.1)\]
Equation
\[W_E = \frac{1}{2} \cdot 17.7 \times 10^{-12} \cdot 10^{14} \cdot 10^{-1}\]
Equation
\[W_E = \frac{1}{2} \cdot 17.7 \times 10^{1}\]
Equation
\[W_E = 8.85 \times 10 = 88.5 \text{ Joules}\]
C
Final Answer
Correct answer: C.