This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Stoke’s Theorem”.
1. Find the value of Stoke’s theorem for y i + z j + x k.
a) i + j
b) j + k
c) i + j + k
d) –i – j – k
Explanation: The curl of y i + z j + x k is i(0-1) – j(1-0) + k(0-1) =
-i –j –k. Since the curl is zero, the value of Stoke’s theorem is zero. The function is said to be irrotational.
2. The Stoke’s theorem uses which of the following operation?
Explanation: ∫A.dl = ∫∫ Curl (A).ds is the expression for Stoke’s theorem. It is clear that the theorem uses curl operation.
3. Which of the following theorem convert line integral to surface integral?
a) Gauss divergence and Stoke’s theorem
b) Stoke’s theorem only
c) Green’ s theorem only
d) Stoke’s and Green’s theorem
Explanation: The Stoke’s theorem is given by ∫A.dl = ∫∫ Curl (A).ds. Green’s theorem is given by, ∫ F dx + G dy = ∫∫ (dG/dx – dF/dy) dx dy. It is clear that both the theorems convert line to surface integral.
4. Find the value of Stoke’s theorem for A = x i + y j + z k. The state of the function will be
d) Curl free
5. The Stoke’s theorem can be used to find which of the following?
a) Area enclosed by a function in the given region
b) Volume enclosed by a function in the given region
c) Linear distance
d) Curl of the function
Explanation: It states that the line integral of a function gives the surface area of the function enclosed by the given region. This is computed using the double integral of the curl of the function.
6. The energy stored in an inductor 2H and current 4A is
Explanation: From Stoke’s theorem, we can calculate energy stored in an inductor as 0.5Li2. E = 0.5 X 2 X 42 = 16 units.
7. The voltage of a capacitor 12F with a rating of 2J energy is
Explanation: We can compute the energy stored in a capacitor from Stoke’s theorem as 0.5Cv2. Thus given energy is 0.5 X 12 X v2. We get v = 0.57 volts.
8. Find the power, given energy E = 2J and current density J = x2 varies from x = 0 and x = 1.
Explanation: From Stoke’s theorem, we can calculate P = E X I = ∫ E. J ds
= 2∫ x2 dx as x = 0->1. We get P = 2/3 units.
9. The conductivity of a material with current density 1 unit and electric field 200 μV is
Explanation: The current density is given by, J = σE. To find conductivity, σ = J/E = 1/200 X 10-6 = 5000.
10. The resistivity of a material with resistance 200 ohm, length 10m and area twice that of the length is
Explanation: Resistance calculated from Ohm’s law and Stoke’s theorem will be R = ρL/A. To get resistivity, ρ = RA/L = 200 X 20/10 = 400.
Sanfoundry Global Education & Learning Series – Electromagnetic Theory.
To practice all areas of Electromagnetic Theory, here is complete set of 1000+ Multiple Choice Questions and Answers.