# Electromagnetic Theory Questions and Answers – Continuity Equation

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This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Continuity Equation”.

1. Find the current when the charge is a time function given by q(t) = 3t + t2 at 2 seconds.
a) 3
b) 5
c) 7
d) 9

Explanation: The current is defined as the rate of change of charge in a circuit ie, I = dq/dt. On differentiating the charge with respect to time, we get 3 + 2t. At time t = 2s, I = 7A.

2. The continuity equation is a combination of which of the two laws?
a) Ohm’s law and Gauss law
b) Ampere law and Gauss law
c) Ohm’s law and Ampere law
d) Maxwell law and Ampere law

Explanation: I = ∫ J.ds is the integral form of Ohm’s law and Div (J) = dq/dt is the Gauss law analogous to D. Through these two equations, we get Div(J) = -dρ/dt. This is the continuity equation.

3. Calculate the charge density for the current density given 20sin x i + ycos z j at the origin.
a) 20t
b) 21t
c) 19t
d) -20t

Explanation: Using continuity equation, the problem can be solved. Div(J) =
– dρ/dt. Div(J) = 20cos x + cos z. At origin, we get 20cos 0 + cos 0 = 21. To get ρ, on integrating the Div(J) with respect to t, the charge density will be 21t.

4. Compute the conductivity when the current density is 12 units and the electric field is 20 units. Also identify the nature of the material.
a) 1.67, dielectric
b) 1.67, conductor
c) 0.6, dielectric
d) 0.6, conductor

Explanation: The current density is the product of conductivity and electric field intensity. J = σE. To get σ, put J = 12 and E = 20. σ = 12/20 = 0.6. Since the conductivity is less than unity, the material is a dielectric.

5. Find the electron density when convection current density is 120 units and the velocity is 5m/s.
a) 12
b) 600
c) 24
d) 720

Explanation: The convection current density is given by J = ρe x v. To get ρe, put J = 120 and v = 5. ρe = 120/5 = 24 units.

6. Calculate the electric field when the conductivity is 20 units, electron density is 2.4 units and the velocity is 10m/s. Assume the conduction and convection current densities are same.
a) 2.4
b) 4.8
c) 3.6
d) 1.2

Explanation: The conduction current density is given by J = σE and the convection current density is J = ρe v. When both are equal, ρe v = σE. To get E, put σ = 20, ρe = 2.4 and v = 10, E = 2.4 x 10/20 = 1.2 units.

7. Find the mobility of the electrons when the drift velocity is 23 units and the electric field is 11 units.
a) 1.1
b) 2.2
c) 3.2
d) 0.9

Explanation: The mobility is defined as the drift velocity per unit electric field. Thus μe = vd/E = 23/11 = 2.1 units.

8. Find the resistance of a cylinder of area 200 units and length 100m with conductivity of 12 units.
a) 1/24
b) 1/48
c) 1/12
d) 1/96

Explanation: The resistance is given by R = ρL/A = L/σA. Put L = 100, σ = 12 and A = 200, we get R = 100/(12 x 200) = 1/24 units.

9. Calculate the potential when a conductor of length 2m is having an electric field of 12.3units.
a) 26.4
b) 42.6
c) 64.2
d) 24.6

Explanation: The electric field is given by E = V/L. To get V, put E = 12.3 and L = 2.Thus we get V = E x L = 12.3 x 2 = 24.6 units.

10. On equating the generic form of current density equation and the point form of Ohm’s law, we can obtain V=IR. State True/False.
a) True
b) False

Explanation: The generic current density equation is J = I/A and the point form of Ohm’s law is J = σ E. On equating both and substituting E = V/L, we get V = IL/σ A = IR which is the Ohm’s law.

Sanfoundry Global Education & Learning Series – Electromagnetic Theory.
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