Test: Continuity Equation

# Test: Continuity Equation

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## 10 Questions MCQ Test Electromagnetic Fields Theory (EMFT) | Test: Continuity Equation

Test: Continuity Equation for Electrical Engineering (EE) 2023 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Continuity Equation questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Continuity Equation MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Continuity Equation below.
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Test: Continuity Equation - Question 1

### Find the current when the charge is a time function given by q(t) = 3t + t2 at 2 seconds.

Detailed Solution for Test: Continuity Equation - Question 1

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.

Test: Continuity Equation - Question 2

### The continuity equation is a combination of which of the two laws?

Detailed Solution for Test: Continuity Equation - Question 2

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.

Test: Continuity Equation - Question 3

### Calculate the charge density for the current density given 20sin x i + ycos z j at the origin.

Detailed Solution for Test: Continuity Equation - Question 3

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.

Test: Continuity Equation - Question 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.

Detailed Solution for Test: Continuity Equation - Question 4

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.

Test: Continuity Equation - Question 5

Find the electron density when convection current density is 120 units and the velocity is 5m/s.

Detailed Solution for Test: Continuity Equation - Question 5

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.

Test: Continuity Equation - Question 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.

Detailed Solution for Test: Continuity Equation - Question 6

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.

Test: Continuity Equation - Question 7

Find the mobility of the electrons when the drift velocity is 23 units and the electric field is 11 units.

Detailed Solution for Test: Continuity Equation - Question 7

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

Test: Continuity Equation - Question 8

Find the resistance of a cylinder of area 200 units and length 100m with conductivity of 12 units.

Detailed Solution for Test: Continuity Equation - Question 8

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.

Test: Continuity Equation - Question 9

Calculate the potential when a conductor of length 2m is having an electric field of 12.3units.

Detailed Solution for Test: Continuity Equation - Question 9

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.

Test: Continuity Equation - Question 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.

Detailed Solution for Test: Continuity Equation - Question 10

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

## Electromagnetic Fields Theory (EMFT)

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## Electromagnetic Fields Theory (EMFT)

11 videos|46 docs|62 tests