Test: Real Time Applications

# Test: Real Time Applications

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## 10 Questions MCQ Test Electromagnetic Fields Theory (EMFT) | Test: Real Time Applications

Test: Real Time Applications for Electrical Engineering (EE) 2023 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Real Time Applications questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Real Time Applications MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Real Time Applications below.
Solutions of Test: Real Time Applications questions in English are available as part of our Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) & Test: Real Time Applications solutions in Hindi for Electromagnetic Fields Theory (EMFT) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Real Time Applications | 10 questions in 10 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) Exam | Download free PDF with solutions
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Test: Real Time Applications - Question 1

### Calculate the capacitance of a material in air with area 20 units and distance between plates is 5m.

Detailed Solution for Test: Real Time Applications - Question 1

Explanation: The capacitance of any material is given by, C = εA/d, where ε = εoεr is the permittivity in air and the material respectively. Thus C = 1 X 8.854 X 10-12 X 20/5 = 35.36pF

Test: Real Time Applications - Question 2

### The resistance of a material with conductivity 2millimho/m2, length 10m and area 50m is

Detailed Solution for Test: Real Time Applications - Question 2

Explanation: The resistance is given by, R = ρL/A, where ρ is the resistivity, the inverse of conductivity. R = 10/(0.002 X 50) = 100 ohm.

Test: Real Time Applications - Question 3

### Find the inductance of a coil with permeability 3.5, turns 100 and length 2m. Assume the area to be thrice the length.

Detailed Solution for Test: Real Time Applications - Question 3

Explanation: The inductance is given by L = μ N2A/l, where μ= μoμr is the permeability of air and the material respectively. N = 100 and Area = 3 X 2 = 6. L = 4π X 10-7 X 1002 X 6/2 = 131.94mH

Test: Real Time Applications - Question 4

Find the current density of a material with resistivity 20 units and electric field intensity 2000 units.

Detailed Solution for Test: Real Time Applications - Question 4

Explanation: The current density is given by J = σ E, where σ is the conductivity. Thus resistivity ρ = 1/σ. J = E/ρ = 2000/20 = 100 units.

Test: Real Time Applications - Question 5

Find the current in a conductor with resistance 2 ohm, electric field 2 units and distance 100cm.

Detailed Solution for Test: Real Time Applications - Question 5

Explanation: We know that E = V/d. To get potential, V = E X d = 2 X 1 = 2 volts. From Ohm’s law, V = IR and current I = V/R = 2/2 = 1A.

Test: Real Time Applications - Question 6

In electric fields, D= ε E. The correct expression which is analogous in magnetic fields will be

Detailed Solution for Test: Real Time Applications - Question 6

Explanation: In electric fields, the flux density is a product of permittivity and field intensity. Similarly, for magnetic fields, the magnetic flux density is the product of permeability and magnetic field intensity, given by B= μ H.

Test: Real Time Applications - Question 7

Find the force on a conductor of length 12m and magnetic flux density 20 units when a current of 0.5A is flowing through it.

Detailed Solution for Test: Real Time Applications - Question 7

Explanation: The force on a conductor is given by F = BIL, where B = 20, I = 0.5 and L = 12. Force F = 20 X 0.5 x 12 = 120 N.

Test: Real Time Applications - Question 8

From the formula F = qE, can prove that work done is a product of force and displacement. State True/False

Detailed Solution for Test: Real Time Applications - Question 8

Explanation: We know that F = qE = qV/d and W = qV. Thus it is clear that qV = W and qV = Fd. On equating both, we get W = Fd, which is the required proof.

Test: Real Time Applications - Question 9

Calculate the power of a material with electric field 100 units at a distance of 10cm with a current of 2A flowing through it.

Detailed Solution for Test: Real Time Applications - Question 9

Explanation: Power is defined as the product of voltage and current.
P = V X I, where V = E X d. Thus P = E X d X I = 100 X 0.1 X 2 = 20 units.

Test: Real Time Applications - Question 10

Compute the power consumed by a material with current density 15 units in an area of 100 units. The potential measured across the material is 20V.

Detailed Solution for Test: Real Time Applications - Question 10

Explanation: Power is given by, P= V X I, where I = J X A is the current.
Thus power P = V X J X A = 20 X 15 X 100 = 30,000 joule = 30kJ.

## Electromagnetic Fields Theory (EMFT)

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## Electromagnetic Fields Theory (EMFT)

11 videos|46 docs|62 tests (Scan QR code)