Test: Line Integral

# Test: Line Integral

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## 10 Questions MCQ Test Electromagnetic Fields Theory (EMFT) | Test: Line Integral

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

### An electric field is given as E = 6y2z i + 12xyz j + 6xy2 k. An incremental path is given by dl = -3 i + 5 j – 2 k mm. The work done in moving a 2mC charge along the path if the location of the path is at p(0,2,5) is (in Joule)

Detailed Solution for Test: Line Integral - Question 1

Explanation: W = -Q E.dl
W = -2 X 10-3 X (6y2z i + 12xyz j + 6xy2 k) . (-3 i + 5 j -2 k)
At p(0,2,5), W = -2(-18.22.5) X 10-3 = 0.72 J.

Test: Line Integral - Question 2

### The integral form of potential and field relation is given by line integral. State True/False

Detailed Solution for Test: Line Integral - Question 2

Explanation: Vab = -∫ E.dl is the relation between potential and field. It is clear that it is given by line integral.

Test: Line Integral - Question 3

### If V = 2x2y – 5z, find its electric field at point (-4,3,6)

Detailed Solution for Test: Line Integral - Question 3

Explanation: E = -Grad (V) = -4xy i – 2×2 j + 5k
At (-4,3,6), E = 48 i – 32 j + 5 k, |E| = √3353 = 57.905 units.

Test: Line Integral - Question 4

Find the potential between two points p(1,-1,0) and q(2,1,3) with E = 40xy i + 20x2 j + 2 k

Detailed Solution for Test: Line Integral - Question 4

Explanation: V = -∫ E.dl = -∫ (40xy dx + 20x2 dy + 2 dz) , from q to p.
On integrating, we get 106 volts.

Test: Line Integral - Question 5

Find the potential between a(-7,2,1) and b(4,1,2). Given E = (-6y/x2 )i + ( 6/x) j + 5 k.

Detailed Solution for Test: Line Integral - Question 5

Explanation: V = -∫ E.dl = -∫ (-6y/x2 )dx + ( 6/x)dy + 5 dz, from b to a.
On integrating, we get -8.214 volts.

Test: Line Integral - Question 6

The potential of a uniformly charged line with density λ is given by,
λ/(2πε) ln(b/a). State True/False.

Detailed Solution for Test: Line Integral - Question 6

Explanation: The electric field intensity is given by, E = λ/(2πεr)
Vab = -∫ E.dr = -∫ λ/(2πεr). On integrating from b to a, we get λ/(2πε) ln(b/a).

Test: Line Integral - Question 7

A field in which a test charge around any closed surface in static path is zero is called

Detailed Solution for Test: Line Integral - Question 7

Explanation: Work done in moving a charge in a closed path is zero. It is expressed as, ∫ E.dl = 0. The field having this property is called conservative or lamellar field.

Test: Line Integral - Question 8

The potential in a lamellar field is

Detailed Solution for Test: Line Integral - Question 8

Explanation: Work done in a lamellar field is zero. ∫ E.dl = 0,thus ∑V = 0. The potential will be zero.

Test: Line Integral - Question 9

Line integral is used to calculate

Detailed Solution for Test: Line Integral - Question 9

Explanation: Length is a linear quantity, whereas area is two dimensional and volume is three dimensional. Thus single or line integral can be used to find length in general.

Test: Line Integral - Question 10

The energy stored in the inductor 100mH with a current of 2A is

Detailed Solution for Test: Line Integral - Question 10

Explanation: dw = ei dt = Li di, W = L∫ i.di
Energy E = 0.5LI2 = 0.5 X 0.1 X 22 = 0.2 Joule.

## Electromagnetic Fields Theory (EMFT)

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## Electromagnetic Fields Theory (EMFT)

11 videos|46 docs|62 tests