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Test: Line Integral - Electrical Engineering (EE) MCQ


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

Test: Line Integral for Electrical Engineering (EE) 2024 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) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Line Integral below.
Solutions of Test: Line Integral questions in English are available as part of our Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) & Test: Line Integral 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: Line Integral | 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
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

Answer: b
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

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

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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

The electric field (E) at a point in space is the force per unit charge experienced by a test charge at that point, hence the unit of E is newton per coulomb (N/C). The electric field is defined as E = - grad V

where grad is the gradient operator and V is the electric potential.

 If V = 2x²y - 5z

the electric field at point (-4, 3, 6) is given as follows:

E = - grad V

From the formula above, the gradient of V = (2x²y - 5z)

is given by grad V = (dV/dx)i + (dV/dy)j + (dV/dz)k

Where i, j and k are the unit vectors along the x, y and z axes respectively.

Thus,grad V = (4xy)i + (2x²)j - 5k

Now, evaluating at (-4, 3, 6)

we get the electric field,E = - grad V

=(-4)(3)(4)i + (2)(16)j - (5)k

= -48i + 32j - 5k

Therefore, the electric field at point (-4,3,6) is  E = -48i + 32j - 5k

which implies that the magniture of the electric field is given by sqrt[(-48)² + (32)² + (-5)²] = 57.905 N/C.

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

Answer: c
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

Answer: c
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

Answer: a
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

Answer: d
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

Answer: b
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

Answer: d
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

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

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