Test: Line Integral


10 Questions MCQ Test Electromagnetic Fields Theory | Test: Line Integral


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This mock test of Test: Line Integral for Electronics and Communication Engineering (ECE) helps you for every Electronics and Communication Engineering (ECE) entrance exam. This contains 10 Multiple Choice Questions for Electronics and Communication Engineering (ECE) Test: Line Integral (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Line Integral quiz give you a good mix of easy questions and tough questions. Electronics and Communication Engineering (ECE) students definitely take this Test: Line Integral exercise for a better result in the exam. You can find other Test: Line Integral extra questions, long questions & short questions for Electronics and Communication Engineering (ECE) on EduRev as well by searching above.
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)

Solution:

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.

QUESTION: 2

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

Solution:

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

QUESTION: 3

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

Solution:

Answer: b
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.

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

Solution:

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

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.

Solution:

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.

QUESTION: 6

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

Solution:

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).

QUESTION: 7

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

Solution:

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.

QUESTION: 8

The potential in a lamellar field is

Solution:

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

QUESTION: 9

Line integral is used to calculate

Solution:

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.

QUESTION: 10

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

Solution:

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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