Line Integral - GATE EE Electromagnetic Fields Theory Free MCQ Test

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)

A.
0.64
B.
0.72
C.
0.78
D.
0.80

Detailed Solution: 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

A.
True
B.
False

Detailed Solution: 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)

A.
47.905
B.
57.905
C.
67.905
D.
77.905

Detailed Solution: 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

A.
104
B.
105
C.
106
D.
107

Detailed Solution: 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.

A.
-8.014
B.
-8.114
C.
-8.214
D.
-8.314

Detailed Solution: 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.

A.
True
B.
False

Detailed Solution: 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

A.
Solenoidal
B.
Rotational
C.
Irrotational
D.
Conservative

Detailed Solution: 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

A.
1
B.
0
C.
-1
D.
∞

Detailed Solution: 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

A.
Force
B.
Area
C.
Volume
D.
Length

Detailed Solution: 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

A.
0.2
B.
0.4
C.
0.6
D.
0.8

Detailed Solution: 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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Line Integral - Free MCQ Practice Test with solutions, GATE EE Electromagnetic

MCQ Practice Test & Solutions: Test: Line Integral (10 Questions)

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)

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

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

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

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

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

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

The potential in a lamellar field is

Line integral is used to calculate

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

Electromagnetic Fields Theory (EMFT)

Electromagnetic Fields Theory (EMFT)

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Line Integral - Free MCQ Practice Test with solutions, GATE EE Electromagnetic

MCQ Practice Test & Solutions: Test: Line Integral (10 Questions)

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)

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

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

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

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

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

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

The potential in a lamellar field is

Line integral is used to calculate

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

Electromagnetic Fields Theory (EMFT)

Electromagnetic Fields Theory (EMFT)

Top Courses for Electrical Engineering (EE)

About this Electrical Engineering (EE) Practice Test

Explore more Electrical Engineering (EE) Study Resources

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)

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

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

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

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