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QUESTION: 1

An electric field is given as E = 6y^{2}z i + 12xyz j + 6xy^{2} 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 (6y^{2}z i + 12xyz j + 6xy^{2} 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 = 2x^{2}y – 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 + 20x^{2} j + 2 k

Solution:

Answer: c

Explanation: V = -∫ E.dl = -∫ (40xy dx + 20x^{2} 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/x^{2} )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 = 0.5LI^{2} = 0.5 X 0.1 X 2^{2} = 0.2 Joule.

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