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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)
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.
The integral form of potential and field relation is given by line integral. State True/False
Answer: a
Explanation: Vab = ∫ E.dl is the relation between potential and field. It is clear that it is given by line integral.
If V = 2x^{2}y – 5z, find its electric field at point (4,3,6)
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.
Find the potential between two points p(1,1,0) and q(2,1,3) with E = 40xy i + 20x^{2} j + 2 k
Answer: c
Explanation: V = ∫ E.dl = ∫ (40xy dx + 20x^{2} dy + 2 dz) , from q to p.
On integrating, we get 106 volts.
Find the potential between a(7,2,1) and b(4,1,2). Given E = (6y/x^{2} )i + ( 6/x) j + 5 k.
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.
The potential of a uniformly charged line with density λ is given by,
λ/(2πε) ln(b/a). State True/False.
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).
A field in which a test charge around any closed surface in static path is zero is called
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.
Answer: b
Explanation: Work done in a lamellar field is zero. ∫ E.dl = 0,thus ∑V = 0. The potential will be zero.
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.
The energy stored in the inductor 100mH with a current of 2A is
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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