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Find the potential between two points p(1,-1,0) and q(2,1,3) with E = 40xy i + 20×2 j + 2 k
Answer: c Explanation: V = -∫ E.dl = -∫ (40xy dx + 20x2 dy + 2 dz) , from q to p. On integrating, we get 106 volts
Answer: c
See lessExplanation: V = -∫ E.dl = -∫ (40xy dx + 20×2 dy + 2 dz) , from q to p.
On integrating, we get 106 volts
If V = 2x2y – 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.
Answer: b
See lessExplanation: 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.
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.
Answer: a
See lessExplanation: Vab = -∫ E.dl is the relation between potential and field. It is clear that it is given by line integral.
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)
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.
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.
See lessWhich of the following Maxwell equations use curl operation?
Answer: a Explanation: Maxwell 1st equation, Curl (H) = J (Ampere law) Maxwell 2nd equation, Curl (E) = -D(B)/Dt (Faraday’s law) Maxwell 3rd equation, Div (D) = Q (Gauss law for electric field) Maxwell 4th equation, Div (B) = 0(Gauss law for magnetic field) It is clear that only 1st and 2nd equationRead more
Answer: a
Explanation: Maxwell 1st equation, Curl (H) = J (Ampere law)
Maxwell 2nd equation, Curl (E) = -D(B)/Dt (Faraday’s law)
Maxwell 3rd equation, Div (D) = Q (Gauss law for electric field)
Maxwell 4th equation, Div (B) = 0(Gauss law for magnetic field)
It is clear that only 1st and 2nd equations use the curl operation.
See lessCurl cannot be empl oyed in which one of the following?
Answer: d Explanation: In the Directional coupler, Magic Tee, Isolator and Terminator the EM waves travel both in linear and angular motion, which involves curl too. But in waveguides, as the name suggests, only guided propagation occurs (no bending or curl of waves).
Answer: d
See lessExplanation: In the Directional coupler, Magic Tee, Isolator and Terminator the EM
waves travel both in linear and angular motion, which involves curl too. But in
waveguides, as the name suggests, only guided propagation occurs (no bending or curl
of waves).
Find the curl of the vector A = yz i + 4xy j + y k
Answer: d Explanation: Curl A = i(Dy(y) – Dz(0)) – j (Dx(0) – Dz(yz)) + k(Dx(4xy) – Dy(yz)) = i + y j + (4y – z)k.
Answer: d
See lessExplanation: Curl A = i(Dy(y) – Dz(0)) – j (Dx(0) – Dz(yz)) + k(Dx(4xy) – Dy(yz)) =
i + y j + (4y – z)k.
Find the curl of A = (y cos ax)i + (y + ex )k
Answer: b Explanation: Curl A = i(Dy(y + ex)) – j (Dx(y + ex) – Dz(y cos ax)) + k(-Dy(y cos ax)) = 1.i – j(ex) – k cos ax = i – ex j – cos ax k.
Answer: b
See lessExplanation: Curl A = i(Dy(y + ex)) – j (Dx(y + ex) – Dz(y cos ax)) + k(-Dy(y cos ax))
= 1.i – j(ex) – k cos ax = i – ex j – cos ax k.
Is the vector is irrotational. E = yz i + xz j + xy k
Answer: a Explanation: Curl E = i(Dy(xy) – Dz(xz)) – j (Dx(xy) – Dz(yz)) + k(Dx(xz) – Dy(yz)) = i(x – x) – j(y – y) + k(z – z) = 0 Since the curl is zero, the vector is irrotational or curl-free.
Answer: a
See lessExplanation: Curl E = i(Dy(xy) – Dz(xz)) – j (Dx(xy) – Dz(yz)) + k(Dx(xz) – Dy(yz)) =
i(x – x) – j(y – y) + k(z – z) = 0
Since the curl is zero, the vector is irrotational or curl-free.
Find the curl of the vector and state its nature at (1,1,-0.2) F = 30 i + 2xy j + 5xz2 k
Answer: d Explanation: Curl F = -5z2j + 2y k. At (1,1,-0.2), Curl F = -0.2 j + 2 k. |Curl F| = √(- 0.22+22) = √4.04.
Answer: d
See lessExplanation: Curl F = -5z2j + 2y k. At (1,1,-0.2), Curl F = -0.2 j + 2 k. |Curl F| = √(-
0.22+22) = √4.04.