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Choose the best definition of a dipole.
Answer: c Explanation: An electric dipole generally refers to two equal and unlike (opposite signs) charges separated by a small distance. It can be anywhere, not necessarily at origin.
Answer: c
See lessExplanation: An electric dipole generally refers to two equal and unlike (opposite signs) charges separated by a small distance. It can be anywhere, not necessarily at origin.
Compute the power consumed by a material with current density 15 units in an area of 100 units. The potential measured across the material is 20V.
Answer: c Explanation: Power is given by, P= V X I, where I = J X A is the current. Thus power P = V X J X A = 20 X 15 X 100 = 30,000 joule = 30kJ.
Answer: c
See lessExplanation: Power is given by, P= V X I, where I = J X A is the current.
Thus power P = V X J X A = 20 X 15 X 100 = 30,000 joule = 30kJ.
Calculate the power of a material with electric field 100 units at a distance of 10cm with a current of 2A flowing through it.
Answer: b Explanation: Power is defined as the product of voltage and current. P = V X I, where V = E X d. Thus P = E X d X I = 100 X 0.1 X 2 = 20 units
Answer: b
See lessExplanation: Power is defined as the product of voltage and current.
P = V X I, where V = E X d. Thus P = E X d X I = 100 X 0.1 X 2 = 20 units
From the formula F = qE, can prove that work done is a product of force and displacement. State True/False
Answer: a Explanation: We know that F = qE = qV/d and W = qV. Thus it is clear that qV = W and qV = Fd. On equating both, we get W = Fd, which is the required proof.
Answer: a
See lessExplanation: We know that F = qE = qV/d and W = qV. Thus it is clear that qV = W and qV = Fd. On equating both, we get W = Fd, which is the required proof.
Find the force on a conductor of length 12m and magnetic flux density 20 units when a current of 0.5A is flowing through it.
Answer: b Explanation: The force on a conductor is given by F = BIL, where B = 20, I = 0.5 and L =12. Force F = 20 X 0.5 x 12 = 120 N
Answer: b
See lessExplanation: The force on a conductor is given by F = BIL, where B = 20, I = 0.5 and L =12. Force F = 20 X 0.5 x 12 = 120 N
Find the current in a conductor with resistance 2 ohm, electric field 2 units and distance 100cm.
Answer: a Explanation: We know that E = V/d. To get potential, V = E X d = 2 X 1 = 2 volts. From Ohm’s law, V = IR and current I = V/R = 2/2 = 1A.
Answer: a
See lessExplanation: We know that E = V/d. To get potential, V = E X d = 2 X 1 = 2 volts. From Ohm’s law, V = IR and current I = V/R = 2/2 = 1A.
If V = 2x2y + 20z – 4/(x2 + y2 ), find the density at A(6, -2.5, 3) in nC/m2 .
Answer: a Explanation: Find E from V, E = -Grad (V). We get E at A(6,-2.5,3) as 59.97i – 71.98j - 20k. Thus D = εE = 8.854 X 10-12 X (59.97i – 71.98j -20k) = (0.531i – 0.6373j – 0.177k) nC/m2 .
Answer: a
Explanation: Find E from V, E = -Grad (V). We get E at A(6,-2.5,3) as 59.97i – 71.98j –
20k. Thus D = εE = 8.854 X 10-12 X
(59.97i – 71.98j -20k) = (0.531i – 0.6373j – 0.177k) nC/m2
See less.
If the potential is given by, V = 10sin θ cosφ/r, find the density at the point P(2, π/2, 0) (in 10-12 units)
Answer: b Explanation: Since V is given find out E.E = -Grad(V) = – Grad(10sin θ cosφ/r). From E, we can easily compute D. D = εE = 8.854 X 10-12 X 5/2 = 22.13 units.
Answer: b
See lessExplanation: Since V is given find out E.E = -Grad(V) = – Grad(10sin θ cosφ/r). From E,
we can easily compute D. D = εE = 8.854 X 10-12 X 5/2 = 22.13 units.
Find the work done moving a charge 2C having potential V = 24volts is
Answer: d Explanation: The work done is the product of charge and potential. W = Q X V = 2 X 24 = 48 units.
Answer: d
See lessExplanation: The work done is the product of charge and potential.
W = Q X V = 2 X 24 = 48 units.
Find the potential of the function V = 60cos θ/r at the point P(3, 60, 25).
Answer: b Explanation: Given V = 60cos θ/r. For r = 3m and θ = 60, we get V = 60cos 60/3 = 20cos 60 = 10 units.
Answer: b
See lessExplanation: Given V = 60cos θ/r. For r = 3m and θ = 60, we get V = 60cos 60/3 = 20cos
60 = 10 units.