Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
Calculate the power of a material with electric field 100 units at a distance of 10cm with a current of 2A flowing through it.
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
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
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.
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.
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
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.
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.
Find the current density of a material with resistivity 20 units and electric field intensity 2000 units
d Explanation: The current density is given by J = σ E, where σ is the conductivity. Thus resistivity ρ = 1/σ. J = E/ρ = 2000/20 = 100 units.
d
See lessExplanation: The current density is given by J = σ E, where σ is the conductivity. Thus
resistivity ρ = 1/σ. J = E/ρ = 2000/20 = 100 units.
Find the inductance of a coil with permeability 3.5, turns 100 and length 2m. Assume the area to be thrice the length
a Explanation: The inductance is given by L = μ N2A/l, where μ= μoμr is the permeability of air and the material respectively. N = 100 and Area = 3 X 2 = 6. L = 4π X 10-7 X 1002 X 6/2 = 131.94mH.
a
See lessExplanation: The inductance is given by L = μ N2A/l, where μ= μoμr is the permeability of
air and the material respectively. N = 100 and Area = 3 X 2 = 6. L = 4π X 10-7 X 1002 X
6/2 = 131.94mH.
Calculate the capacitance of a material in air with area 20 units and distance between plates is 5m.
a Explanation: The capacitance of any material is given by, C = εA/d, where ε = εoεr is the permittivity in air and the material respectively. Thus C = 1 X 8.854 X 10-12 X 20/5 = 35.36pF
a
See lessExplanation: The capacitance of any material is given by, C = εA/d, where ε = εoεr is the
permittivity in air and the material respectively. Thus C = 1 X 8.854 X 10-12 X 20/5 =
35.36pF
Find the electric flux density surrounding a material with field intensity of 2xyz placed in transformer oil ( εr = 2.2) at the point P(1,2,3) is (in 10-10 units)
c Explanation: D = εE, where ε = εo εr. The flux density is given by, D = 8.854 X 10-12 X 2.2 X 2(1)(2)(3) = 2.33 X 10-10 units.
c
See lessExplanation: D = εE, where ε = εo εr. The flux density is given by,
D = 8.854 X 10-12 X 2.2 X 2(1)(2)(3) = 2.33 X 10-10 units.
The electric flux density and electric field intensity have which of the following relation?
a Explanation: The electric flux density is directly proportional to electric field intensity. The proportionality constant is permittivity. D=ε E. It is clear that both are in linear relationship.
a
See lessExplanation: The electric flux density is directly proportional to electric field intensity. The
proportionality constant is permittivity. D=ε E. It is clear that both are in linear
relationship.
Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 6m.
d Explanation: The radius R = 6m encloses all the three Gaussian cylinders. By Gauss law, ψ = Q D(2πRL) = σ(2πRL), D(2π X 6) = Q1 + Q2 + Q3 = σ1(2π X 2) + σ2(2π X 4) + σ3(2π X 5), here σ1 = 5, σ2 = -2 and σ3 = -3. We get D = -13/6 units
d
See lessExplanation: The radius R = 6m encloses all the three Gaussian cylinders.
By Gauss law, ψ = Q
D(2πRL) = σ(2πRL), D(2π X 6) = Q1 + Q2 + Q3 = σ1(2π X 2) + σ2(2π X 4) + σ3(2π X
5), here σ1 = 5, σ2 = -2 and σ3 = -3. We get D = -13/6 units
Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ =-3 at R = 5m. Find the flux density at R = 4.5m
c Explanation: The Gaussian cylinder of R = 4.5m encloses sum of charges of two cylinders (R = 2m and R = 4m). By Gauss law, ψ = Q D(2πRL) = σ(2πRL), D(2π X 4.5) = Q1 + Q2 = σ1(2π X 2) + σ2(2π X 4), here σ1 = 5 and σ2 = -2. We get D = 2/4.5 units
c
See lessExplanation: The Gaussian cylinder of R = 4.5m encloses sum of charges of two
cylinders (R = 2m and R = 4m).
By Gauss law, ψ = Q
D(2πRL) = σ(2πRL), D(2π X 4.5) = Q1 + Q2 = σ1(2π X 2) + σ2(2π X 4), here σ1 = 5
and σ2 = -2. We get D = 2/4.5 units