Answer:
C. 0.25J
Explanation:
Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;
L is the inductance
I is the current flowing in the inductor
Given parameters
L = 20mH = 20×10^-3H
I = 5A
Required
Energy stored in the magnetic field.
E = 1/2 × 20×10^-3 × 5²
E = 1/2 × 20×10^-3 × 25
E = 10×10^-3 × 25
E = 0.01 × 25
E = 0.25Joules.
Hence the energy stored in the magnetic field of this inductor is 0.25Joules
Answer:
Following are the solution to the given question:
Explanation:
For charging plates that are connected in a similar manner:
Calculating the total charge:
Calculating the common potential:
Calculating the charge after redistribution:
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Answer:
The linear charge density is 5.19 X 10⁻⁶ C/m
Explanation:
The potential difference between two cylinders, is given as
V = (λ/2πε)ln(b/a)
where;
λ is the line charge density on the power line.
b is the distance between the power line = 1 m
a is the radius of the wire = 1.5 cm = 0.015 m
ε is the permittivity of free space = 8.9 X 10⁻¹² C
V*2πε = λ* ln(b/a)
3900 *(2π*8.9 x10⁻¹²)= λ *ln(1/0.015)
2.1812 X 10⁻⁷ = 4.1997* λ
λ = 5.19 X 10⁻⁶ C/m
Therefore, the linear charge density is 5.19 X 10⁻⁶ C/m
