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What is the energy stored in the magnetic field at a solenoid of 40m long and 4m diameter wound with 100 turns of wire carrying a current at 20 A?
- a)0.705 Joule
- b)0.789 Joule
- c)0.587 Joule
- d)0.658 Joule
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
What is the energy stored in the magnetic field at a solenoid of 40m l...
The inductance of a solenoid is given by,
Energy stored in an inductor is given by,
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What is the energy stored in the magnetic field at a solenoid of 40m l...
To find the energy stored in the magnetic field of a solenoid, we can use the formula:
E = (1/2) * L * I^2,
where E is the energy stored, L is the inductance of the solenoid, and I is the current flowing through the solenoid.
First, let's calculate the inductance of the solenoid using the formula:
L = (μ0 * N^2 * A) / l,
where L is the inductance, μ0 is the permeability of free space (4π x 10^-7 H/m), N is the number of turns of wire, A is the cross-sectional area of the solenoid, and l is the length of the solenoid.
Given:
Number of turns, N = 100
Length, l = 40 m
Diameter, d = 4 m
1. Finding the cross-sectional area (A):
The diameter of the solenoid is given as 4 m, so the radius (r) will be half of that, i.e., 2 m.
A = π * r^2
= π * (2 m)^2
= 4π m^2
2. Finding the inductance (L):
L = (μ0 * N^2 * A) / l
= (4π x 10^-7 H/m) * (100^2 turns^2) * (4π m^2) / 40 m
= 16π^2 x 10^-7 H
≈ 1.591 x 10^-5 H
3. Finding the energy stored (E):
E = (1/2) * L * I^2
= (1/2) * (1.591 x 10^-5 H) * (20 A)^2
= 0.789 J
Therefore, the energy stored in the magnetic field of the solenoid is approximately 0.789 Joules.
E = (1/2) * L * I^2,
where E is the energy stored, L is the inductance of the solenoid, and I is the current flowing through the solenoid.
First, let's calculate the inductance of the solenoid using the formula:
L = (μ0 * N^2 * A) / l,
where L is the inductance, μ0 is the permeability of free space (4π x 10^-7 H/m), N is the number of turns of wire, A is the cross-sectional area of the solenoid, and l is the length of the solenoid.
Given:
Number of turns, N = 100
Length, l = 40 m
Diameter, d = 4 m
1. Finding the cross-sectional area (A):
The diameter of the solenoid is given as 4 m, so the radius (r) will be half of that, i.e., 2 m.
A = π * r^2
= π * (2 m)^2
= 4π m^2
2. Finding the inductance (L):
L = (μ0 * N^2 * A) / l
= (4π x 10^-7 H/m) * (100^2 turns^2) * (4π m^2) / 40 m
= 16π^2 x 10^-7 H
≈ 1.591 x 10^-5 H
3. Finding the energy stored (E):
E = (1/2) * L * I^2
= (1/2) * (1.591 x 10^-5 H) * (20 A)^2
= 0.789 J
Therefore, the energy stored in the magnetic field of the solenoid is approximately 0.789 Joules.
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What is the energy stored in the magnetic field at a solenoid of 40m long and 4m diameter wound with 100 turns of wire carrying a current at 20 A?a)0.705 Jouleb)0.789 Joulec)0.587 Jouled)0.658 JouleCorrect answer is option 'B'. Can you explain this answer?
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