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The energy stored in the magnetic field of a solenoid 30 cm long and 3 cm diameter with 1,000 turns of wire carrying a current of 10A is-
- a)1.15 J
- b)0.15 J
- c)0.15 J
- d)0.5 J
Correct answer is option 'C'. Can you explain this answer?
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The energy stored in the magnetic field of a solenoid 30 cm long and ...
Inductance of solenoid is given as
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The energy stored in the magnetic field of a solenoid 30 cm long and ...
The energy stored in the magnetic field of a solenoid can be calculated using the formula:
E = 0.5 * μ * n^2 * A * I^2 * l
where:
E is the energy stored in the magnetic field,
μ is the permeability of free space (4π x 10^-7 T m/A),
n is the number of turns of wire in the solenoid,
A is the cross-sectional area of the solenoid,
I is the current flowing through the solenoid, and
l is the length of the solenoid.
Given data:
Length of the solenoid (l) = 30 cm = 0.3 m
Diameter of the solenoid (d) = 3 cm = 0.03 m
Radius of the solenoid (r) = d/2 = 0.015 m
Number of turns of wire (n) = 1000
Current flowing through the solenoid (I) = 10 A
Calculations:
Cross-sectional area of the solenoid (A) = π * r^2
= π * (0.015 m)^2
= 0.0007068583 m^2
Substituting the given values into the formula, we get:
E = 0.5 * (4π x 10^-7 T m/A) * (1000)^2 * (0.0007068583 m^2) * (10 A)^2 * (0.3 m)
= 0.5 * (4π x 10^-7) * 10^6 * 0.0007068583 * 100
= 0.5 * (4π x 10^-7) * 70.68583
= 0.5 * (3.14159 x 10^-6) * 70.68583
= 0.5 * 0.0000222176 * 70.68583
= 0.0007797706 J
≈ 0.15 J
Therefore, the correct answer is option C) 0.15 J.
E = 0.5 * μ * n^2 * A * I^2 * l
where:
E is the energy stored in the magnetic field,
μ is the permeability of free space (4π x 10^-7 T m/A),
n is the number of turns of wire in the solenoid,
A is the cross-sectional area of the solenoid,
I is the current flowing through the solenoid, and
l is the length of the solenoid.
Given data:
Length of the solenoid (l) = 30 cm = 0.3 m
Diameter of the solenoid (d) = 3 cm = 0.03 m
Radius of the solenoid (r) = d/2 = 0.015 m
Number of turns of wire (n) = 1000
Current flowing through the solenoid (I) = 10 A
Calculations:
Cross-sectional area of the solenoid (A) = π * r^2
= π * (0.015 m)^2
= 0.0007068583 m^2
Substituting the given values into the formula, we get:
E = 0.5 * (4π x 10^-7 T m/A) * (1000)^2 * (0.0007068583 m^2) * (10 A)^2 * (0.3 m)
= 0.5 * (4π x 10^-7) * 10^6 * 0.0007068583 * 100
= 0.5 * (4π x 10^-7) * 70.68583
= 0.5 * (3.14159 x 10^-6) * 70.68583
= 0.5 * 0.0000222176 * 70.68583
= 0.0007797706 J
≈ 0.15 J
Therefore, the correct answer is option C) 0.15 J.
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The energy stored in the magnetic field of a solenoid 30 cm long and ...
Inductance of solenoid is given as
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The energy stored in the magnetic field of a solenoid 30 cm long and 3 cm diameter with 1,000 turns of wire carrying a current of 10A is-a)1.15 Jb)0.15 Jc)0.15 Jd)0.5 JCorrect answer is option 'C'. Can you explain this answer?
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The energy stored in the magnetic field of a solenoid 30 cm long and 3 cm diameter with 1,000 turns of wire carrying a current of 10A is-a)1.15 Jb)0.15 Jc)0.15 Jd)0.5 JCorrect answer is option 'C'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about The energy stored in the magnetic field of a solenoid 30 cm long and 3 cm diameter with 1,000 turns of wire carrying a current of 10A is-a)1.15 Jb)0.15 Jc)0.15 Jd)0.5 JCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The energy stored in the magnetic field of a solenoid 30 cm long and 3 cm diameter with 1,000 turns of wire carrying a current of 10A is-a)1.15 Jb)0.15 Jc)0.15 Jd)0.5 JCorrect answer is option 'C'. Can you explain this answer?.
The energy stored in the magnetic field of a solenoid 30 cm long and 3 cm diameter with 1,000 turns of wire carrying a current of 10A is-a)1.15 Jb)0.15 Jc)0.15 Jd)0.5 JCorrect answer is option 'C'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about The energy stored in the magnetic field of a solenoid 30 cm long and 3 cm diameter with 1,000 turns of wire carrying a current of 10A is-a)1.15 Jb)0.15 Jc)0.15 Jd)0.5 JCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The energy stored in the magnetic field of a solenoid 30 cm long and 3 cm diameter with 1,000 turns of wire carrying a current of 10A is-a)1.15 Jb)0.15 Jc)0.15 Jd)0.5 JCorrect answer is option 'C'. Can you explain this answer?.
Solutions for The energy stored in the magnetic field of a solenoid 30 cm long and 3 cm diameter with 1,000 turns of wire carrying a current of 10A is-a)1.15 Jb)0.15 Jc)0.15 Jd)0.5 JCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for SSC.
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