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A copper wire of length 10 m and radius (10–2 /π)m has electrical resistance of 10 Ω. The current density in the wire for an electric field strength of 10 (V/m) is:
- a)106 A/m2
- b)10–5 A/m2
- c)105 A/m2
- d)104 A/m2
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A copper wire of length 10 m and radius (10–2 /π)m has electr...
M) is used to make a coil. The coil is wound uniformly and tightly to form a solenoid with 1000 turns. Find the magnitude of the magnetic field inside the solenoid when a current of 5 A passes through it.
The magnetic field inside a solenoid can be calculated using the formula:
B = μ₀ * n * I
Where B is the magnetic field, μ₀ is the permeability of free space (4π*10^-7 T*m/A), n is the number of turns per unit length, and I is the current.
First, we need to find the number of turns per unit length (n) of the solenoid. This can be calculated using the formula:
n = N / L
Where N is the total number of turns and L is the length of the solenoid.
Given that the solenoid has 1000 turns and a length of 10 m:
n = 1000 / 10 = 100 turns/m
Now we can calculate the magnetic field inside the solenoid:
B = μ₀ * n * I
B = (4π*10^-7 T*m/A) * (100 turns/m) * (5 A)
B = 2π*10^-5 T
Therefore, the magnitude of the magnetic field inside the solenoid when a current of 5 A passes through it is 2π*10^-5 T.
The magnetic field inside a solenoid can be calculated using the formula:
B = μ₀ * n * I
Where B is the magnetic field, μ₀ is the permeability of free space (4π*10^-7 T*m/A), n is the number of turns per unit length, and I is the current.
First, we need to find the number of turns per unit length (n) of the solenoid. This can be calculated using the formula:
n = N / L
Where N is the total number of turns and L is the length of the solenoid.
Given that the solenoid has 1000 turns and a length of 10 m:
n = 1000 / 10 = 100 turns/m
Now we can calculate the magnetic field inside the solenoid:
B = μ₀ * n * I
B = (4π*10^-7 T*m/A) * (100 turns/m) * (5 A)
B = 2π*10^-5 T
Therefore, the magnitude of the magnetic field inside the solenoid when a current of 5 A passes through it is 2π*10^-5 T.
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Community Answer
A copper wire of length 10 m and radius (10–2 /π)m has electr...
Radius of wire = 10–2 /√π
Cross sectional area A = πr2 = 10–4 m2
Cross sectional area A = πr2 = 10–4 m2
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A copper wire of length 10 m and radius (10–2 /π)m has electrical resistance of 10 Ω. The current density in the wire for an electric field strength of 10 (V/m) is:a)106 A/m2b)10–5 A/m2c)105 A/m2d)104 A/m2Correct answer is option 'C'. Can you explain this answer?
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A copper wire of length 10 m and radius (10–2 /π)m has electrical resistance of 10 Ω. The current density in the wire for an electric field strength of 10 (V/m) is:a)106 A/m2b)10–5 A/m2c)105 A/m2d)104 A/m2Correct answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A copper wire of length 10 m and radius (10–2 /π)m has electrical resistance of 10 Ω. The current density in the wire for an electric field strength of 10 (V/m) is:a)106 A/m2b)10–5 A/m2c)105 A/m2d)104 A/m2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A copper wire of length 10 m and radius (10–2 /π)m has electrical resistance of 10 Ω. The current density in the wire for an electric field strength of 10 (V/m) is:a)106 A/m2b)10–5 A/m2c)105 A/m2d)104 A/m2Correct answer is option 'C'. Can you explain this answer?.
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