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A solenoid of 20 cm long and 1 cm diameter has a uniform winding of 1000 turns. If the solenoid is placed in a uniform field of 2 Wb/m2 flux density and a current of 10 amps, is passed through the solenoid winding, then the maximum torque on the solenoid will be
- a)10N-m
- b)20N-m
- c)30N-m
- d)60N-m
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A solenoid of 20 cm long and 1 cm diameter has a uniform winding of 10...
Given, l = 20 cm = 0.2 m,
r - 0.5 cm = radius = 0.005 m
N = 1000,
B = 2 Wb/m2,
I = 10 A
The torque acting on the solenoid is
T = NIAB sinθ
T will be maximum when
θ = 90°, i.e. sin90° = 1
∴ Tmax = 1000 x 10 (0.005 x 0.2) x 2
= 20 N-m
r - 0.5 cm = radius = 0.005 m
N = 1000,
B = 2 Wb/m2,
I = 10 A
The torque acting on the solenoid is
T = NIAB sinθ
T will be maximum when
θ = 90°, i.e. sin90° = 1
∴ Tmax = 1000 x 10 (0.005 x 0.2) x 2
= 20 N-m
Most Upvoted Answer
A solenoid of 20 cm long and 1 cm diameter has a uniform winding of 10...
To find the maximum torque on the solenoid, we can use the equation:
τ = N * B * A * I
Where:
τ is the torque
N is the number of turns
B is the magnetic flux density
A is the cross-sectional area of the solenoid
I is the current passing through the solenoid winding
Let's calculate each parameter step by step:
1. Cross-sectional Area (A):
The diameter of the solenoid is given as 1 cm, which means the radius is 0.5 cm or 0.005 m.
The cross-sectional area of a solenoid is given by the formula: A = π * r^2
Substituting the values, we get:
A = π * (0.005)^2 ≈ 0.00007854 m^2
2. Magnetic Flux Density (B):
The solenoid is placed in a uniform field with a flux density of 2 Wb/m^2.
Therefore, B = 2 Wb/m^2
3. Number of Turns (N):
The solenoid has 1000 turns.
Therefore, N = 1000
4. Current (I):
The current passing through the solenoid winding is given as 10 Amps.
Therefore, I = 10 Amps
Now, we can substitute the values into the torque equation:
τ = N * B * A * I
= 1000 * 2 Wb/m^2 * 0.00007854 m^2 * 10 Amps
≈ 0.157 N-m
Therefore, the maximum torque on the solenoid is approximately 0.157 N-m.
Since none of the given options match this value, it seems that there might be an error in the question or the options provided. Without additional information or clarification, it is not possible to determine the correct answer.
τ = N * B * A * I
Where:
τ is the torque
N is the number of turns
B is the magnetic flux density
A is the cross-sectional area of the solenoid
I is the current passing through the solenoid winding
Let's calculate each parameter step by step:
1. Cross-sectional Area (A):
The diameter of the solenoid is given as 1 cm, which means the radius is 0.5 cm or 0.005 m.
The cross-sectional area of a solenoid is given by the formula: A = π * r^2
Substituting the values, we get:
A = π * (0.005)^2 ≈ 0.00007854 m^2
2. Magnetic Flux Density (B):
The solenoid is placed in a uniform field with a flux density of 2 Wb/m^2.
Therefore, B = 2 Wb/m^2
3. Number of Turns (N):
The solenoid has 1000 turns.
Therefore, N = 1000
4. Current (I):
The current passing through the solenoid winding is given as 10 Amps.
Therefore, I = 10 Amps
Now, we can substitute the values into the torque equation:
τ = N * B * A * I
= 1000 * 2 Wb/m^2 * 0.00007854 m^2 * 10 Amps
≈ 0.157 N-m
Therefore, the maximum torque on the solenoid is approximately 0.157 N-m.
Since none of the given options match this value, it seems that there might be an error in the question or the options provided. Without additional information or clarification, it is not possible to determine the correct answer.
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A solenoid of 20 cm long and 1 cm diameter has a uniform winding of 1000 turns. If the solenoid is placed in a uniform field of 2 Wb/m2 flux density and a current of 10 amps, is passed through the solenoid winding, then the maximum torque on the solenoid will bea)10N-mb)20N-mc)30N-md)60N-mCorrect answer is option 'B'. Can you explain this answer?
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A solenoid of 20 cm long and 1 cm diameter has a uniform winding of 1000 turns. If the solenoid is placed in a uniform field of 2 Wb/m2 flux density and a current of 10 amps, is passed through the solenoid winding, then the maximum torque on the solenoid will bea)10N-mb)20N-mc)30N-md)60N-mCorrect answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A solenoid of 20 cm long and 1 cm diameter has a uniform winding of 1000 turns. If the solenoid is placed in a uniform field of 2 Wb/m2 flux density and a current of 10 amps, is passed through the solenoid winding, then the maximum torque on the solenoid will bea)10N-mb)20N-mc)30N-md)60N-mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solenoid of 20 cm long and 1 cm diameter has a uniform winding of 1000 turns. If the solenoid is placed in a uniform field of 2 Wb/m2 flux density and a current of 10 amps, is passed through the solenoid winding, then the maximum torque on the solenoid will bea)10N-mb)20N-mc)30N-md)60N-mCorrect answer is option 'B'. Can you explain this answer?.
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