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A movable part of an electromagnet is displaced axially by 1.2 mm, as depicted in Fig. 2.22. Pole w idth is 5 mm, flux density is 1 Wb/m2 and gap length is 4 mm. If the gap length is kept constant, determine the force that acts in direction to bring them into axial alignment.?
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A movable part of an electromagnet is displaced axially by 1.2 mm, as ...
Calculation of Force to bring the movable part into axial alignment:
Given:
- Displacement of the movable part (Δx) = 1.2 mm
- Pole width (w) = 5 mm
- Flux density (B) = 1 Wb/m²
- Gap length (g) = 4 mm
Calculating the Area of the Pole Face:
The area of the pole face (A) can be calculated using the formula:
A = w * Δx
Substituting the given values, we get:
A = 5 mm * 1.2 mm
A = 6 mm²
Calculating the Force:
The force (F) acting to bring the movable part into axial alignment can be calculated using the formula:
F = B * A
Substituting the given values, we get:
F = 1 Wb/m² * 6 mm²
F = 6 N
Explanation:
When a current passes through an electromagnet, it creates a magnetic field. This magnetic field interacts with the magnetic field of the movable part, resulting in a force that tries to bring them into axial alignment.
In this scenario, the movable part is displaced axially by 1.2 mm. The pole face, which is the part of the electromagnet that interacts with the movable part, has a width of 5 mm. By calculating the area of the pole face, we determine the area over which the magnetic field interacts with the movable part.
The force acting on the movable part can be calculated by multiplying the flux density (1 Wb/m²) with the area of the pole face (6 mm²). This gives us a force of 6 N.
Therefore, the force acting in the direction to bring the movable part into axial alignment is 6 N.
Given:
- Displacement of the movable part (Δx) = 1.2 mm
- Pole width (w) = 5 mm
- Flux density (B) = 1 Wb/m²
- Gap length (g) = 4 mm
Calculating the Area of the Pole Face:
The area of the pole face (A) can be calculated using the formula:
A = w * Δx
Substituting the given values, we get:
A = 5 mm * 1.2 mm
A = 6 mm²
Calculating the Force:
The force (F) acting to bring the movable part into axial alignment can be calculated using the formula:
F = B * A
Substituting the given values, we get:
F = 1 Wb/m² * 6 mm²
F = 6 N
Explanation:
When a current passes through an electromagnet, it creates a magnetic field. This magnetic field interacts with the magnetic field of the movable part, resulting in a force that tries to bring them into axial alignment.
In this scenario, the movable part is displaced axially by 1.2 mm. The pole face, which is the part of the electromagnet that interacts with the movable part, has a width of 5 mm. By calculating the area of the pole face, we determine the area over which the magnetic field interacts with the movable part.
The force acting on the movable part can be calculated by multiplying the flux density (1 Wb/m²) with the area of the pole face (6 mm²). This gives us a force of 6 N.
Therefore, the force acting in the direction to bring the movable part into axial alignment is 6 N.
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A movable part of an electromagnet is displaced axially by 1.2 mm, as depicted in Fig. 2.22. Pole w idth is 5 mm, flux density is 1 Wb/m2 and gap length is 4 mm. If the gap length is kept constant, determine the force that acts in direction to bring them into axial alignment.?
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A movable part of an electromagnet is displaced axially by 1.2 mm, as depicted in Fig. 2.22. Pole w idth is 5 mm, flux density is 1 Wb/m2 and gap length is 4 mm. If the gap length is kept constant, determine the force that acts in direction to bring them into axial alignment.? 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 movable part of an electromagnet is displaced axially by 1.2 mm, as depicted in Fig. 2.22. Pole w idth is 5 mm, flux density is 1 Wb/m2 and gap length is 4 mm. If the gap length is kept constant, determine the force that acts in direction to bring them into axial alignment.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A movable part of an electromagnet is displaced axially by 1.2 mm, as depicted in Fig. 2.22. Pole w idth is 5 mm, flux density is 1 Wb/m2 and gap length is 4 mm. If the gap length is kept constant, determine the force that acts in direction to bring them into axial alignment.?.
A movable part of an electromagnet is displaced axially by 1.2 mm, as depicted in Fig. 2.22. Pole w idth is 5 mm, flux density is 1 Wb/m2 and gap length is 4 mm. If the gap length is kept constant, determine the force that acts in direction to bring them into axial alignment.? 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 movable part of an electromagnet is displaced axially by 1.2 mm, as depicted in Fig. 2.22. Pole w idth is 5 mm, flux density is 1 Wb/m2 and gap length is 4 mm. If the gap length is kept constant, determine the force that acts in direction to bring them into axial alignment.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A movable part of an electromagnet is displaced axially by 1.2 mm, as depicted in Fig. 2.22. Pole w idth is 5 mm, flux density is 1 Wb/m2 and gap length is 4 mm. If the gap length is kept constant, determine the force that acts in direction to bring them into axial alignment.?.
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