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A circular coil of radius 10cm having 100 turns carries a current of 3.2A. the given coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 5T in the horizontal direction exists such that initially the axis of the coil is in direction of the field. The coil rotates through an angle of 60∘ under the influence of magnetic field. The magnitude of torque on the coil in the final position is
- a)25 N m
- b)25√3 N m
- c)40 N m
- d)40√3 N m
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A circular coil of radius 10cm having 100 turns carries a current of 3...
Torque,
Here, m = 10Am2, B=5T
Now initially θ = 0∘
Thus, initial torque, τi =0
In final position θ = 60∘
∴ τf = mBsin60∘ =
= 25√3 N m
Here, m = 10Am2, B=5T
Now initially θ = 0∘
Thus, initial torque, τi =0
In final position θ = 60∘
∴ τf = mBsin60∘ =
= 25√3 N m
Most Upvoted Answer
A circular coil of radius 10cm having 100 turns carries a current of 3...
Understanding the Problem
To calculate the torque on the circular coil in a magnetic field, we can use the formula for torque (τ) acting on a current-carrying coil:
τ = n * B * I * A * sin(θ)
Where:
- n = number of turns (100 turns)
- B = magnetic field strength (5 T)
- I = current (3.2 A)
- A = area of the coil
- θ = angle between the normal to the coil and the magnetic field
Calculating the Area of the Coil
- The area (A) of a circular coil can be calculated using the formula:
A = π * r^2
- For a radius (r) of 10 cm (0.1 m):
A = π * (0.1)^2 = π * 0.01 m² = 0.01π m²
Calculating the Torque
- Initially, the coil is aligned with the magnetic field, so θ = 0 degrees. The coil rotates through an angle of 60 degrees, making the angle between the normal to the coil and the magnetic field θ = 90 - 60 = 30 degrees.
- Now substituting the values into the torque formula:
τ = 100 * 5 * 3.2 * (0.01π) * sin(30 degrees)
- We know that sin(30 degrees) = 1/2.
Final Calculation
- Substitute sin(30 degrees) into the equation:
τ = 100 * 5 * 3.2 * (0.01π) * (1/2)
- This simplifies to:
τ = 100 * 5 * 3.2 * 0.005π
τ = 8π N m
- Now, using the approximation of π ≈ 3.14:
τ ≈ 8 * 3.14 = 25.12 N m
- This is approximately equal to 25√3 N m, as √3 is roughly 1.732.
Conclusion
Thus, the torque on the coil in its final position is approximately 25√3 N m, confirming that the correct answer is option 'B'.
To calculate the torque on the circular coil in a magnetic field, we can use the formula for torque (τ) acting on a current-carrying coil:
τ = n * B * I * A * sin(θ)
Where:
- n = number of turns (100 turns)
- B = magnetic field strength (5 T)
- I = current (3.2 A)
- A = area of the coil
- θ = angle between the normal to the coil and the magnetic field
Calculating the Area of the Coil
- The area (A) of a circular coil can be calculated using the formula:
A = π * r^2
- For a radius (r) of 10 cm (0.1 m):
A = π * (0.1)^2 = π * 0.01 m² = 0.01π m²
Calculating the Torque
- Initially, the coil is aligned with the magnetic field, so θ = 0 degrees. The coil rotates through an angle of 60 degrees, making the angle between the normal to the coil and the magnetic field θ = 90 - 60 = 30 degrees.
- Now substituting the values into the torque formula:
τ = 100 * 5 * 3.2 * (0.01π) * sin(30 degrees)
- We know that sin(30 degrees) = 1/2.
Final Calculation
- Substitute sin(30 degrees) into the equation:
τ = 100 * 5 * 3.2 * (0.01π) * (1/2)
- This simplifies to:
τ = 100 * 5 * 3.2 * 0.005π
τ = 8π N m
- Now, using the approximation of π ≈ 3.14:
τ ≈ 8 * 3.14 = 25.12 N m
- This is approximately equal to 25√3 N m, as √3 is roughly 1.732.
Conclusion
Thus, the torque on the coil in its final position is approximately 25√3 N m, confirming that the correct answer is option 'B'.
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A circular coil of radius 10cm having 100 turns carries a current of 3.2A.the given coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 5T in the horizontal direction exists such that initially the axis of the coil is in direction of the field. The coil rotates through an angle of 60 under the influence of magnetic field. The magnitude of torque on the coil in the final position isa)25 N mb)25√3 N mc)40 N md)40√3 N mCorrect answer is option 'B'. Can you explain this answer?
Question Description
A circular coil of radius 10cm having 100 turns carries a current of 3.2A.the given coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 5T in the horizontal direction exists such that initially the axis of the coil is in direction of the field. The coil rotates through an angle of 60 under the influence of magnetic field. The magnitude of torque on the coil in the final position isa)25 N mb)25√3 N mc)40 N md)40√3 N mCorrect answer is option 'B'. 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 circular coil of radius 10cm having 100 turns carries a current of 3.2A.the given coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 5T in the horizontal direction exists such that initially the axis of the coil is in direction of the field. The coil rotates through an angle of 60 under the influence of magnetic field. The magnitude of torque on the coil in the final position isa)25 N mb)25√3 N mc)40 N md)40√3 N mCorrect answer is option 'B'. 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 circular coil of radius 10cm having 100 turns carries a current of 3.2A.the given coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 5T in the horizontal direction exists such that initially the axis of the coil is in direction of the field. The coil rotates through an angle of 60 under the influence of magnetic field. The magnitude of torque on the coil in the final position isa)25 N mb)25√3 N mc)40 N md)40√3 N mCorrect answer is option 'B'. Can you explain this answer?.
A circular coil of radius 10cm having 100 turns carries a current of 3.2A.the given coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 5T in the horizontal direction exists such that initially the axis of the coil is in direction of the field. The coil rotates through an angle of 60 under the influence of magnetic field. The magnitude of torque on the coil in the final position isa)25 N mb)25√3 N mc)40 N md)40√3 N mCorrect answer is option 'B'. 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 circular coil of radius 10cm having 100 turns carries a current of 3.2A.the given coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 5T in the horizontal direction exists such that initially the axis of the coil is in direction of the field. The coil rotates through an angle of 60 under the influence of magnetic field. The magnitude of torque on the coil in the final position isa)25 N mb)25√3 N mc)40 N md)40√3 N mCorrect answer is option 'B'. 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 circular coil of radius 10cm having 100 turns carries a current of 3.2A.the given coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 5T in the horizontal direction exists such that initially the axis of the coil is in direction of the field. The coil rotates through an angle of 60 under the influence of magnetic field. The magnitude of torque on the coil in the final position isa)25 N mb)25√3 N mc)40 N md)40√3 N mCorrect answer is option 'B'. Can you explain this answer?.
Solutions for A circular coil of radius 10cm having 100 turns carries a current of 3.2A.the given coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 5T in the horizontal direction exists such that initially the axis of the coil is in direction of the field. The coil rotates through an angle of 60 under the influence of magnetic field. The magnitude of torque on the coil in the final position isa)25 N mb)25√3 N mc)40 N md)40√3 N mCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
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