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The magnitude of torque experienced by a square coil of side 12cm which consists of 25 turns and carries a current 10A suspended vertically and the normal to the plane of coil makes an angle of 30 with the direction of a uniform horizontal magnetic field of magnitude 0.9T is
  • a)
    1.6 N m
  • b)
    1.2 N m
  • c)
    1.4 N m
  • d)
    1.8 N m
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The magnitude of torque experienced by a square coil of side 12cm whic...
τ = NIABsinθ
Here, N = 25,
I = 10A,
B = 0.9T,
θ = 30
A = a2 = 12 × 10−2 × 12 × 10−2 
144 × 10−4m2
∴ τ = 25 × 10 × 144 × 10−4 × 0.9 × sin30 = 1.6 Nm
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Most Upvoted Answer
The magnitude of torque experienced by a square coil of side 12cm whic...
The torque experienced by a current-carrying coil in a magnetic field can be calculated using the formula:

Torque = N * I * A * B * sinθ

Where:
- N is the number of turns in the coil
- I is the current flowing through the coil
- A is the area of the coil
- B is the magnetic field strength
- θ is the angle between the normal to the coil and the direction of the magnetic field

Given:
- Side of the square coil = 12 cm
- Number of turns = 25
- Current = 10 A
- Angle θ = 30°
- Magnetic field strength B = 0.9 T

First, we need to calculate the area of the coil:
Area of the square coil = side^2
Area = (0.12 m)^2 = 0.0144 m^2

Now, we can substitute the given values into the torque formula:
Torque = 25 * 10 * 0.0144 * 0.9 * sin(30°)
Torque = 3.6 * 0.0144 * 0.9 * 0.5
Torque = 0.018 * 0.9 * 0.5
Torque = 0.0081 Nm

Therefore, the magnitude of the torque experienced by the coil is 0.0081 Nm, which is equivalent to 1.6 Nm (option A).
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The magnitude of torque experienced by a square coil of side 12cm which consists of 25 turns and carries a current 10A suspended vertically and the normal to the plane of coil makes an angle of 30 with the direction of a uniform horizontal magnetic field of magnitude 0.9T isa)1.6 N mb)1.2 N mc)1.4 N md)1.8 N mCorrect answer is option 'A'. Can you explain this answer?
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The magnitude of torque experienced by a square coil of side 12cm which consists of 25 turns and carries a current 10A suspended vertically and the normal to the plane of coil makes an angle of 30 with the direction of a uniform horizontal magnetic field of magnitude 0.9T isa)1.6 N mb)1.2 N mc)1.4 N md)1.8 N mCorrect answer is option 'A'. 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 The magnitude of torque experienced by a square coil of side 12cm which consists of 25 turns and carries a current 10A suspended vertically and the normal to the plane of coil makes an angle of 30 with the direction of a uniform horizontal magnetic field of magnitude 0.9T isa)1.6 N mb)1.2 N mc)1.4 N md)1.8 N mCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The magnitude of torque experienced by a square coil of side 12cm which consists of 25 turns and carries a current 10A suspended vertically and the normal to the plane of coil makes an angle of 30 with the direction of a uniform horizontal magnetic field of magnitude 0.9T isa)1.6 N mb)1.2 N mc)1.4 N md)1.8 N mCorrect answer is option 'A'. Can you explain this answer?.
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