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A moving coil of a meter has 250 turns and a length and depth of 40 mm and 30 mm respectively. It is positioned in a uniform radial flux density of 450 mT. The coil carries a current of 160 mA. The torque on the coil is?
  • a)
    0.0216 N-m
  • b)
    0.0456 N-m
  • c)
    0.1448 N-m
  • d)
    1 N-m
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A moving coil of a meter has 250 turns and a length and depth of 40 mm...
Given, N = 250, L = 40 x 10-3, d = 30 x 10-3m, I = 160 x 10-3A, B = 450 x 10-3 T
Torque = 250 x 450 x 10-3 x 40 x 10-3 × 30 x 10-3 x 160 x 10-3
= 200 x 10-6 N-m = 0.0216 N-m.
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Community Answer
A moving coil of a meter has 250 turns and a length and depth of 40 mm...
To calculate the torque on the coil, we can use the formula:

Torque = NIABsinθ

where N is the number of turns in the coil, I is the current passing through the coil, A is the area of the coil, B is the magnetic field strength, and θ is the angle between the normal to the coil and the magnetic field lines.

Let's calculate each parameter step by step:

1. Number of turns (N): Given as 250.

2. Current (I): Given as 160 mA. Convert it to Amperes by dividing by 1000:
I = 160 mA / 1000 = 0.16 A.

3. Area of the coil (A): The coil has a rectangular shape with length and depth given as 40 mm and 30 mm respectively. Convert the dimensions to meters:
Length = 40 mm = 40/1000 = 0.04 m
Depth = 30 mm = 30/1000 = 0.03 m
The area can be calculated as the product of length and depth:
A = 0.04 m * 0.03 m = 0.0012 m^2.

4. Magnetic field strength (B): Given as 450 mT. Convert it to Tesla by dividing by 1000:
B = 450 mT / 1000 = 0.45 T.

5. Angle (θ): The coil is positioned in a uniform radial flux density, which means the angle between the normal to the coil and the magnetic field lines is 90 degrees. Since sin(90) = 1, we can ignore this term in the calculation.

Now, we can substitute the values into the torque formula:

Torque = (250 turns) * (0.16 A) * (0.0012 m^2) * (0.45 T) * 1
= 0.0216 N-m

Therefore, the torque on the coil is 0.0216 N-m, which corresponds to option A.
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A moving coil of a meter has 250 turns and a length and depth of 40 mm and 30 mm respectively. It is positioned in a uniform radial flux density of 450 mT. The coil carries a current of 160 mA. The torque on the coil is?a)0.0216 N-mb)0.0456 N-mc)0.1448 N-md)1 N-mCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2025 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 moving coil of a meter has 250 turns and a length and depth of 40 mm and 30 mm respectively. It is positioned in a uniform radial flux density of 450 mT. The coil carries a current of 160 mA. The torque on the coil is?a)0.0216 N-mb)0.0456 N-mc)0.1448 N-md)1 N-mCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A moving coil of a meter has 250 turns and a length and depth of 40 mm and 30 mm respectively. It is positioned in a uniform radial flux density of 450 mT. The coil carries a current of 160 mA. The torque on the coil is?a)0.0216 N-mb)0.0456 N-mc)0.1448 N-md)1 N-mCorrect answer is option 'A'. Can you explain this answer?.
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