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A 45.5-turn circular coil of radius 4.85 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.490 T. If the coil carries a current of 26.7 mA, find the magnitude of the maximum possible torque exerted on the coil.

User Jaegow
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2 Answers

6 votes

Step-by-step explanation:

Below is an attachment containing the solution.

A 45.5-turn circular coil of radius 4.85 cm can be oriented in any direction in a-example-1
User Simon David
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5 votes

Answer:

4.399 Nm

Step-by-step explanation:

The maximum Torque on a coil is given as,

τ = BNIA...................... Equation 1

Where τ = Maximum torque exerted on the coil, B = Magnetic Field, N = Number of turns, I = Current, A = Area.

Given: N = 45.5 Turns, B = 0.49 T, I = 26.7 mA = 0.0267 A,

A = πr², Where r = radius of the coil, r= 4.85 cm = 0.0485 m

A = 3.14(0.0485)²

A = 7.39×10⁻³ m².

Substitute into equation 1

τ = 45.5×0.49×26.7×7.39×10⁻³

τ = 4.399 Nm

User Erv Walter
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