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aev [14]
3 years ago
5

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.

Physics
2 answers:
il63 [147K]3 years ago
3 0

Answer:

4.399 Nm

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

igomit [66]3 years ago
3 0

Explanation:

Below is an attachment containing the solution.

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Explanation:

Given,

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The gravitational force between the two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between them. It is given by the formula

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Hence, the force between the two bodies is, F = 2.6692 X 10⁻⁹ N

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Learn more about Newton laws of motion:

brainly.com/question/3820012

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Answer:45 m/s north

Explanation:

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