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Veronika [31]
4 years ago
10

A flat circular loop of wire of radius 0.50 m that is carrying a 2.0-A current is in a uniform magnetic field of 0.30 T. What is

the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field
Physics
1 answer:
Luba_88 [7]4 years ago
4 0

Answer:

The magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J

Explanation:

Given;

radius of the circular loop of wire = 0.5 m

current in circular loop of wire = 2 A

strength of magnetic field in the wire = 0.3 T

τ = μ x Bsinθ

where;

τ is the magnitude of the magnetic torque

μ is the dipole moment of the magnetic field

θ is the inclination angle, for a plane area perpendicular to the magnetic field, θ = 90

μ = IA

where;

I is current in circular loop of wire

A is area of the circular loop = πr² = π(0.5)² = 0.7855 m²

μ = 2 x 0.7885 = 1.571 A.m²

τ = μ x Bsinθ =  1.571 x 0.3 sin(90)

τ = 0.4713 J

Therefore, the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J

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Answer:

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Explanation:

To solve these types of problems we must remember that the average of the speeds is determined by dividing the distance over time.

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Now with the second velocity and  6 [s], we can calculate the second distance.

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5 0
3 years ago
Express 79 m in units of (a) centimeters
dem82 [27]

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Explanation:

(a) We know that 1 meter = 100 centimeters

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(b)Since 1 meter = 3.28084 feet

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79\ m=259.19\text{ feet}=259.19\times12=3110.28\text{ inches}

(d) \text{Since 1 feet= }\dfrac{1}{5280}\text{ mile}

79\ m=259.19\text{ feet}=\dfrac{259.19}{5280}= 0.0490890151515\approx0.049\text{ miles}

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