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soldier1979 [14.2K]
3 years ago
9

A circular loop of radius 0.10 m is rotating in a uniform external magnetic field of 0.20 T. Find the magnetic flux through the

loop due to the external field when the plane of the loop and the magnetic field vector are parellel.
Physics
1 answer:
laiz [17]3 years ago
4 0

Answer:

The magnetic flux through the loop is 0.006284 m².T

Explanation:

Given;

radius of circular loop, r = 0.1 m

magnetic field strength, B = 0.2 T

Magnetic flux through the loop due to the external field is given as;

Φ = BAcosθ

where;

Φ is magnetic flux

B is the magnetic field strength

A is the area of the loop

θ is the angle of inclination of between the plane of the loop and external magnetic field

A = πr² = π(0.1)² = 0.03142 m²

θ = 0, since they are parallel

Φ = BAcosθ

Φ = 0.2 x 0.03142 x cos0

Φ = 0.006284 m².T

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5 0
4 years ago
The water passing over victoria falls, located along the zambezi river on the border of zimbabwe and zabia drops about 105 m. Ho
7nadin3 [17]

Answer:

1029 Jkg⁻¹

Explanation:

h = height dropped by the water falling = 105 m

g = acceleration due to gravity = 9.8 ms⁻²

m = mass of water

Using conservation of energy, the internal energy produced is same as the gravitational potential energy lost, hence

Internal energy = Gravitational potential energy

U = mgh \\\frac{U}{m} = gh\\\frac{U}{m} = (9.8) (105)\\\frac{U}{m} = 1029 Jkg^{-1}

7 0
4 years ago
Is there water on Uranus? And what temperature is it
stiks02 [169]
It atmosphere is compsosed promarly of hyrogen and helium

5 0
3 years ago
What is the magnitude of the resultant vector? Round
Naddika [18.5K]

Answer: 13.9 m

Explanation:

4 0
3 years ago
The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 c
Reika [66]

Answer:

the knee extensors must exert 15.87 N

Explanation:

Given the data in the question;

mass m = 4.5 kg

radius of gyration k = 23 cm = 0.23 m

angle ∅ = 30°

∝ = 1 rad/s²

distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m

using the expression;

ζ = I∝

ζ = mk²∝

we substitute

ζ = 4.5 × (0.23)² × 1

ζ  = 0.23805 N-m

so

from; ζ = rFsin∅

F = ζ / rsin∅

we substitute

F = 0.23805 / (0.03 × sin( 30 ° )

F = 0.23805 / (0.03 × 0.5)

F F = 0.23805 / 0.015

F = 15.87 N

Therefore, the knee extensors must exert 15.87 N

7 0
3 years ago
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