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Law Incorporation [45]

3 years ago

14

What is the magnetic force of a 30m long power line that carries a current of 50A. The magnetic field runs perpendicular to the

power line and has a strength of 30T.
A. 45000 T/N

B. 45 V

C. 45 N

D. 45000 N

1 answer:

Ne4ueva [31]3 years ago

8 0

Answer:

F=BILsin90 when perpendicular sin90 =1 30T x50x30 so you can get 45000N

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What is the magnetic flux linkage, in units of Weber, for a coil of 360 turns and cross sectional area of 0.133 m^2 when the mag

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Answer:

83.3 Wb

Explanation:

The magnetic flux linkage through the coil is given by:

$N\phi = BAN sin \theta$

where

B is the magnetic field strength

A is the cross sectional area

N is the number of turns in the coil

$\theta$ is the angle between the direction of the field and the normal to the coil

In this problem:

B = 1.74 T

A = 0.133 m^2

N = 360

$\theta=90^{\circ}$

Therefore, the magnetic flux linkage is

$N\phi = (1.74 T)(0.133 m^2)(360) sin 90^{\circ}=83.3 Wb$

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3 years ago

An unknown fluid has a specific gravity of 0.750. What is the volume of 22.5 kg of this fluid?

Andru [333]

<h2>Option C is the correct answer.</h2>

Explanation:

Specific gravity of fluid = 0.750

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Option C is the correct answer.

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