Answer:
Current, I = 1000 A
Explanation:
It is given that,
Length of the copper wire, l = 7300 m
Resistance of copper line, R = 10 ohms
Magnetic field, B = 0.1 T

Resistivity, 
We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :




r = 0.00199 m
or

The magnetic field on a current carrying wire is given by :



I = 1000 A
So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.
Answer:
we have to find out the critical resolved shear stress. As it it given in the question
Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.
a) Slip will occur in the direction where cosФ cosλ are maximum. Cosine for all possible λ values are given as follows.
cos(62.4°) = 0.46
cos(72.0°) = 0.31
cos(81.1°) = 0.15
Thus, the slip direction is at the angle of 62.4° along the tensile axis.
b) now the critical resolved shear stress can be find out by the following equation.
τ
= σ
( cosФ cosλ)
now by putting values,
= (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23
The force due to gravity is equal to the product of the mass and acceleration due to gravity. Since the acceleration due to gravity is constant, we just have to rely on the comparison of their masses to determine the mass effect. Thus, the answer to this item is Wrestler C weighing 171 pounds.
b) between poles M1 and M2
Explanation:
From the expression, we can deduce that r is the distance between two magnetic poles M1 and M2.
The law of attraction between two magnetic poles states that:
<em> the force of attraction or repulsion between two magnetic poles is a function of the product of the strength of the magnetic poles and the square of the distance between the pole</em>s
Mathematically:
FM = K 
here r is the distance between the poles
FM is the magnetic force between the poles
M1 is the strength of the first magnetic pole
M2 is the strength of the second pole
K is the magnetic field constant
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magnetic pole brainly.com/question/2191993
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magnetic field due to a finite straight conductor is given by

here since it forms an equilateral triangle so we will have

also the perpendicular distance of the point from the wire is

now from the above equation magnetic field due to one wire is given by



now since in equilateral triangle there are three such wires so net magnetic field will be
