A 0.130 m radius, 485-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a
2 answers:
Answer:
The magnetic field strength needed is 1.619 T
Explanation:
Given;
Number of turns, N = 485-turn
Radius of coil, r = 0.130 m
time of revolution, t = 4.17 ms = 0.00417 s
average induced emf, V = 10,000 V.
Average induced emf is given as;
V = -ΔФ/Δt
where;
ΔФ is change in flux
Δt is change in time
ΔФ
where;
N is the number of turns
B is the magnetic field strength
A is the area of the coil = πr²
θ is the angle of inclination of the coil and the magnetic field,
V = NBACos0/t
V = NBA/t
B = (Vt)/NA
B = (10,000 x 0.00417) / (485 x π x 0.13²)
B =1.619 T
Thus, the magnetic field strength needed is 1.619 T
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Answer:
1.117935:1
Explanation:
Since the wires are of the same material, they will have the same resistivity .
The cross-sectional area of the of a wire is given by;
where d is the diameter of the wire.
Also, the relationship between resistance R, cross-sectional area A and length l of a wire is given as follows;
Since the resistivity same for both wires, say wire 1 and wire 2, we can wreite the following;
Hence from eqaution (3), the ration of wire 1 to 2 is expressed as;
Given;
We then use equation (1) to fine the ratio of the area to
bearing in mind that
This ratio gives 0.8281. Substituting this into equation (5), we get the following;