Question

A generator uses a coil that has 60 turns and a 0.50-T magnetic field. The frequency of this generator is 60.0 Hz, and its emf has an rms value of 120 V. Assuming that each turn of the coil is a square (an approximation), determine the length of the wire from which the coil is made.

Answer 1

120Vrms = 120*1.414 V peak = 169.7 Vpk

So the area of each coil is A = 169.7V / (2pi * 60Hz * 60turns * 0.5T) = 0.015 m^2

So each turn of the coil is sqrt(0.015) m square = 0.122m * 0.122m

So the total length of wire is 4*0.122m per turn * 60 turns = 29.4 m --> Answer

Answer 2

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

A generator uses a coil that has 100 turns and a .50 T magnetic field. The frequency of the generator is 60 Hz, and its emf has an rms value of 120 V. Assuming that each turn of the coil is a square (an approximation), determine the length of the wire from which the coil is made.

Answer

Peak emf ?_max = NBA?

Given,  ?_max = 120x?2 = 169.44 V

N = 100; B = 0.50 T; f = 60Hz;

? = 2?x60

Area of cross -section, A = ?_max /NB?

Sorry, El storm; will return later. Returning -too late;

A = a²; perimeter, p = 4a;

Length of the wire, L = N.p = N.4.a

                    A = ?_max /NB? =169.44/100x0.5x2?x60

                                             = 0.8988x10^-2 m²

The side of the square, a = ?A =?0.8988x10^-2 = 0.0948 m

Perimeter of the square, p = 4a = 0.3792 m

Length of the wire,

                      L = N.p = 100x 0.3792m = 37.92 m

Answer 3

emf = N*B*omega*A*sin(omega*t)
A = emf / [N*B*omega*sin(omega*t)]

emf = 120V * sqrt2 = 169.7V
omega = 2*pi*f = 377 rad/s

A = L^2 since it's square, however, the length of one turn is 4L.

A = 169.7 / [60*0.5*377*sin(377*t)]
= 169.7 / [60*0.5*377] max val of sin = 1
A = 0.1125m
L = sqrt A = (0.1125m)^1/2 = 0.1061m
Length of one turn = 4*0.1061m = 0.4243m

Length of 80 turns = 60*0.4243m = 25.45m