Time-Varying Magnetic Fields Q#2

[BroIIy]

Hello, i just got done ranting on Q#1 (its in another post, and I am still in a ranting mode) and was about to go to bed for the day, when i figured "well heck i have been working on these 2 questions for 9 hours and have made almost no progress, asking the people in my class for help is like throwing a brick into the ocean and waiting for it to float, and my 1st Question (that i posted here) already had someone actually try and help me (which is a lot more help than what i have received through the past 30 chapters so far...) " so... i figure before i pass out for the day i will post this 2nd question, for 2 reasons... 1) there is a very high chance that by myself i will miss this question, even though each chapter has several questions and point wise it won't really hurt which leds to reason #2) after i do miss it, i will already be on the next deadline to complete the next chapter (we go through a chapter a day) and i will more than likely never figure out WHY i missed it (again referring to the brick story) i guess i can add a 3rd reason being that, well heck you guys are actually very helpfull

Homework Statement

A long, straight wire carries a current I = I0 cos(220πt), where t is time in seconds. Two sides of a fixed rectangular loop are 8.5 cm long and are parallel to the wire; the other sides are 0.80 cm long. The nearest long side is 2.0 cm from the wire. What is I0 if the maximum emf induced in the loop is 1.3 µV? (Ignore the small variation of the magnetic field across the loop and calculate any B values at the location where the loop is nearest the wire.)

Homework Equations

sry...

The Attempt at a Solution

there is a similar question in one of my books but it is slightly different. in which case i attempted to use it to obtain the answer but it didnt work out (here is what i tried)
R = .085
L = .008
x = .02
E = 1.3E-6
w = 220*pi
E / (( u_o * R ) / (2*pi) * ( w * ln( (x+L) / x ) * cos(w) ) = I

other than that, i am at a loss of thought

 

[Doc Al]

Homework Statement

A long, straight wire carries a current I = I0 cos(220πt), where t is time in seconds. Two sides of a fixed rectangular loop are 8.5 cm long and are parallel to the wire; the other sides are 0.80 cm long. The nearest long side is 2.0 cm from the wire. What is I0 if the maximum emf induced in the loop is 1.3 µV? (Ignore the small variation of the magnetic field across the loop and calculate any B values at the location where the loop is nearest the wire.)

I will assume that the farthest long side is 2.8 cm from the wire. (The magnetic field is perpendicular to the loop.)

First things first: What's the value of B within the loop? What's the formula for the field from a long straight wire carrying some current?

Then apply Faraday's law. What's the flux through that loop? What's the rate of change of that flux? (Take the derivative.) Set the max value of that to the given value for max EMF.

 

[BroIIy]

here is the work

N = 1
Length = .085
Width = .008
Distance = .02
Emax = 1.3E-6
W = 220*pi (this is my symbol for Omega)
A = Length * Width
u_o = 4 * pi * 10^-7

Emax = N * A * B * W
find 'B'
B = Emax / ( N * A * W )
solve for 'B' and plug it into...
B for a straight wire ==
B = u_o * I / (2 * pi * Distance )
find 'I'
I = B * (2 * pi * Distance) / u_o
solve for 'I'

thank you again Doc Al, now i shall go back to that other question and try to solve it.

 

[Doc Al]

Emax = N * A * B * W

I'm curious how you arrived at this.

Here's how I would do this one.
First find B:

$$B = \frac{\mu_0 I}{2 \pi d} = \frac{\mu_0 I_0 \cos\omega t}{2 \pi d}$$

The flux is just B*A, and the EMF is the rate of change of the flux (ignoring signs):

$$E = d\phi/dt = \frac{\mu_0 I_0 A \omega \sin\omega t}{2 \pi d}$$

The max value is when sin = 1, thus:

$$E_{max} = \frac{\mu_0 I_0 A \omega}{2 \pi d}$$

Just solve for $$I_0$$.

 

[BroIIy]

i have several notebooks and as we run through the chapters i just write down all notes i can. in the one notebook for chapter 30 (this chapter) i have for 'E'
E = N * A * B * W * sin(Wt)
Emax = N * A * B * W
E = -dIB / dt
E = -B*L*Velocity
E = I * Resistance

now like i said before I am not too good with the concepts, and i haven't had any math classes yet (which hurts me on intergrals and derivitives and the like...especially since I am in a calc based class lol ) but i am good at puzzles. i solve the problems by identifing the "pieces" that i am given, and what other pieces i can obtain through them. then relating them all together and solving for the missing parts. which is prolly why everyone in the class takes ~4 hrs a chapter to answer all 40+ questions and i end up taking 15+ hrs...however... i get more of them correct =)

i don't know (in your equation) what the 'd' represents, but i pluged in
Emax = 1.36E-6
d = .02 ??
A = Area of the square Length * width
W = 220*pi
and the answer came out to be
.2893726238
the correct answer (from the key) is
0.277
and using my method i received
.2766061844
(we are allowed a 1% error ratio)
your answer prolly is correct but i think i screwed up the calculations somewhere sorry

 

[Doc Al]

i don't know (in your equation) what the 'd' represents, but i pluged in
Emax = 1.36E-6
d = .02 ??

The "d" stands for distance from the wire. (See where it appears in the formula for B.) So d = .02.

A = Area of the square Length * width
W = 220*pi
and the answer came out to be
.2893726238

Not sure how you got that.

the correct answer (from the key) is
0.277
and using my method i received
.2766061844

That's the same answer I get using my equation. My point was not that your answer was wrong, but whether you understood where the answer comes from.

(we are allowed a 1% error ratio)
your answer prolly is correct but i think i screwed up the calculations somewhere sorry

Yep, you made an error somewhere.