Asked by Annoyingmous

An infinite straight wire carrying a current I=5 A flowing to the right is placed above a rectangular loop of wire with width w=11 cm and length L=27 cm, as shown in the figure below. The distance from the infinite wire to the closest side of the rectangle is h=1.7 cm. The loop of wire has resistance R=0.31 Ohm.


(a) What is the magnitude (in Tesla) of the magnetic field due to the infinite wire at the point P in the rectangular loop, a distance r=3.2 cm from the wire (see figure).


In which direction does it point?


(b) Calculate the magnitude of the magnetic flux (in Tesla  m2) through the rectangular loop due to the magnetic field created by the infinite wire.

(c) Suppose the current in the infinite wire starts increasing in time according to I=bt, with b=50 Amps/sec. What is the magnitude (in Amps) of the induced current in the loop? Neglect any contribution to the magnetic flux through the loop due to the magnetic field created by the induced current.


(d) What is the direction of this current flow?
Answered by ^_ ^
a) B = mu_0*I/(2*Pi*r)

b) phi = (mu_0*I*L/(2*Pi))*ln(1+w/h)

c) i = (-1/R)*(mu_0*L)/(2*Pi)*ln(1+w/h)*(di/dt)
Answered by Phy
What is di/dt ??
Answered by Anonymous
Phy, it's the b in the variation I=bt. Just use the b value in the problem for (di/dt)
Answered by Phy
Thanks, can u tell me the solutions of 1,4,6,9,10,11,12?
Answered by dd
the C is wrong for me using the b in di/dt
Answered by robbo
Have you guys gt 3,5,7,8,9,10,11,12...????? Let's help each other and get this done guys!!
Answered by Sat
C is wrong ...
Answered by boss
Some people haven't done even one question.
Answered by boss
I wonder if some people understand the subject at all. It is an insult to the lecturer.
Answered by Sat
Hey boss, how did u come across this topic if you have not googled the question? It's obvious that you cheat also!

BTW - for C just use the value for b (the one in the question) b=(di/dt)
Answered by Anonymous
Hey guys, what abput prob. 10?
Answered by BG
10, b) how to solve
Answered by Jack
Please, problem 8!!!
Answered by robbo
Problems 8-12 guys, please???/
Answered by P
Please tell ans of any question please
Q3 a=0 last part =0
Answered by BG
Anyone, can help with questions:

11 c)
10 b)
8 d,e,f)
5 b,d,e)


Answered by BG
11 a)
q/(4*pi*E_0*(r^2)*k)

11 b)
q*k/(r^2)

11 d) 0
Answered by me
anyone for question 7 please
Answered by Jack
Hi BG, can you put the results for 8 a, b,c?
Answered by AAA
Please, Q10?
Answered by AAA
anyone for question 3 please??????
Answered by Jack
5 e)
v= m*g*R/(B^2*W^2)
Answered by BG
8 a)
E = (st)/(S*E_0)
S = Area

8 b) E*d

8 c)
C = E*E_0*S/d
Answered by BG
S = pi(b^2 - a^2)
Answered by BG
Jack, did you solve anything else from this list:
11 c)
10 b)
8 d,e,f)
5 b,d)
Answered by BG
me, I will exchange solution 8 on 7)
Answered by BG
me, a = lambda/0.5
d = lambda/0.125
Answered by BG
denaminator may be different, it depends on drawing
Answered by AAA
How I can know my denominator watching my grahp BG?
Answered by BG
prob 8
Answered by BG
I think drawings at all identical so don't worry
Answered by me
BG sorry don't have 8 yet, i'll post it as soon as i solve it and thanks for 7
Answered by me
i have 5 do you still need 5b and 5d
Answered by Jack
BG,

8 d)

B= mu_0*(r^2-a^2)*s/(2*r*pi*(b^2-a^2))
Answered by Jack
BG, do you have 10 c and d?
Answered by AAA
Please me, question 5
Answered by OVNI
8 e)
t*s^2/2*PI^2*(b^2-a^2)

Please anyone 10 and 9c)
Answered by me
5) a)=B(l-W)z(t)
b)=(B*V*W)/R
c)=(B^2*V*W^2)/R
d)= (M*g*R)/(B^2*W^2)
Answered by Jack
Please OVNI, 11 c)
Answered by Jack
8 f)
Result of e multiplied by 2*pi*b*d
Answered by robbo
i'm conufsed about the parentheses that might havve been omitted in OVNI's answer to 8e....

t*s^2/(2*PI^2*(b^2-a^2) )


Is this correct?
Answered by Jack
I think a b is missing!

t*s^2/(2*PI^2*b*(b^2-a^2) )
Answered by OVNI
Sorry, it's

(t*s^2)/(2*PI^2*b*(b^2-a^2))

Please, anyone 9c) and 10c)
Answered by robbo
8f, 8e solutions aren't working out as given by OVNI and Jack... help guys?
Answered by Jack
Please 11 c) and 10!
Answered by robbo
Jack, 11c is 0.5*L*I^2
Answered by AAA
Thank you so much!
Question 10??
Answered by OVNI
Sorry, but I posted as I worked and I got a green check mark.
Please any one 9 and 10
Answered by robbo
Sorry I meant 10c is 0.5*L*I^2.... the LC circuit sum, does anybody have 10 b and d??
Answered by robbo
how do you get 11 b and c???
Answered by OVNI
Roboo, why in your 11c you use L, and I?????
Answered by robbo
I meant 10c OVNI! Mistake... SOrry!
Answered by OVNI
Please 10d) and 9c)
Answered by robbo
9b????
Answered by Jack
robbo, the same that 9 a)
Answered by 124
9b = 9a, E1 = V/d1 and E2 = V/d2

9c)try with charge conservation
Answered by Jack
robbo 10 b)
Answered by BG
11 b)
q*9*(10^9)/r^2
r = distance from the origin
Answered by Jack
10 b) 10 d) and 11 c) please!!!!
Answered by w
5A doesn't seem to work
Answered by P
Please give ans for Q5 for values I am getting them last chance please

What is the total magnetic flux through the loop (in Tesla m ) when is 62 cm. Include only the magnetic flux associated with the external field (i.e. ignore the flux associated with the magnetic self-field generated by the current in the wire loop). Note that you do not need to calculate or know at what time the loop is at this location.

incorrect

(b) Using Faraday's Law and Ohm's Law, find the magnitude (in Amps) of the induced current in the bar at the time when 1.00 m/sec. Note that you do not need to calculate or know at what time the loop has this speed.

incorrect

(c) Which way does the current flow around the loop, clockwise or counterclockwise?

Status: correct

(d) What is the total magnetic force (in Newtons) on the rigid wire loop when 1.00 m/sec? Again, ignore any effects due to the self magnetic field.


direction:
Status: correct


magnitude (in Newtons):

incorrect

(e) What is the magnitude of the terminal speed (in m/sec) of the loop (i.e. the speed at which the loop will be moving when it no longer accelerates)?
Answered by BG
8 e)

poling vector = E*B/mu_0 = ??
Answered by BG
it's correct ?
Answered by 124
P, phi=B*w*(L-z)
|em|=|dphi/dt|=B*w*V
i=|em|/R=B*w*V/R
F=watt/ms^-1=(|em|^2/R)*1/V
Answered by 124
BG, OK
Answered by Jack
10 b) please!
Answered by 124
10b)= frequency that abs(i(t))=max
Answered by Jack
f= 1/(2*pi*sqrt(L*C))?
Answered by Jack
and 10 d), please
Answered by 124
i(t)=I0*cos(w*t) and abs(i(t))=max if
w*t=k*pi, for k=1=>t1=pi/w
f=1/t1=w/pi=1/(pi*sqrt(L*C))
Answered by BG
124, i have E and B correct
but

pointing vector with formula E*B/mu_0, mu_0 = 4*pi*10^(-7), incorrect.

pliz, tell where is error?
Answered by 124
10d)=>average p=v*i on [0,3*t1]
Answered by 124
BG, S=E*B/mu_0
= 1/(2*pi^2*b)*s^2*t/(b^2-a^2)
Answered by 124
10b) f=1/t1=w/pi=1/(pi*sqrt(L*C))
Answered by 124
BG, B(r=b) = mu_0*s/(2*pi*b)is not the the previous B(r)
Answered by BG
124, thanks a lot, you the man. )
Answered by BG
124, 11 c) - Could you please help and give any hints on it?
Answered by Jack
11 c) please!
Answered by robbo
in 10d for P=V*I, where do we get V from???
Answered by 124
robbo, v=Ldi/dt
P_mean=1/(3T-0)*int(v*i,0,3*T)
T=2*pi*sqrt(L*C)

11c)e)=>Q_net+ q = Q_encl in the gauss Law
Answered by robbo
S=E*B/mu_0
= 1/(2*pi^2*b)*(s^2)*(t/(b^2-a^2))


What's wrong with this formula guys??

Please help?
Answered by Jack
(t*s^2)/(2*(PI^2)*b*(b^2-a^2))
Answered by 124
robbo, formula is right, check s^2*t
for example s=24u and t=5u
s^2*t = 24*24*5*1u^3
Answered by Jack
10 d) Positive or negative?
Answered by Jack
Is this correct for 10 d)=

P=1/3T * int(-L*w*(Imax)^2*sin(wt)*cos(wt), 0 , 3T)
Thanks!!!
Answered by robbo
Jack, and 124, for the formula:(t*s^2)/(2*(PI^2)*b*(b^2-a^2)) = 5.27300181668e-11


at s=12uC/sec, t=1usec, b=0.0053m, a=0.0014m


I've followed the formula, but the answer shows wrong... I hope you guys can help?
Answered by robbo
Jack, the formula you've used is correct :)
Answered by robbo
Q11 formulae, anybody??
Answered by 124
Jack, yes the formula is OK
Answered by robbo
124, I have no idea what on earth I'm doing wrong, followed your formula... and input values and got the same answer everytime I tried, apparently its incorrect! :( Can you input my values and see what answer you get?
Answered by 124
robbo,sorry i forgot e_0
(1u*12u^2)/(2*(PI^2)*5.3m*(5.3m^2-1.4m^2)*8.85e-12) =5.95215457341
E =|E|er
|E|=1/(4*pi*e_0*kappa)*q/r^2 for (a<r<b) |E|=1/(4*pi*e_0)*q/r^2 for r>=b
Answered by 124
S=E*B/mu_0
= 1/(2*pi^2*b*epsilon_0)*(s^2)*(t/(b^2-a^2))
Answered by 124
guys, you can subscribe in the new course ReView (8.MReVx)
Answered by 124
10d)
P=1/3T * int(-L*w*(Imax)^2*sin(wt)*cos(wt), 0 , 3T) = 0!!!!!

T=2*pi*sqrt(L*C)
Answered by robbo
124, i think its supposed to be t^2/(b^2-a^2) instead of t to the power of 1.
Answered by robbo
9c, 11c anybody?
Answered by Jack
11 c) Is the net charge positive or negative?
Answered by MIT
THIS IS CHEATING!!! YOU ARE NOT FOLLOWING THE HONOUR CODE!!
Answered by HarvardX
Dear MIT, you got here by???? Googling for what?
Answered by l
1c please
Answered by Jack
11 c) please!!!
Answered by 124
it seems that we cheated. So I leave. Good luck to all!