Posted by Annoyingmous on Saturday, June 15, 2013 at 4:35am.
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)
What is di/dt ??
Phy, it's the b in the variation I=bt. Just use the b value in the problem for (di/dt)
Thanks, can u tell me the solutions of 1,4,6,9,10,11,12?
the C is wrong for me using the b in di/dt
Have you guys gt 3,5,7,8,9,10,11,12...????? Let's help each other and get this done guys!!
C is wrong ...
Some people haven't done even one question.
I wonder if some people understand the subject at all. It is an insult to the lecturer.
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)
Hey guys, what abput prob. 10?
10, b) how to solve
Please, problem 8!!!
Problems 8-12 guys, please???/
Please tell ans of any question please
Q3 a=0 last part =0
Anyone, can help with questions:
11 c)
10 b)
8 d,e,f)
5 b,d,e)
11 a)
q/(4*pi*E_0*(r^2)*k)
11 b)
q*k/(r^2)
11 d) 0
anyone for question 7 please
Hi BG, can you put the results for 8 a, b,c?
Please, Q10?
anyone for question 3 please??????
5 e)
v= m*g*R/(B^2*W^2)
8 a)
E = (st)/(S*E_0)
S = Area
8 b) E*d
8 c)
C = E*E_0*S/d
S = pi(b^2 - a^2)
Jack, did you solve anything else from this list:
11 c)
10 b)
8 d,e,f)
5 b,d)
me, I will exchange solution 8 on 7)
me, a = lambda/0.5
d = lambda/0.125
denaminator may be different, it depends on drawing
How I can know my denominator watching my grahp BG?
prob 8
I think drawings at all identical so don't worry
BG sorry don't have 8 yet, i'll post it as soon as i solve it and thanks for 7
i have 5 do you still need 5b and 5d
BG,
8 d)
B= mu_0*(r^2-a^2)*s/(2*r*pi*(b^2-a^2))
BG, do you have 10 c and d?
Please me, question 5
8 e)
t*s^2/2*PI^2*(b^2-a^2)
Please anyone 10 and 9c)
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)
Please OVNI, 11 c)
8 f)
Result of e multiplied by 2*pi*b*d
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?
I think a b is missing!
t*s^2/(2*PI^2*b*(b^2-a^2) )
Sorry, it's
(t*s^2)/(2*PI^2*b*(b^2-a^2))
Please, anyone 9c) and 10c)
8f, 8e solutions aren't working out as given by OVNI and Jack... help guys?
Please 11 c) and 10!
Jack, 11c is 0.5*L*I^2
Thank you so much!
Question 10??
Sorry, but I posted as I worked and I got a green check mark.
Please any one 9 and 10
Sorry I meant 10c is 0.5*L*I^2.... the LC circuit sum, does anybody have 10 b and d??
how do you get 11 b and c???
Roboo, why in your 11c you use L, and I?????
I meant 10c OVNI! Mistake... SOrry!
Please 10d) and 9c)
9b????
robbo, the same that 9 a)
9b = 9a, E1 = V/d1 and E2 = V/d2
9c)try with charge conservation
robbo 10 b)
11 b)
q*9*(10^9)/r^2
r = distance from the origin
10 b) 10 d) and 11 c) please!!!!
5A doesn't seem to work
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)?
8 e)
poling vector = E*B/mu_0 = ??
it's correct ?
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
BG, OK
10 b) please!
10b)= frequency that abs(i(t))=max
f= 1/(2*pi*sqrt(L*C))?
and 10 d), please
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))
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?
10d)=>average p=v*i on [0,3*t1]
BG, S=E*B/mu_0
= 1/(2*pi^2*b)*s^2*t/(b^2-a^2)
10b) f=1/t1=w/pi=1/(pi*sqrt(L*C))
BG, B(r=b) = mu_0*s/(2*pi*b)is not the the previous B(r)
124, thanks a lot, you the man. )
124, 11 c) - Could you please help and give any hints on it?
11 c) please!
in 10d for P=V*I, where do we get V from???
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
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?
(t*s^2)/(2*(PI^2)*b*(b^2-a^2))
robbo, formula is right, check s^2*t
for example s=24u and t=5u
s^2*t = 24*24*5*1u^3
10 d) Positive or negative?
Is this correct for 10 d)=
P=1/3T * int(-L*w*(Imax)^2*sin(wt)*cos(wt), 0 , 3T)
Thanks!!!
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?
Jack, the formula you've used is correct :)
Q11 formulae, anybody??
Jack, yes the formula is OK
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?
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
S=E*B/mu_0
= 1/(2*pi^2*b*epsilon_0)*(s^2)*(t/(b^2-a^2))
guys, you can subscribe in the new course ReView (8.MReVx)
10d)
P=1/3T * int(-L*w*(Imax)^2*sin(wt)*cos(wt), 0 , 3T) = 0!!!!!
T=2*pi*sqrt(L*C)
124, i think its supposed to be t^2/(b^2-a^2) instead of t to the power of 1.
9c, 11c anybody?
11 c) Is the net charge positive or negative?
THIS IS CHEATING!!! YOU ARE NOT FOLLOWING THE HONOUR CODE!!
Dear MIT, you got here by???? Googling for what?
1c please
11 c) please!!!
it seems that we cheated. So I leave. Good luck to all!