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?

User Icon for ^_ ^ ^_ ^ answered
11 years ago

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)

User Icon for Phy Phy answered
11 years ago

What is di/dt ??

User Icon for Anonymous Anonymous answered
11 years ago

Phy, it's the b in the variation I=bt. Just use the b value in the problem for (di/dt)

User Icon for Phy Phy answered
11 years ago

Thanks, can u tell me the solutions of 1,4,6,9,10,11,12?

User Icon for dd dd answered
11 years ago

the C is wrong for me using the b in di/dt

User Icon for robbo robbo answered
11 years ago

Have you guys gt 3,5,7,8,9,10,11,12...????? Let's help each other and get this done guys!!

User Icon for Sat Sat answered
11 years ago

C is wrong ...

User Icon for boss boss answered
11 years ago

Some people haven't done even one question.

User Icon for boss boss answered
11 years ago

I wonder if some people understand the subject at all. It is an insult to the lecturer.

User Icon for Sat Sat answered
11 years ago

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)

User Icon for Anonymous Anonymous answered
11 years ago

Hey guys, what abput prob. 10?

User Icon for BG BG answered
11 years ago

10, b) how to solve

User Icon for Jack Jack answered
11 years ago

Please, problem 8!!!

User Icon for robbo robbo answered
11 years ago

Problems 8-12 guys, please???/

User Icon for P P answered
11 years ago

Please tell ans of any question please

Q3 a=0 last part =0

User Icon for BG BG answered
11 years ago

Anyone, can help with questions:

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

User Icon for BG BG answered
11 years ago

11 a)

q/(4*pi*E_0*(r^2)*k)

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

11 d) 0

User Icon for me me answered
11 years ago

anyone for question 7 please

User Icon for Jack Jack answered
11 years ago

Hi BG, can you put the results for 8 a, b,c?

User Icon for AAA AAA answered
11 years ago

Please, Q10?

User Icon for AAA AAA answered
11 years ago

anyone for question 3 please??????

User Icon for Jack Jack answered
11 years ago

5 e)

v= m*g*R/(B^2*W^2)

User Icon for BG BG answered
11 years ago

8 a)

E = (st)/(S*E_0)
S = Area

8 b) E*d

8 c)
C = E*E_0*S/d

User Icon for BG BG answered
11 years ago

S = pi(b^2 - a^2)

User Icon for BG BG answered
11 years ago

Jack, did you solve anything else from this list:

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

User Icon for BG BG answered
11 years ago

me, I will exchange solution 8 on 7)

User Icon for BG BG answered
11 years ago

me, a = lambda/0.5

d = lambda/0.125

User Icon for BG BG answered
11 years ago

denaminator may be different, it depends on drawing

User Icon for AAA AAA answered
11 years ago

How I can know my denominator watching my grahp BG?

User Icon for BG BG answered
11 years ago

prob 8

User Icon for BG BG answered
11 years ago

I think drawings at all identical so don't worry

User Icon for me me answered
11 years ago

BG sorry don't have 8 yet, i'll post it as soon as i solve it and thanks for 7

User Icon for me me answered
11 years ago

i have 5 do you still need 5b and 5d

User Icon for Jack Jack answered
11 years ago

BG,

8 d)

B= mu_0*(r^2-a^2)*s/(2*r*pi*(b^2-a^2))

User Icon for Jack Jack answered
11 years ago

BG, do you have 10 c and d?

User Icon for AAA AAA answered
11 years ago

Please me, question 5

User Icon for OVNI OVNI answered
11 years ago

8 e)

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

Please anyone 10 and 9c)

User Icon for me me answered
11 years ago

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)

User Icon for Jack Jack answered
11 years ago

Please OVNI, 11 c)

User Icon for Jack Jack answered
11 years ago

8 f)

Result of e multiplied by 2*pi*b*d

User Icon for robbo robbo answered
11 years ago

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?

User Icon for Jack Jack answered
11 years ago

I think a b is missing!

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

User Icon for OVNI OVNI answered
11 years ago

Sorry, it's

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

Please, anyone 9c) and 10c)

User Icon for robbo robbo answered
11 years ago

8f, 8e solutions aren't working out as given by OVNI and Jack... help guys?

User Icon for Jack Jack answered
11 years ago

Please 11 c) and 10!

User Icon for robbo robbo answered
11 years ago

Jack, 11c is 0.5*L*I^2

User Icon for AAA AAA answered
11 years ago

Thank you so much!

Question 10??

User Icon for OVNI OVNI answered
11 years ago

Sorry, but I posted as I worked and I got a green check mark.

Please any one 9 and 10

User Icon for robbo robbo answered
11 years ago

Sorry I meant 10c is 0.5*L*I^2.... the LC circuit sum, does anybody have 10 b and d??

User Icon for robbo robbo answered
11 years ago

how do you get 11 b and c???

User Icon for OVNI OVNI answered
11 years ago

Roboo, why in your 11c you use L, and I?????

User Icon for robbo robbo answered
11 years ago

I meant 10c OVNI! Mistake... SOrry!

User Icon for OVNI OVNI answered
11 years ago

Please 10d) and 9c)

User Icon for robbo robbo answered
11 years ago

9b????

User Icon for Jack Jack answered
11 years ago

robbo, the same that 9 a)

User Icon for 124 124 answered
11 years ago

9b = 9a, E1 = V/d1 and E2 = V/d2

9c)try with charge conservation

User Icon for Jack Jack answered
11 years ago

robbo 10 b)

User Icon for BG BG answered
11 years ago

11 b)

q*9*(10^9)/r^2
r = distance from the origin

User Icon for Jack Jack answered
11 years ago

10 b) 10 d) and 11 c) please!!!!

User Icon for w w answered
11 years ago

5A doesn't seem to work

User Icon for P P answered
11 years ago

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)?

User Icon for BG BG answered
11 years ago

8 e)

poling vector = E*B/mu_0 = ??

User Icon for BG BG answered
11 years ago

it's correct ?

User Icon for 124 124 answered
11 years ago

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

User Icon for 124 124 answered
11 years ago

BG, OK

User Icon for Jack Jack answered
11 years ago

10 b) please!

User Icon for 124 124 answered
11 years ago

10b)= frequency that abs(i(t))=max

User Icon for Jack Jack answered
11 years ago

f= 1/(2*pi*sqrt(L*C))?

User Icon for Jack Jack answered
11 years ago

and 10 d), please

User Icon for 124 124 answered
11 years ago

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))

User Icon for BG BG answered
11 years ago

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?

User Icon for 124 124 answered
11 years ago

10d)=>average p=v*i on [0,3*t1]

User Icon for 124 124 answered
11 years ago

BG, S=E*B/mu_0

= 1/(2*pi^2*b)*s^2*t/(b^2-a^2)

User Icon for 124 124 answered
11 years ago

10b) f=1/t1=w/pi=1/(pi*sqrt(L*C))

User Icon for 124 124 answered
11 years ago

BG, B(r=b) = mu_0*s/(2*pi*b)is not the the previous B(r)

User Icon for BG BG answered
11 years ago

124, thanks a lot, you the man. )

User Icon for BG BG answered
11 years ago

124, 11 c) - Could you please help and give any hints on it?

User Icon for Jack Jack answered
11 years ago

11 c) please!

User Icon for robbo robbo answered
11 years ago

in 10d for P=V*I, where do we get V from???

User Icon for 124 124 answered
11 years ago

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

User Icon for robbo robbo answered
11 years ago

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?

User Icon for Jack Jack answered
11 years ago

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

User Icon for 124 124 answered
11 years ago

robbo, formula is right, check s^2*t

for example s=24u and t=5u
s^2*t = 24*24*5*1u^3

User Icon for Jack Jack answered
11 years ago

10 d) Positive or negative?

User Icon for Jack Jack answered
11 years ago

Is this correct for 10 d)=

P=1/3T * int(-L*w*(Imax)^2*sin(wt)*cos(wt), 0 , 3T)
Thanks!!!

User Icon for robbo robbo answered
11 years ago

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?

User Icon for robbo robbo answered
11 years ago

Jack, the formula you've used is correct :)

User Icon for robbo robbo answered
11 years ago

Q11 formulae, anybody??

User Icon for 124 124 answered
11 years ago

Jack, yes the formula is OK

User Icon for robbo robbo answered
11 years ago

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?

User Icon for 124 124 answered
11 years ago

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

User Icon for 124 124 answered
11 years ago

S=E*B/mu_0

= 1/(2*pi^2*b*epsilon_0)*(s^2)*(t/(b^2-a^2))

User Icon for 124 124 answered
11 years ago

guys, you can subscribe in the new course ReView (8.MReVx)

User Icon for 124 124 answered
11 years ago

10d)

P=1/3T * int(-L*w*(Imax)^2*sin(wt)*cos(wt), 0 , 3T) = 0!!!!!

T=2*pi*sqrt(L*C)

User Icon for robbo robbo answered
11 years ago

124, i think its supposed to be t^2/(b^2-a^2) instead of t to the power of 1.

User Icon for robbo robbo answered
11 years ago

9c, 11c anybody?

User Icon for Jack Jack answered
11 years ago

11 c) Is the net charge positive or negative?

User Icon for MIT MIT answered
11 years ago

THIS IS CHEATING!!! YOU ARE NOT FOLLOWING THE HONOUR CODE!!

User Icon for HarvardX HarvardX answered
11 years ago

Dear MIT, you got here by???? Googling for what?

User Icon for l l answered
11 years ago

1c please

User Icon for Jack Jack answered
11 years ago

11 c) please!!!

User Icon for 124 124 answered
11 years ago

it seems that we cheated. So I leave. Good luck to all!