Thursday

August 25, 2016
Posted by **s** on Friday, April 26, 2013 at 1:20pm.

(a) Calculate the magnitude of the total current (in A) carried by the rotating sphere for the following values of σ, ω and R:

σ= 6 10−4C m−2, ω= 7 rad⋅s−1 and R=1m

unanswered

(b) Calculate the magnitude of the magnetic field B(z) (in T) that is generated by the circular current of the rotating shell at a point P on the z-axis for the following values of σ, ω , z and R:

σ= 6 10−4C m−2, ω= 7 rad⋅s−1, z= 1.6 m and R=1m

unanswered

- Physics -
**Anonymous**, Saturday, April 27, 2013 at 1:20pma) 0

b) 0 - Physics -
**Anonymous**, Saturday, April 27, 2013 at 1:50pma) I = 52.78 e-3

b) B = 5.53 e-8 T - Physics -
**FLu**, Saturday, April 27, 2013 at 3:41pmAnonymous, can you provide formula because of different values please?

- Physics -
**Mag**, Saturday, April 27, 2013 at 4:45pma) Calculate the magnitude of the total current (in A) carried by the rotating sphere for the following values of , and R:

= 6 ~10−4C m−2, = 7 rad⋅s−1 and R=1m

(b) Calculate the magnitude of the magnetic field B(z) (in T) that is generated by the circular current of the rotating shell at a point P on the z-axis for the following values of , , z and R:

= 6 ~10−4C m−2, = 5 rad⋅s−1, z= 1.92 m and R=1m - Physics -
**Mag**, Saturday, April 27, 2013 at 4:48pm(a) Calculate the magnitude of the total current (in A) carried by the rotating sphere for the following values of , and R:

sigma= 6x10−4C m−2, w= 5 rad⋅s−1 and R=1m

b) Calculate the magnitude of the magnetic field B(z) (in T) that is generated by the circular current of the rotating shell at a point P on the z-axis for the following values of , , z and R:

sigma= 6x10−4C m−2, w= 7 rad⋅s−1, z= 1.6 m and R=1m - Physics -
**Mag**, Saturday, April 27, 2013 at 4:49pmb) Calculate the magnitude of the magnetic field B(z) (in T) that is generated by the circular current of the rotating shell at a point P on the z-axis for the following values of , , z and R:

sigma= 6x10−4C m−2, w= 5 rad⋅s−1, z= 1.6 m and R=1m - Physics -
**Mag**, Saturday, April 27, 2013 at 4:50pmb) Calculate the magnitude of the magnetic field B(z) (in T) that is generated by the circular current of the rotating shell at a point P on the z-axis for the following values of , , z and R:

sigma= 6x10−4C m−2, w= 5 rad⋅s−1, z= 1.92 m and R=1m - Physics -
**Phy**, Saturday, April 27, 2013 at 6:13pma) (a) Calculate the magnitude of the total current (in A) carried by the rotating sphere for the following values of sigma,omega and R:

sigma = 5 times 10^{-4},C/m^2, omega = 4 rad/sec and R = 1 m

b)(b) Calculate the magnitude of the magnetic field B(z) (in T) that is generated by the circular current of the rotating shell at a point P on the z-axis for the following values of sigma, omega , z and R:

sigma = 5 times 10^{-4} {C m}^{-2}, \omega = 4 rad/sec , z= 2.1 m and R =1m - Physics -
**Anonymous**, Saturday, April 27, 2013 at 7:59pma) I = Sigma*A*omega

b) B = mu*I/(2*(z-R)) - Physics -
**Phy**, Sunday, April 28, 2013 at 12:47amHow to calculate 'A'???

- Physics -
**Rubens**, Sunday, April 28, 2013 at 12:57am@Phy

with spherical surface - Physics -
**Phy**, Sunday, April 28, 2013 at 1:13amThat is a wrong formula, i m getting my answers wrong

- Physics -
**FLu**, Sunday, April 28, 2013 at 6:16amHave the same problem, how to calculate for A?

- Physics -
**Mag**, Sunday, April 28, 2013 at 6:19amMe too please help!

(a) Calculate the magnitude of the total current (in A) carried by the rotating sphere for the following values of , and R:

sigma= 6x10−4C m−2, w= 5 rad⋅s−1 and R=1m

b) Calculate the magnitude of the magnetic field B(z) (in T) that is generated by the circular current of the rotating shell at a point P on the z-axis for the following values of , , z and R:

sigma= 6x10−4C m−2, w= 5 rad⋅s−1, z= 1.92 m and R=1m - Physics -
**Phy**, Sunday, April 28, 2013 at 6:20amA=4*pi*r*r

- Physics -
**FLu**, Sunday, April 28, 2013 at 6:29amThanks Phy:

is this right?

a) I = Sigma*A*omega

6*10^-4*4*pi*1*1*5

I don't seem to get it right.

b) mu*I/(2*(z-R))

(6*10^-4)/(2*(1.92-1))

I am not sure, if I found I though. Please could someone help?

thanks - Physics -
**Phy**, Sunday, April 28, 2013 at 6:37amYeah even i got it wrong, the formula itself is wrong

- Physics -
**FLu**, Sunday, April 28, 2013 at 6:46amOh, I see.

Anonymous or someone else, could you maybe provide an example with real numbers? That would also help others. - Physics -
**Hur**, Sunday, April 28, 2013 at 6:59amYes, please if possible provide answer with one example,thankyou!

- Physics -
**Gise**, Sunday, April 28, 2013 at 7:10amRubens can show example with real number please?

- Physics -
**FLu**, Sunday, April 28, 2013 at 8:17amPhy, did you by any chance manage Problems?

Bainbridge mass spectrometer

Magnetic Field of a current-carrying ribbon

Second RL Circuit

Magnetic Field of a loop: I got Bx and By but Bz could not figure out Bz.

I have Problems 1 (RL Circuit) and 2. - Physics -
**Linda**, Sunday, April 28, 2013 at 9:47amPls Flu post 1 and 2.

- Physics -
**FLu**, Sunday, April 28, 2013 at 10:55amProblem 1 as follows:

a)

I1:(R1+R2)/V

I2:(R1+R2)/V

I3:0

b)

I1:V/R1

I2:0

I3:V/R1

Problem 2:

I got right like this:

if V right reads -0.5

then multiply by two and drop the minus, so a) 1

b) is the same value without minus, so b) 0.5

Anyone for the other problems please? - Physics -
**Mag**, Sunday, April 28, 2013 at 10:58amThanks FLu, I got it now.

Please someone for the other problems? - Physics -
**Glin**, Sunday, April 28, 2013 at 11:02amMe too thanks FLu!

Please explanation for other problems? - Physics -
**FLu**, Sunday, April 28, 2013 at 11:58amAnyone?

- Physics -
**Sag**, Sunday, April 28, 2013 at 11:59amHas anybody got the explanation for other problems please?

- Physics -
**Nusa**, Sunday, April 28, 2013 at 12:57pmAnybody!

- Physics -
**Anonymous**, Sunday, April 28, 2013 at 1:16pma) I = Sigma*A*f

f = omega/(2pi)

b) B = mu*I/(2*(z-R)) - Physics -
**ANONYMOUS**, Sunday, April 28, 2013 at 1:52pmValue of mu? Please

- Physics -
**Anonymous**, Sunday, April 28, 2013 at 1:59pmThe other problems?

Bainbridge mss spectrometer

Magnetic Field of a current-carrying ribbon

Second Rl circuit

magnetic Field of a loop - Bz. - Physics -
**FLu**, Sunday, April 28, 2013 at 2:07pmIs value for mu not 1.25663706*10^-6? Is it different to mu_0 then?

can you show us an example with the above numbers please? - Physics -
**FLu**, Sunday, April 28, 2013 at 2:14pmAnonymous can you tell me what the value for A is then?

Something does not add up.

This calculation below looks odd to me:

(6*10^-4*4*pi*1*1*5)*(omega/2*pi)

Anyone did figure out yet? - Physics -
**FLu**, Sunday, April 28, 2013 at 2:23pmAnonymous, the formula is wrong. Did you get the answer right by the way?

If somebody got the answer could they provide it step by step with numbers though?

thanks - Physics -
**ANONYMOUS!!**, Sunday, April 28, 2013 at 2:24pmI got the a) right, the formula is:

I = Sigma*A*f

f = omega/(2pi)

i got the b) wrong, i guess Mu is not 4piE-7 - Physics -
**FLu**, Sunday, April 28, 2013 at 2:39pmoh I see, so there is a difference between mu and mu_0. I cannot find any values as well.

could you just help me with calculation a) please?

If my values are:

sigma= 6x10−4C m−2, w= 5 rad⋅s−1 and R=1m

Is A= 4*pi*r*r ? if not could you correct my values below Anonymous please?

Formula for a)

(6x10-4*4*pi*1*1)*(5/2*pi)

I think I am making mistakes with value A, please correct me Anonymous. Thanks! - Physics -
**Anon**, Sunday, April 28, 2013 at 2:43pmYou should be using mu/4pi for answer in b

- Physics -
**FLu**, Sunday, April 28, 2013 at 2:47pmyou mean this is the formula for b) Anon?(mu/4*pi)/(2*(z-R))

Is mu same as mu_0?

thanks

could you also check if my procedure for a) is right Anon?

thanks - Physics -
**Anon**, Sunday, April 28, 2013 at 2:50pmyes it is good

- Physics -
**Naseng**, Sunday, April 28, 2013 at 2:54pmThis procedure is right Anon really?

I have just one try left can somebody check please?

a)(6x10-4*4*pi*1*1)*(5/2*pi)

b)(mu/4*pi)/(2*(z-R)) - Physics -
**Anon**, Sunday, April 28, 2013 at 2:57pmyou are missing the current I in b

- Physics -
**Naseng**, Sunday, April 28, 2013 at 3:01pmI is the result from problem a) right?

Could you give me the complete formula or correct the above please Anon? - Physics -
**hmmmm**, Sunday, April 28, 2013 at 3:01pmanyone for part b please,clearly

- Physics -
**Anon**, Sunday, April 28, 2013 at 3:02pmcorrect

- Physics -
**Naseng**, Sunday, April 28, 2013 at 3:02pmbefore I forget mu is the value this:

1.25663706*10^-6 ? - Physics -
**Anon**, Sunday, April 28, 2013 at 3:03pmyes but you need mu/4pi

- Physics -
**FLu**, Sunday, April 28, 2013 at 3:04pmAnon, is it this formula for b) then please?

b)(mu*I/4*pi)/(2*(z-R))

and as Naseng said is the value for mu=

1.25663706*10^-6

if not what is it? - Physics -
**Anonymous**, Sunday, April 28, 2013 at 3:13pmmu ia mu_0

- Physics -
**FLu**, Sunday, April 28, 2013 at 3:13pmAnon, many thanks now a) I got but stuck with b)

I would be forever greatful, if I can figure b) out with your help.

a) 0.007 so that is my I

b)(mu*I/4*pi)/(2*(z-R))

(1.25663706*10^-6*0.007/4*pi)/(2*(2.2-1)

I followed your advice, but there seems to be an error in my calculation, could you help please? That would also help others to figure out. I am wondering if it is a bracket missing or if one of the values are wrong.

Thanks for time Anon! - Physics -
**Anon**, Sunday, April 28, 2013 at 3:16pmmu = 4pi * 10^(-7)

- Physics -
**hmmmm**, Sunday, April 28, 2013 at 3:19pmi used this getting wrong answer

- Physics -
**FLu**, Sunday, April 28, 2013 at 3:21pmThanks Anon but 1.25663706*10^-6 is 4*pi*10^-7, please check if you like.

It still does not work, any other suggestion, am I missing a bracket somewhere? - Physics -
**hmmmm**, Sunday, April 28, 2013 at 3:21pmdid u get it right FLU?

- Physics -
**FLu**, Sunday, April 28, 2013 at 3:22pmhmmmm do not use for b) yet as there seems to be missing but I got a) right so can assure you the formula must be right.

- Physics -
**FLu**, Sunday, April 28, 2013 at 3:23pmNo, but I think we are near there is something missing but cannot figure it out, maybe a bracket issue!? Anon any suggestion or even Anonymous? You are usually great help guys! thanks

- Physics -
**FLu**, Sunday, April 28, 2013 at 3:25pmAnon could you write the formula for b) please with the above numbers? would be very helpful.

- Physics -
**Anon**, Sunday, April 28, 2013 at 3:32pm(I*10^(-7))/(2*(z-R))

- Physics -
**FLu**, Sunday, April 28, 2013 at 3:37pmAnon thanks, I thought it is

(I*4*pi*10^(-7))/(2*(z-R))

did you not say mu is=

4pi * 10^(-7) ? - Physics -
**X10**, Sunday, April 28, 2013 at 3:40pm@anon this formula works for my friend but not for me I don't know what is the problem going on perhaps the formula isn't derived correctly...:(

- Physics -
**Anon**, Sunday, April 28, 2013 at 3:41pmyes i did and i also said you need to use

mu/4pi - Physics -
**X10**, Sunday, April 28, 2013 at 3:41pm@FLu let me know if u got the correct answer with this formula and what it is ?

- Physics -
**Anon**, Sunday, April 28, 2013 at 3:42pmyou need to make sure you use the right value for I

- Physics -
**ind0**, Sunday, April 28, 2013 at 3:45pmanyone has answer for other problems

- Physics -
**X10**, Sunday, April 28, 2013 at 3:46pmYup I've got I correct it is: 8.4*10^-3

sigma= 7 10−4C m−2, ω= 6 rad⋅s−1, z= 1.9 m and R=1m

and B as i calculated: 4.66*10^-10 but the grader is not accepting it - Physics -
**ind0**, Sunday, April 28, 2013 at 3:46pmThe other problems?

Bainbridge mss spectrometer

Magnetic Field of a current-carrying ribbon

Second Rl circuit

magnetic Field of a loop - Bz. - Physics -
**X10**, Sunday, April 28, 2013 at 3:47pmI've solved all problems just got stuck with this one...

- Physics -
**Any**, Sunday, April 28, 2013 at 3:48pmConsider a thin, infinitely long conducting ribbon that carries a uniform current density j (current per unit area). The width of the ribbon is w and its thickness s is extremely small (s≪w). P is a point in the plane of the ribbon, at a large distance (x≫s) from the ribbon edge. (See the figure below)

What is the magnitude of the magnetic field B (in T) at point P for the following values of w , j, s and x?

w= 8 cm; s= 0.1 cm; j=1A/m2 and x= 22 cm.

Please help with formula to solve it - Physics -
**ind0**, Sunday, April 28, 2013 at 3:49pmcan you share formulas or aproach

- Physics -
**X10**, Sunday, April 28, 2013 at 3:50pmQ:3

for part a: B=E/v

for part b: (L^2+h^2)/2L

for part c: m=rqB/v - Physics -
**X10**, Sunday, April 28, 2013 at 3:50pmQ6: ((mu_0*I)/2*pi*w)*ln(1+(w/x)) .... (I = J*w*s) and convert cm to m.

- Physics -
**Any**, Sunday, April 28, 2013 at 3:54pmIn the circuit shown in the figure below, the switch closes at t=0, R=5.5 Ohm, ε=9 V, L=0.08 H.

(a) What are the currents (in A) through the two bottom branches at t=0+ (just after the switch is closed)?

I1 and I2 ?

(b) What are the currents (in A) through the two bottom branches at a much later time t≈∞?

I1 and I2 ?

Please help with the formula - Physics -
**ind0**, Sunday, April 28, 2013 at 4:13pmX10

Q3 part c and Q6 are wrong could check formulas - Physics -
**X10**, Sunday, April 28, 2013 at 4:15pmfor q3 part c use the Bo not the one derived in part a...and for q6 make sure u have converted the values in m from cm

- Physics -
**ind0**, Sunday, April 28, 2013 at 4:21pmQ6: ((mu_0*I)/2*pi*w)*ln(1+(w/x))

do u divide by 2 * pi * w - Physics -
**ind0**, Sunday, April 28, 2013 at 4:31pmThanks

what the formula for Bz in Q5 - Physics -
**Anon**, Sunday, April 28, 2013 at 4:32pmX10 are done now

- Physics -
**FLu**, Sunday, April 28, 2013 at 4:39pmX10 there must be a problem with the grader as you suggested your friend got the right answer and a friend of mine got the right answer too with the provided formula by Anton.

- Physics -
**FLu**, Sunday, April 28, 2013 at 4:46pmIn the circuit shown in the figure below, the switch closes at t=0, R=5.5 Ohm, =9 V, L=0.08 H.

(a) What are the currents (in A) through the two bottom branches at t=0+ (just after the switch is closed)?

I1 and I2 ?

(b) What are the currents (in A) through the two bottom branches at a much later time t֡?

I1 and I2 ?

Any the first I1 and last I2 is value 0 both in mine. Any formula to solve the middle ones? I2 and I1 - Physics -
**FLu**, Sunday, April 28, 2013 at 5:04pmX10 thanks for your help with Q6:((mu_0*I)/2*pi*w)*ln(1+(w/x)) (I = J*w*s) and convert cm to m.

I am having issues with the numbers could you check if this is right please?

If my values are:

w=8cm; s=0.1cm; j=1A/m(squareroot) and x=22cm

Is this formula correct:

I=1*0.1*8=0.8

((1.25663706*10^-6*0.8)/2*pi*8)*ln(1+(8/22))

Thanks X10! - Physics -
**DOnny**, Sunday, April 28, 2013 at 5:24pmX10 do you know how to solve for the magnetic field of a loop? i've been cracking my head

- Physics -
**Mag**, Sunday, April 28, 2013 at 6:03pmHave same problem like Flu.

I=1*0.1*8=0.8

((1.25663706*10^-6*0.8)/2*pi*8)*ln(1+(8/22))

This formula did not work, could somebody check it please?

Thankyou - Physics -
**Saga**, Sunday, April 28, 2013 at 6:04pmYes, same here, is there an issue with converting it from cm to m maybe? can somebody help where the mistake lie. X10 maybe. thanks!

- Physics -
**DOnny**, Sunday, April 28, 2013 at 6:08pmit has to be in meters

I=J*w*s (w and s are in meters so change them)

then B= (mu_0*(J*w*s))/(2*pi*w)*(ln(1+w/x))

all are in meters. should get something like a 6.++e-7 - Physics -
**DOnny**, Sunday, April 28, 2013 at 6:11pmnow someone please help me with this question

A current I=2.9 A flows around a continuous path that consists of portions of two concentric circles of radii a and a/2, respectively, where a=4 cm, and two straight radial segments. The point P is at the common center of the two circle segments.

stuck. literally - Physics -
**Dr. Lewin**, Sunday, April 28, 2013 at 6:11pmYou dont have any HONOR !!

- Physics -
**FLu**, Sunday, April 28, 2013 at 6:13pmThanks DOnny, did I convert it right?

I want to be sure, as this is my last try.

((1.25663706*10^-6*(1*0.08*0.001))/2*pi*0.08)*ln(1+(0.08/0.22))

Thanks again. - Physics -
**DOnny**, Sunday, April 28, 2013 at 6:16pmFlu our values may vary so i am not certain. but the formulas are fine

if you could assist me on Q5 Flu ? - Physics -
**Saga**, Sunday, April 28, 2013 at 6:17pmDOnny, I am getting 3.91822*10^-12 and that is marked wrong, could you tell me the full value? 6.++e-7

please - Physics -
**FLu**, Sunday, April 28, 2013 at 6:20pmDOnny my values are like this below:

w=8cm; s=0.1cm; j=1A/m(squareroot) and x=22cm

However, I am uncertain as Saga seems to got a wrong answer. - Physics -
**Saga**, Sunday, April 28, 2013 at 6:21pmYes, have the same value as Flue but marked wrong when I used the formula. There must be something missing, a bracket maybe? Could somebody check.

- Physics -
**DOnny**, Sunday, April 28, 2013 at 6:23pmI have only the Bx and By, I assume you have them? it is both 0 with mine. I could not figure Bz out yet but will let you know when I do.

- Physics -
**FLu**, Sunday, April 28, 2013 at 6:25pmSorry DOnny, have accidentally used your username. will let you know about Bz when I have figured it out. but have issues with the last problem still. Cannot figure out where the issue lies even though the numbers are converted to meters now.

- Physics -
**DOnny**, Sunday, April 28, 2013 at 6:32pmsorry its supposed to be 6.++e-11

if you have your fx-570MS calculator you should be able to find mu_0 there. dont bother expanding the values - Physics -
**DOnny**, Sunday, April 28, 2013 at 6:34pm((1.256637061e-6)*(1*0.08*0.001))/(2*pi*0.08)*(ln(1+(0.08/0.22))

- Physics -
**FLu**, Sunday, April 28, 2013 at 6:36pmUnfortunaley, can only use wolframalpha but that should work.

Could you help me with figuring this problem out?

my values are:

w=8cm; s=0.1cm; j=1A/m(squareroot) and x=22cm

Is this formula correct:

((1.25663706*10^-6*(1*0.08*0.001))/2*pi*0.08)*ln(1+(0.08/0.22))

If you could help me with this I would eternally be greatful! - Physics -
**DOnny**, Sunday, April 28, 2013 at 6:38pmthe brackets... BRACKETS

((1.25663706*10^-6*(1*0.08*0.001))/2*pi*0.08)*ln(1+(0.08/0.22)) = 3.91822*10^-12

((1.25663706e-6)*(1*0.08*0.001))/(2*pi*0.08)*ln(1+(0.08/0.22))= 6.2030986e-11 - Physics -
**FLu**, Sunday, April 28, 2013 at 6:38pmThanks will try out!

- Physics -
**Niznkl**, Sunday, April 28, 2013 at 6:40pm@FLu.. No!!! We have the same values for all the test !!

And I got it right (carring- ribbon !! Give me the B part of the last problem.. and i give this result.. - Physics -
**FLu**, Sunday, April 28, 2013 at 6:41pmDOnny you are a star! It worked after a huge struggle! I will try to figure Bz out and will let you know, if I get something.

Thanks again! - Physics -
**FLu**, Sunday, April 28, 2013 at 6:43pmNiznkl, thanks I got it now with DOnny's help. Which b part do you mean?

- Physics -
**Niznkl**, Sunday, April 28, 2013 at 6:45pmthe problem of the sphere .. the magnetic field at the point p .. is all that I need to finish... we got the same values

- Physics -
**FLu**, Sunday, April 28, 2013 at 6:54pmNiznkl, it did not work for me but a friend of mine solved it with this formula provided by Anton so it should work. X10 stated that the friend solved it too. However, there was an issue with ours.

Use this, it should work and report back: remember I is the value from your answer a) so add that below.

(I*10^(-7))/(2*(z-R))

If you are finished could you provide by any chance the formula for second RL circuit problem 4? The first is 0 and last one too. However I could not figure out the middle part for a)I2 and b)I1

I would appreciate if you could provide the formula.

thanks - Physics -
**FLu**, Sunday, April 28, 2013 at 6:58pmOH and before I forget the formula for Magnetic field of a loop last part Bz?

Most of us had difficulties with that, so if you could provide us with the formula, I, DOnny, X10 and others would be greatful. - Physics -
**DOnny**, Sunday, April 28, 2013 at 7:23pmmagnetic loop is the only one i can't solve

- Physics -
**Anonymous**, Sunday, April 28, 2013 at 7:28pmB=[(2*mu_0)/(3*z^3)]*(R^4)*sigma*omega

- Physics -
**Niznkl**, Sunday, April 28, 2013 at 7:50pmMagnetic field in a loop----

B= B_1 +B_2

yo have to calculate 2 magnetic fields for a current negative(see Y axis) ( - ) plus current positive

For a current I=0.2 and a=6cm

Bz= 4.7125*10^(-6) - Physics -
**Spartacus**, Sunday, April 28, 2013 at 8:30pm@Niznkl can you give us the formula that you use to get that result for Bz?

- Physics -
**John**, Sunday, April 28, 2013 at 10:18pmI need the formulas for Q5 , Q6 and Q7

- Physics -
**Anonymous**, Monday, April 29, 2013 at 1:00amA spherical shell of radius R carries a uniform surface charge density (charge per unit area) . The center of the sphere is at the origin and the shell rotates with angular velocity (in rad/sec) around the z-axis (z=0 at the origin). Seen from below, the sphere rotates clockwise. (See the figure below)

(a) Calculate the magnitude of the total current (in A) carried by the rotating sphere for the following values of , and R:

= 6 ~10−4C m−2, = 5 rad⋅s−1 and R=1m

unanswered

(b) Calculate the magnitude of the magnetic field B(z) (in T) that is generated by the circular current of the rotating shell at a point P on the z-axis for the following values of , , z and R:

= 6 ~10−4C m−2, = 5 rad⋅s−1, z= 1.9 m and R=1m

Please answer - Physics -
**Dr. Slump**, Monday, April 29, 2013 at 4:28amHonor code????

- Physics -
**FLu**, Monday, April 29, 2013 at 5:56amNiznkl, thanks I seem to have different values for I at least it is, I=0.75 and a=6 cm.

COuld you give us the formula please so we can calculate it, or maybe a step by step guide? I could not figure it out.

thanks!

Anyone have formula for second RL circuit, Problem 4 too? - Physics -
**Nitra**, Monday, April 29, 2013 at 5:59amPlease Niznkl, provide step by step formula with the value so we can calculate with different values! For magnetic field of a loop

- Physics -
**My**, Monday, April 29, 2013 at 6:00amPlease formula for magnetic field of a loop last part Bz?

- Physics -
**FLu**, Monday, April 29, 2013 at 6:01amCorrection it is I=0.85 and a=6cm

- Physics -
**Saga**, Monday, April 29, 2013 at 6:02amMagnetic field of a loop formula for Bz please?

- Physics -
**Ortum**, Monday, April 29, 2013 at 6:21amYes, how to calculate Bz for different values please?

- Physics -
**X10**, Monday, April 29, 2013 at 7:46amQ5:

mu_0*i/4r where m_0= 4*pi*10^-7 and r=a please convert cm to m - Physics -
**X10**, Monday, April 29, 2013 at 8:12amThankyou guys...specially that Anonymous person who provided the correct formula for calculating B in the last question and so mid is over now..

last quest part b: B=[(2*mu_0)/(3*z^3)]*(R^4)*sigma*omega - Physics -
**zerocool**, Monday, April 29, 2013 at 8:32amhonour code breached

- Physics -
**Ortum**, Monday, April 29, 2013 at 8:55amX10 and Anonymous thanks. Could you tell what are the values of z, R, sigma and omega?

I cannot identify them. - Physics -
**Saga**, Monday, April 29, 2013 at 8:59amThanks guys. Having problems identifying the variables associated with z, R, sigma and omega too. Is that z=I and R=a? What is sigma and omega value please?

- Physics -
**My**, Monday, April 29, 2013 at 9:19amX10 please formula for second RL circuit!

And the values for omega, sigma, z and R? - Physics -
**FLu**, Monday, April 29, 2013 at 9:24amGuys, I think the formula what X10 provided is for last problem on charged sphere not magnetic field of loop. The values can be found on it.

X10 did you figure out Bz for magnetic field of a loop?

and anyone for second RL circuit formula please? - Physics -
**Ortum**, Monday, April 29, 2013 at 9:27amOh thanks Flu!

anyone formula for second RL circuit question and formula for Bz magnetic field loop? - Physics -
**X10**, Monday, April 29, 2013 at 9:29amFor Bz quest:5

0

0

mu_0*i/4r where m_0= 4*pi*10^-7 and r=a please convert cm to m - Physics -
**FLu**, Monday, April 29, 2013 at 9:54amGreat thanks X10!

For Problem 4, second RL Circuit question, could you help with formula?

I know that first and last are 0 but the middle one I cannot get. - Physics -
**Ortum**, Monday, April 29, 2013 at 10:17amThanks X10, Anyone for second RL circuit? formula to calculate

- Physics -
**X10**, Monday, April 29, 2013 at 10:45amI've solved RL circuits using circuit simulator software. Anyhow for Q4

part - a) I1=0,I2=V/(2R+R)

part - b) I1=V/2R,I2=0

These formula can be used but I haven't used these.. - Physics -
**Anonymous**, Monday, April 29, 2013 at 11:56amQ3 pls!!!

A Bainbridge mass spectrometer is shown in the figure. A charged particle with mass m, charge |q|=4.8 10−19C and speed v=3 106 m/s enters from the bottom of the figure and traces out the trajectory shown in the fields shown. The only electric field E=9 103 V/m is in the region where the trajectory of the charge is a straight line.

(a) When the particle is moving through the first (straight-line) segment of its trajectory, what is the magnitude of the magnetic field B in Tesla?

(b) The charge hits the left wall of the spectrometer at a vertical distance h=0.179 m above where it entered the upper region and a horizontal distance L=0.425 m to the left of where it entered the upper region (see sketch). What is the radius r of the trajectory in m?

(c) The mass of the particle can be determined using the radius r, the charge q, the speed v, and the magnetic field B0. Using a value of B0=0.4 T, evaluate the mass of the particle in kg. (Note that the magnitude of the field in the curved section, B0, is NOT the same as the magnitude in the straight section, B, found in part a). - Physics -
**Spartacus**, Monday, April 29, 2013 at 12:27pmSomeone get the last question right using this formula: [(2*mu_0)/(3*z^3)]*(R^4)*sigma*omega ??

- Physics -
**FLu**, Monday, April 29, 2013 at 1:19pmAnonymous, I think that is question 3 and was provided by X10. Just use the below formula, it works I tried.

Q:3

for part a: B=E/v

for part b: (L^2+h^2)/2L

for part c: m=rqB/v - Physics -
**Anonymous**, Monday, April 29, 2013 at 1:41pmThanks, FLu!

- Physics -
**Dope Feeder**, Monday, April 29, 2013 at 2:43pm//electron9.phys.utk.edu/phys513/Modules/module13/problems13.htm