Asked by s

A spherical shell of radius R carries a uniform surface charge density (charge per unit area) σ. The center of the sphere is at the origin and the shell rotates with angular velocity ω (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 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
Answered by Anonymous
a) 0
b) 0
Answered by Anonymous
a) I = 52.78 e-3

b) B = 5.53 e-8 T
Answered by FLu
Anonymous, can you provide formula because of different values please?
Answered by Mag
a) Calculate the magnitude of the total current (in A) carried by the rotating sphere for the following values of ƒÐ, ƒÖ and R:
ƒÐ= 6 ~10−4C m−2, ƒÖ= 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 ƒÐ, ƒÖ , z and R:
ƒÐ= 6 ~10−4C m−2, ƒÖ= 5 rad⋅s−1, z= 1.92 m and R=1m



Answered by Mag
(a) Calculate the magnitude of the total current (in A) carried by the rotating sphere for the following values of ó, ù 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 ó, ù , z and R:
sigma= 6x10−4C m−2, w= 7 rad⋅s−1, z= 1.6 m and R=1m

Answered by Mag
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 ó, ù , z and R:
sigma= 6x10−4C m−2, w= 5 rad⋅s−1, z= 1.6 m and R=1m


Answered by Mag
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 ó, ù , z and R:
sigma= 6x10−4C m−2, w= 5 rad⋅s−1, z= 1.92 m and R=1m


Answered by Phy
a) (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
Answered by Anonymous
a) I = Sigma*A*omega
b) B = mu*I/(2*(z-R))
Answered by Phy
How to calculate 'A'???
Answered by Rubens
@Phy
with spherical surface
Answered by Phy
That is a wrong formula, i m getting my answers wrong
Answered by FLu
Have the same problem, how to calculate for A?
Answered by Mag
Me too please help!

(a) Calculate the magnitude of the total current (in A) carried by the rotating sphere for the following values of ó, ù 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 ó, ù , z and R:
sigma= 6x10−4C m−2, w= 5 rad⋅s−1, z= 1.92 m and R=1m


Answered by Phy
A=4*pi*r*r
Answered by FLu
Thanks 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


Answered by Phy
Yeah even i got it wrong, the formula itself is wrong
Answered by FLu
Oh, I see.
Anonymous or someone else, could you maybe provide an example with real numbers? That would also help others.
Answered by Hur
Yes, please if possible provide answer with one example,thankyou!
Answered by Gise
Rubens can show example with real number please?
Answered by FLu
Phy, 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.
Answered by Linda
Pls Flu post 1 and 2.
Answered by FLu
Problem 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?
Answered by Mag
Thanks FLu, I got it now.
Please someone for the other problems?
Answered by Glin
Me too thanks FLu!

Please explanation for other problems?
Answered by FLu
Anyone?
Answered by Sag
Has anybody got the explanation for other problems please?
Answered by Nusa
Anybody!
Answered by Anonymous
a) I = Sigma*A*f
f = omega/(2pi)

b) B = mu*I/(2*(z-R))
Answered by ANONYMOUS
Value of mu? Please
Answered by Anonymous
The other problems?

Bainbridge mss spectrometer

Magnetic Field of a current-carrying ribbon

Second Rl circuit

magnetic Field of a loop - Bz.
Answered by FLu


Is 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?
Answered by FLu
Anonymous 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?
Answered by FLu
Anonymous, 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
Answered by ANONYMOUS!!
I 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
Answered by FLu
oh 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!
Answered by Anon
You should be using mu/4pi for answer in b
Answered by FLu
you 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


Answered by Anon
yes it is good
Answered by Naseng
This 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))


Answered by Anon
you are missing the current I in b
Answered by Naseng
I is the result from problem a) right?

Could you give me the complete formula or correct the above please Anon?
Answered by hmmmm
anyone for part b please,clearly
Answered by Anon
correct
Answered by Naseng
before I forget mu is the value this:
1.25663706*10^-6 ?
Answered by Anon
yes but you need mu/4pi
Answered by FLu
Anon, 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?
Answered by Anonymous
mu ia mu_0
Answered by FLu
Anon, 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!


Answered by Anon
mu = 4pi * 10^(-7)
Answered by hmmmm
i used this getting wrong answer
Answered by FLu
Thanks 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?
Answered by hmmmm
did u get it right FLU?
Answered by FLu
hmmmm 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.
Answered by FLu
No, 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
Answered by FLu
Anon could you write the formula for b) please with the above numbers? would be very helpful.
Answered by Anon
(I*10^(-7))/(2*(z-R))
Answered by FLu
Anon thanks, I thought it is
(I*4*pi*10^(-7))/(2*(z-R))
did you not say mu is=
4pi * 10^(-7) ?




Answered by X10
@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...:(
Answered by Anon
yes i did and i also said you need to use
mu/4pi
Answered by X10
@FLu let me know if u got the correct answer with this formula and what it is ?
Answered by Anon
you need to make sure you use the right value for I
Answered by ind0
anyone has answer for other problems
Answered by X10
Yup 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
Answered by ind0
The other problems?

Bainbridge mss spectrometer

Magnetic Field of a current-carrying ribbon

Second Rl circuit

magnetic Field of a loop - Bz.
Answered by X10
I've solved all problems just got stuck with this one...
Answered by Any
Consider 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
Answered by ind0
can you share formulas or aproach
Answered by X10
Q:3
for part a: B=E/v
for part b: (L^2+h^2)/2L
for part c: m=rqB/v
Answered by X10
Q6: ((mu_0*I)/2*pi*w)*ln(1+(w/x)) .... (I = J*w*s) and convert cm to m.
Answered by Any
In 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
Answered by ind0
X10
Q3 part c and Q6 are wrong could check formulas
Answered by X10
for 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
Answered by ind0
Q6: ((mu_0*I)/2*pi*w)*ln(1+(w/x))

do u divide by 2 * pi * w
Answered by ind0
Thanks
what the formula for Bz in Q5
Answered by Anon
X10 are done now
Answered by FLu
X10 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.
Answered by FLu

In the circuit shown in the figure below, the switch closes at t=0, R=5.5 Ohm, ¦Å=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¡Ö¡Þ?
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
Answered by FLu
X10 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!


Answered by DOnny
X10 do you know how to solve for the magnetic field of a loop? i've been cracking my head
Answered by Mag
Have 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
Answered by Saga
Yes, same here, is there an issue with converting it from cm to m maybe? can somebody help where the mistake lie. X10 maybe. thanks!
Answered by DOnny
it 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
Answered by DOnny
now 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
Answered by Dr. Lewin
You don't have any HONOR !!
Answered by FLu
Thanks 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.
Answered by DOnny
Flu our values may vary so i am not certain. but the formulas are fine

if you could assist me on Q5 Flu ?
Answered by Saga
DOnny, I am getting 3.91822*10^-12 and that is marked wrong, could you tell me the full value? 6.++e-7
please
Answered by FLu
DOnny 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.
Answered by Saga
Yes, 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.
Answered by DOnny
I 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.
Answered by FLu
Sorry 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.
Answered by DOnny
sorry its supposed to be 6.++e-11

if you have your fx-570MS calculator you should be able to find mu_0 there. don't bother expanding the values
Answered by DOnny
((1.256637061e-6)*(1*0.08*0.001))/(2*pi*0.08)*(ln(1+(0.08/0.22))
Answered by FLu
Unfortunaley, 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!

Answered by DOnny
the 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
Answered by FLu
Thanks will try out!
Answered by Niznkl
@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..
Answered by FLu
DOnny 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!
Answered by FLu
Niznkl, thanks I got it now with DOnny's help. Which b part do you mean?
Answered by Niznkl
the problem of the sphere .. the magnetic field at the point p .. is all that I need to finish... we got the same values
Answered by FLu
Niznkl, 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


Answered by FLu
OH 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.
Answered by DOnny
magnetic loop is the only one i can't solve
Answered by Anonymous
B=[(2*mu_0)/(3*z^3)]*(R^4)*sigma*omega
Answered by Niznkl
Magnetic 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)
Answered by Spartacus
@Niznkl can you give us the formula that you use to get that result for Bz?
Answered by John
I need the formulas for Q5 , Q6 and Q7
Answered by Anonymous
A spherical shell of radius R carries a uniform surface charge density (charge per unit area) ƒÐ. The center of the sphere is at the origin and the shell rotates with angular velocity ƒÖ (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 ƒÐ, ƒÖ and R:
ƒÐ= 6 ~10−4C m−2, ƒÖ= 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 ƒÐ, ƒÖ , z and R:
ƒÐ= 6 ~10−4C m−2, ƒÖ= 5 rad⋅s−1, z= 1.9 m and R=1m

Please answer
Answered by Dr. Slump
Honor code????
Answered by FLu
Niznkl, 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?
Answered by Nitra
Please Niznkl, provide step by step formula with the value so we can calculate with different values! For magnetic field of a loop
Answered by My
Please formula for magnetic field of a loop last part Bz?
Answered by FLu
Correction it is I=0.85 and a=6cm
Answered by Saga
Magnetic field of a loop formula for Bz please?
Answered by Ortum
Yes, how to calculate Bz for different values please?
Answered by X10
Q5:
mu_0*i/4r where m_0= 4*pi*10^-7 and r=a please convert cm to m
Answered by X10
Thankyou 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
Answered by zerocool
honour code breached
Answered by Ortum
X10 and Anonymous thanks. Could you tell what are the values of z, R, sigma and omega?
I cannot identify them.
Answered by Saga
Thanks 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?
Answered by My
X10 please formula for second RL circuit!
And the values for omega, sigma, z and R?
Answered by FLu
Guys, 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?
Answered by Ortum
Oh thanks Flu!
anyone formula for second RL circuit question and formula for Bz magnetic field loop?
Answered by X10
For Bz quest:5
0
0
mu_0*i/4r where m_0= 4*pi*10^-7 and r=a please convert cm to m
Answered by FLu
Great 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.
Answered by Ortum
Thanks X10, Anyone for second RL circuit? formula to calculate
Answered by X10
I'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..
Answered by Anonymous
Q3 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).

Answered by Spartacus
Someone get the last question right using this formula: [(2*mu_0)/(3*z^3)]*(R^4)*sigma*omega ??
Answered by FLu
Anonymous, 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
Answered by Anonymous
Thanks, FLu!
Answered by Dope Feeder
//electron9.phys.utk.edu/phys513/Modules/module13/problems13.htm
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