Homework Statement
A sphere magnet of radius a has permanent uniform magnetization, z-axis. So we were trying to find the magnetic field at the center of the magnet in class.
Homework Equations
j_M(θ)=Msin(θ)
dI=j_M(θ)adθ
circular loop
r_{loop}=asin(θ)The Attempt at a Solution
When using...
.._
./..\ ....| R=radius
|→ | → →→ I ...|
.\_/
dI=j(r)dr? j(r) would be in current density per unit-length, or J(r) like the question
Sorry I attempted to draw a circle with current going perpendicular to R
The direction would be counterclockwise
Okay, how would that affect my calculation?
I'm sorry if I confused you by writing big J instead of small j. Here J is not current per unit area
Homework Statement
J=J(r)(θhat), J(r)=(3ITr2)/R3
a)total current of the disk?
b)J(r) was brought by a charge disk spinning at ang vel. ω, find the surface charge dens σ(r)?
Homework Equations
J=I/πr2
I=∫Jda
J=vσ
The Attempt at a Solution
a)I assumed I would have to integrate J(r)da from 0 to...
Homework Statement
I have the equation
Z=1/N!h3N∫∫d3qid3pie-βH(q,p)
How can I get the entropy from this equation assuming a classical gas of N identical, noninteracting atoms inside a volume V in equilibrium at T where it has an internal degree of freedom with energies 0 and ε
What...
Homework Statement
If you have atoms that are normally located at the normal lattice positions or at an interstitial position where energy >ε, how can I find the probability that n interstitial positions are occupied by atoms?
If we were to use large N how can I find the fraction of...
Part a) I got it
Since the sum of both charges are 0, so
0=Qa/4πa+Qb/4πb
Solving for Qa I get -Qb(a/b)
Part b and c I don't get it
If I integrate from b to r ∫Q/4πr^2 + from r to a ∫Q/4πr^2= -V(x)
I don't get the same results
even if If integrate from ∞ to b ∫Q/4πr^2 + from...
Yes I can
Since we have the boundary conditions
V = Va at x = a
V = Vb at x = b
Va=Ca+D
Vb=Cb+D
Set D=Va-Ca
Solve for C on Vb
Then solve for D. Then just plug in Vx
Thanks a lot!
Sorry, I totally forgot that is a charge free slab. Therefore I'm still stuck on the boundary conditions
How come your results is equal to the answer? Va is not part of it
So if at we have Q+ at a and Q- at b, from a to b Vb=0 and at a the potential will be Va or we have a distance of d from a to b, Va=Qd/Aε0.
So V(x)=C(x)+D
at V(0)=0+D
D=Vb
but you got D as Va?
How did start solving for C? What boundary conditions?
I got it. Since Sn=X1+X2+X3, all I have to do is to calculate the mean from Sn and the variance from Sn
Sn=1/4(P{A})+1/4+(P{G})...
Sn=(1/4)^2...
than just plug in the mean and variance into the probability density function