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Magnetic Field due to an Asymmetric Hollow Cylinder - Assignment | PHYS 212, Assignments of Physics

Material Type: Assignment; Class: University Physics: Elec & Mag; Subject: Physics; University: University of Illinois - Urbana-Champaign; Term: Fall 2008;

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Discussion Question 9B
P212, Week 9
Magnetic Field due to an Asymmetric Hollow Cylinder
A conducting cylinder is oriented parallel to the z axis and carries a uniform current I in the
negative z direction (into the page). This cylinder is hollow, however, with a cylindrical bore
centered on the point Q shown in the figure. The radius of this bore is R1, while the outer radius
of the cylinder is R2.
(a) Calculate the magnetic field at the point P on the y axis.
(i) This appears to be a horrendous problem, with no symmetry! But really, the hollow
cylinder is a superposition of two solid cylinders ... and a solid cylinder of current is
something we can deal with. How could you treat this system so that it consists of two solid
cylinders?
Two separate cylinders, one w/ I into the page that has radius R2, and one centered at Q, w/
radius R1 and current I out of the page
(ii) Calculate the field at point P due to each of the two cylinders, then add the two
contributions together. Remember to work by components since you are adding field
vectors.
x
y
P
Q
R1
R2
I = 2.5 A into page
R1 = 5 cm
R2 = 12 cm
point P: y = 15 cm
point Q: y = -3 cm
2
22
21
2
outer 2
The current density is 66 87 A/m We think of this as outer
area
cylinder with current I 3 025 amps into the plane and an inner cylinder
with 2 5 3 025 0 5252
inner outer
II
J..
RR)
RJ .
II-I .. .
π
π
== =
(−
==
===
()()( )()( )
or +0 5252 out of page .
We now use expression for infinite wire:
2 7 2e-7 3 025 2e-7 0 5252
B= 3 450
0.15 0.15+0.03
.
e- I . .
ˆˆ ˆˆ
xx x.xT
r
μ
=− =
G
pf2
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Discussion Question 9B P212, Week 9 Magnetic Field due to an Asymmetric Hollow Cylinder

A conducting cylinder is oriented parallel to the z axis and carries a uniform current I in the negative z direction (into the page). This cylinder is hollow, however, with a cylindrical bore centered on the point Q shown in the figure. The radius of this bore is R 1 , while the outer radius of the cylinder is R 2.

(a) Calculate the magnetic field at the point P on the y axis.

(i) This appears to be a horrendous problem, with no symmetry! But really, the hollow cylinder is a superposition of two solid cylinders ... and a solid cylinder of current is something we can deal with. How could you treat this system so that it consists of two solid cylinders? Two separate cylinders, one w/ I into the page that has radius R2, and one centered at Q, w/ radius R 1 and current I out of the page (ii) Calculate the field at point P due to each of the two cylinders, then add the two contributions together. Remember to work by components since you are adding field vectors.

x

y

P

R 1 Q

R 2

I = 2.5 A into page

R 1 = 5 cm

R 2 = 12 cm

point P : y = 15 cm

point Q : y = -3 cm

2 2 2 2 1 2 outer 2

The current density is 66 87 A/m We think of this as outer area

cylinder with current I 3 025 amps into the plane and an inner cylinder

with (^) inner outer 2 5 3 025 0 5252

I I

J..

R R )

R J.

I I - I...

( ) ( )( ) ( )( )

or +0 5252 out of page. We now use expression for infinite wire: 2 7 2e-7 3 025 2e-7 0 5252 B= 3 450 0.15 0.15+0.

e - I.. ˆx ˆx ˆx. ˆx T r ∑ =^ −^ =^ μ

G

(b) Calculate the magnetic field at the point Q at the center of the hollow bore.

Again, compute the field due to each of the two cylinders separately, then add the fields together. This time, one of the contributions will be very easy .... but one of them requires a little more thought. What is the enclosed current that goes into Ampere's Law?

x

y

P

R 1 Q

R 2

I = 2.5 A into page

R 1 = 5 cm

R 2 = 12 cm

point P : y = 15 cm

point Q : y = -3 cm

2

The inner cylinder contributes no B. The current from the outer cylinder is 0 03 0 1891 A. We next use 2 7 2 7 0 1891 1 26 0 03

I (^) outer J.. e - I e -. B ˆx^ ˆx^. ˆx r.

G