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SP2015 Exam 2 grade distribution (regrades not included)
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Homework note: use the Biot-Savart Law if you must calculate the magnetic field of a ring or coil. Don’t start with a “random” equation from the text or lecture notes. Example: Special Homework #7. For the loop, start with the Biot-Savart Law.
Special Homework #7 is due tomorrow. Download it from the Physics 24 web site if lost it.
Homework Hints
ˆμ
4π
02
I d rdB=
r
The Biot-Savart law gives the magnetic field due to an infinitesimal current-carrying wire:
Integrate over the entire wire to get the total magnetic field. If the wire is long and straight:
μ
π0 IB=
2 r
If you want to calculate the field of a tiny segment of a wire (“tiny”= wire segment length is much less than other lengths/distances in the problem) you can use the Biot-Savart law directly. No need to integrate.
Today’s agenda:
Magnetic Field Due To A Current Loop.You must be able to apply the Biot-Savart Law to calculate the magnetic field of a current loop.
Ampere’s Law.You must be able to use Ampere’s Law to calculate the magnetic field for high-symmetry current configurations.
Solenoids.You must be able to use Ampere’s Law to calculate the magnetic field of solenoids and toroids. You must be able to use the magnetic field equations derived with Ampere’s Law to make numerical magnetic field calculations for solenoids and toroids.
Magnetic Field of a Current Loop
P
a
xx
A circular ring of radius a carries a current I as shown. Calculate the magnetic field at a point P along the axis of the ring at a distance x from its center.
I
y
z
Draw a figure. Write down the starting equation. It tells you what to do next.
ˆμ
4π
02
I d rdB=
rDraw a figure. Write down the starting equation. It tells you what to do next.
Magnetic Field of a Current Loop
P
a
x
dB
x
A circular ring of radius a carries a current I as shown. Calculate the magnetic field at a point P along the axis of the ring at a distance x from its center.
dl r̂r
90-
I
y
dBy
dBx
90-
z
Complicated diagram! You are supposed to visualize the ring lying in the yz plane.dl is in the yz plane. r is in the xy plane and is perpendicular to dl.* Thus
r̂
ˆ . d r =d
Also, dB must lie in the xy plane* (perpendicular to dl) and is perpendicular to r.
*Only when dl is centered on the y-axis!
ˆμ
4π
02
I d rdB=
r
P
a
x
dB
x
dl r̂r
90-
I
y
dBy
dBx
90-
z
ˆμ
4π
02
I d rdB=
r
μ
4π
02
I ddB=
r
μ
4π 0
2 2
I ddB=
x a
μ μ
4π 4π
0 0x 1/22 2 2 2 2 2
I d I d adB = cos =
x a x a x a
μ μ
4π 4π
0 0y 1/22 2 2 2 2 2
I d I d xdB = sin =
x a x a x a
By symmetry, By will be 0. Do you see why? Use symmetry to find By. Don’t try to integrate dBy to get By. See here for the reason.
PxdBz x
dl
r̂r
I
y
dBy
dBxz
When dl is not centered at z=0, there will be a z-component to the magnetic field, but by symmetry Bz will be zero.
x xring
B = dB
μ
4π
0x 3/22 2
I a ddB =
x a
I, x, and a are constant as you integrate around the ring!
P
a
x
dB
x
dl r̂r
90-
I
y
dBy
dBx
90-
z
μ μ
4π 4π
0 0
x 3/2 3/22 2 2 2ring
I a I a B = d = 2 a
x a x a
μ
20
x 3/22 2
I a B =
2 x aThis is not on your starting equation sheet. I will add it if we have homework or a test problem where I judge that you need it.
At the center of the ring, x=0.
P
a
x
dB
x
dl r̂r
90-
I
y
dBy
dBx
90-
z
μ 2
0x,center 3/22
I a B =
2 a
This is not on your starting equation sheet, but I will add it if we have homework or a test problem where I judge that you need it. For homework, if a problem requires this equation, you need to derive it!
μ μ20 0
x,center 3
I a IB = =
2a 2a
For N tightly packed concentric rings (a tight coil)…
μ0x,center
N IB =
2a
Magnetic Field at the center of a Current Loop
A circular ring of radius a lies in the xy plane and carries a current I as shown. Calculate the magnetic field at the center of the loop.
a
z
dlI
x
y
One-page derivation, so you won’t have to go through the general derivation (I will work it at the blackboard)!
The direction of the magnetic field will be different if the plane of the loop is not in the xy plane.
Homework hint: use this derivation in Special Homework
#7.
I
B
r
Ampere’s Law
Just for kicks, let’s evaluate the line integral along the direction of B over a closed circular path around a current-carrying wire.
ds π
B ds=B ds=B 2 r
B ds
μπ μ
π
0
0
IB ds= 2 r = I
2 r
The above calculation is only for the special case of a long straight wire, but you can show that the result is valid in general.
μ
0 B ds= I Ampere’s Law
I is the total current that passes through a surface bounded by the closed (and not necessarily circular) path of integration.Ampere’s Law is useful for calculating the magnetic field due to current configurations that have high symmetry.
I
B
r
ds
The current I passing through a loop is positive if the direction of integration is the same as the direction of B from the right hand rule.
I
B
r
dspositive I negative I
μ
0 enclB ds= IYour text writes
Your starting equation sheet has
I1
ds
If your path includes more than one source of current, add all the currents (with correct sign).
I2
because the current that you use is the current “enclosed” by the closed path over which you integrate.
μ
0 1 2B ds= I - I
μ
E0 encl
dB ds= I
dtThe reason for the 2nd term on the right will become apparent later. Set it equal to
zero for now.
0
Example: a cylindrical wire of radius R carries a current I that is uniformly distributed over the wire’s cross section. Calculate the magnetic field inside and outside the wire.
I
R
Cross-section of the wire:
Rr
direction of I
B
μ
0 enclB ds= I
Choose a path that “matches” the symmetry of the magnetic field (so that the dot product and integral are easy to evaluate); in this case, the field is tangent to the path.
μ μ
0 encl 0
A enclosed by rB ds= I = I
A enclosed by R
πμ μ μ
π
2
0 encl 0 0 2
rA enclosed by rB ds= I = I = I
A enclosed by R R
πμ μ μ μ
π
2 2
0 encl 0 0 0 22
rA enclosed by r rB ds= I = I = I = I
A enclosed by R RR
Rr
direction of I
B
Over the closed circular path r:
π
B ds=B ds=2 rB
Combine results and solve for B:
π μ2
0 2
r2 rB= I
R
μμ μ
π π π
20
0 0 2 2 2
Ir rB= I = I = r
2 rR 2 R 2 R
B is linear in r.
R
r
direction of I
B
Outside the wire:
πr μ
0 B ds=B ds=2 B= I
μ
π0 IB=
2 r
(as expected).
B
rR
Plot:
A lot easier than using the Biot-Savart Law!
Calculating Electric and Magnetic Fields
Electric Field
in general: Coulomb’s Law
for high symmetry configurations: Gauss’ Law
Magnetic Field
in general: Biot-Savart Law
for high symmetry configurations: Ampere’s
Law
This analogy is rather flawed because Ampere’s Law is not really the “Gauss’ Law of magnetism.”
Today’s agenda:
Magnetic Field Due To A Current Loop.You must be able to apply the Biot-Savart Law to calculate the magnetic field of a current loop.
Ampere’s Law.You must be able to use Ampere’s Law to calculate the magnetic field for high-symmetry current configurations.
Solenoids.You must be able to use Ampere’s Law to calculate the magnetic field of solenoids and toroids. You must be able to use the magnetic field equations derived with Ampere’s Law to make numerical magnetic field calculations for solenoids and toroids.
Magnetic Field of a Solenoid
A solenoid is made of many loops of wire, packed closely* together. Here’s the magnetic field from a loop of wire:
Some images in this section are from hyperphysics.
*But not so closely that you can useμ0 N I
B=2a
Stack many loops to make a solenoid:
Ought to remind you of the magnetic field of a bar magnet.
You can use Ampere’s law to calculate the magnetic field of a solenoid.
B
I
1 2 3 4
B ds= B ds B ds B ds B ds
l
0 0 0 μ
0 enclosedB ds= B = I
μ 0 B = N I N is the number of loops enclosed by our surface.
B
I
l
μ0
NB= I Magnetic field of a solenoid
of length l , N loops, current I. n=N/l (number of turns per unit length).μ0 B= n I
The magnetic field inside a long solenoid does not depend on the position inside the solenoid (if end effects are neglected).
A toroid* is just a solenoid “hooked up” to itself.
π
B ds=B ds=B 2 r
μ μ
0 enclosed 0 B ds= I = N I
π μ0 B 2 r = N I
μ
π0 N I
B= 2 r
Magnetic field inside a toroid of N loops, current I.
The magnetic field inside a toroid is not subject to end effects, but is not constant inside (because it depends on r).
*Your text calls this a “toroidal solenoid.”
B
0
NB = μ I
π
-7 400T mB = 4 ×10 2 A
A 0.1 m
B = 0.01 T
Example: a thin 10-cm long solenoid has a total of 400 turns of wire and carries a current of 2 A. Calculate the magnetic field inside near the center.
“Help! Too many similar starting equations!”
μ
π0 IB=
2 rlong straight wire
μ0 N IB=
2acenter of N loops of radius a
μ0
NB= I
μ0 B= n I
solenoid, length l, N turns
solenoid, n turns per unit length
μ
π0 N I
B= 2 r
toroid, N loops
“How am I going to know which is which on the next exam?”
use Ampere’s law (or note the lack of N)
probably not a starting equation
field inside a solenoid is constant
field inside a solenoid is constant
field inside a toroid depends on position (r)
