1820 Hans Christian Oersted - Physics and Astronomy...
Transcript of 1820 Hans Christian Oersted - Physics and Astronomy...
Magnetic Field Produced by a Moving Charge
03
( )4
q v rBr
μπ
×=
r rr
Most convenient view for determining the direction of the magnetic field
produced by a moving charge.
Magnetic Field Produced by a Moving Charge
03
( )4
q v rBr
μπ
×=
r rr
Most convenient view for determining the direction of the magnetic field
produced by a moving charge.
Battery A
Units
2
0 0
1cμ ε
=
70
7 2 2
7 2
4 10 /
4 10 /4 10 /
T m A
N s CN A
μ π
π
π
−
−
−
= × ⋅
= × ⋅
= ×
7 2 2 2
0
1 (10 / )4
N s C cπε
−= ⋅
permeability of the vacuum
sec)//(][ meterCoulombNewtonsB ⋅=
mANsmCNteslaT ⋅=⋅= /1//1)(1
03
( )4
q v rBr
μπ
×=
r rr
BvqFrrr
×=
Forces Produced by Moving Charges
Force generated by magnetic interaction for oppositely moving equal-sign charges has repulsive sign.
22
0 0 2B
E
F vvF c
μ ε= =
Congratulation: we arrive to the idea of relativity!
Biot-Savart Law
Infinitesimally small element of a current carrying wire produces an infinitesimally small magnetic field
0μ is called permeability of free space
0 03 3
( ) ( )4 4
q v r I d l rB dBr r
μ μπ π
× ×= ⇒ =
r rr rr r
dQ nqSdl vdQ vnqSdl Idl= ⇒ = =r r uur
70
7 2 2
7 2
4 10 /
4 10 /4 10 /
T m A
N s CN A
μ π
π
π
−
−
−
= × ⋅
= × ⋅
= ×
once again vector’s direction is delegated from v to dl
( )
( )
03
03/ 20 2 2
/0
3/ 20 2
max 2 2
02 2
0max
( )4
24
24 1
/ sin
24
1 tan2
a
z
a x
z
I d l rdBr
I xB dyx y
I dzx z
az y x tgx a
I aBx x a
IR
μπμπ
μπ
α α
μπ
μ απ
×=
= − =+
= −+
= = =+
= −+
= −
∫
∫
r rr
Magnetic Field Produced by a Current
y x z× = −
( )
( )
2 2 3/ 20
3/ 20 2 20
/
3/ 20 20
max 2 2
2 2 2 20 0
4 ( )2
4
24 1
sin
1 2 14 4
x
a
x
a x
x
xdyd Ex y
xE dyx y
dzz
az tgx a
a QEx x a x x a
λπελπε
λπε
α α
λπε πε
=+
= =+
=+
= =+
= =+ +
∫
∫
ur
Electric Field Produced by a Line of Charge, Chapter 21,Example 21.11
Magnetic Field Produced by a Linear Current:again Flatlandia
0| |2
IBr
μπ
=02rEr
λπε
=
E versus Bsimilarity: axial symmetry in both casesdifferences: 1) E is radial; B is circular2) instead of Gauss’s law -- Ampere’s law
Flatlandia: when dependence on one of the coordinates drops out; cylinders or a world confined/reduced to a plane
Magnetic Field Produced by a Straight Current Carrying Wire:
0| |2
IBr
μπ
=
70
7 2 2
7 2
4 10 /
4 10 /4 10 /
T m A
N s CN A
μ π
π
π
−
−
−
= × ⋅
= × ⋅
= ×
70 ( ) 4 10| |
2 ( ) 2I I Amper T mBr r meter Amper
μ ππ π
−× ⋅= = ×
Magnetic Field of Two Parallel Wires with Oppositely Directed Currents
Principle of superposition:Magnetic field created by few current-carrying wires should be summed according to the vector summation rules.
Forces between Two Parallel Wires with Oppositely Directed Currents :
Force per unit length:
0 1 2
2F I IL r
μπ
=
Currents are not like charges:Oppositely directed currents repel each other!Same directed currents attract each other.
7 22 74 10 // (1 ) 2 10 /
2 1N AF L A N mm
ππ
−−×
= = ××
Example 28.5
Forces between Two Parallel Wires
0 1 2
7 22
3
4
2
4.5 15,0004 10 // (15,000 )2 4.5 10
10 /
F I IL r
r mm I AN AF L A
mN m
μπ
ππ
−
−
=
= =
×=
× ×=
Pulse currents break magnets!
Electromagnets (most of magnets used in industry are the electromagnets)
An electromagnet contains a current carrying wire with numerous turns
Magnetic Field along the Axis of a Current Loop
0 03 2 2 2 2 1/ 2
20
2 2 3/ 2
2 27
2 2 3/ 2 3
( )4 4 ( ) ( )
2 ( )( )2 10 ( ) /
( ) ( )
xI d l r I d l adB dB
r x a x aIaB
x aIa mampers T m amper
x a m
μ μπ π
μ
π −
×= ⇒ =
+ +
= =+
= × ⋅+
r rrr
Charge Q
Current I
Electro-magnet: Magnetic Field created by an N-loops coil
20
2 2 3/ 2
0 magnet02 2 3/ 2 2 2 3/ 2
2 ( )
2 ( ) 2 ( )
NIaBx aINS
x a x a
μ
μ μμπ π
= =+
= =+ +
magnetitem item magnet
potential energy ( )
cos
magnetic
U B Bμ μ φ= − ⋅ = −ur ur
magnetic field created by the current magnetizes iron items
Gauss’s law for Magnetism
magnetic charges do not exist
magnetic lines have no ends
∫ =⋅S
enclosedQSdE0ε
rr
0∫ =⋅ SdBrr
Some magnetic field lines remind electric dipole. Where is the difference?.
Do we have something instead of the Gauss’s law? Ampere’s law
∫contour
0B dl Iμ=ur r
orientation of the contour !
Caution: Sign of the currents enclosed by the contour are determined by the orientation of the contour.
For this orientation of the contour (anticlockwise), currents I1 and I3 are positive while I3 is negative.
Contour may be not in plain, and currents may be not perpendicular to the contour plane.
Ampere’s law (Example 28.8)
Most effective when it is applied for symmetrical configurations:
cylindrical conductor ∫contour
0B dl Iμ=ur r
Ampere’s law (Example 28.9)
Most effective when it is applied for symmetrical configurations:
solenoid
Find similarity between solenoid and capacitor
∫contour
0B dl Iμ=ur r
∫contour
0
0
B dl BL nLI
B nI
μ
μ
= =
=
ur r
Ampere’s law (Example 28.10)
Most effective when it is applied for symmetrical configurations:
Toroidal solenoid∫
contour0B dl Iμ=
ur r
∫contour
0
0
2
2
B dl rB NI
NIBr
π μ
μπ
= =
=
ur r
What information (may be questions for the Exam) produces this picture?
Magnetic field produced by a clock-like current in the coils (look on the picture from the left side) is directed from left to right. A positive charge responding to the magnetic field rotates counter-clock wise ( i.e., opposite to the current which induces this magnetic field). Magnetic field produced by the rotating charge tries to compensate the original magnetic field.
Current-loops with oppositely directed, i.e., anti-parallel, currents repulse. (Current loops with parallel currents attract.) This is why rotating charge repels from coil 1 and goes toward the middle of the magnetic bottle.Check that you fully understand all these statements.