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‫א‪ .‬נקבע ציר ‪ x‬כך שראשית הצירים ממוקמת על התיל השמאלי‪ .‬השדה המגנטי בין התילים הוא‪:‬‬
‫‪μ0 I‬‬
‫‪μ0 I‬‬
‫‪μ I ⎛1‬‬
‫⎞ ‪1‬‬
‫⎟‬
‫‪−‬‬
‫‪= 0 ⎜⎜ −‬‬
‫⎠⎟ )‪2πx 2π (d − x) 2π ⎝ x (d − x‬‬
‫=‪B‬‬
‫‪d‬‬
‫שימו לב שבתחום‬
‫‪2‬‬
‫‪d‬‬
‫ובתחום ‪ < x < d‬כיוון השדה הוא יוצא מהדף ) • (‪.‬‬
‫‪2‬‬
‫‪I‬‬
‫• • • • •‬
‫• • • • •‬
‫• × × × × ×‬
‫• × × × × ×‬
‫‪2R‬‬
‫‪R‬‬
‫‪v‬‬
‫< ‪ 0 < x‬כיוון השדה הוא לתוך הדף ) × (‬
‫‪I‬‬
‫‪a‬‬
‫‪a‬‬
‫‪x‬‬
‫נחשב את השטף המגנטי עבור העניבה השמאלית )ניתן לחשב‬
‫גם עבור העניבה הימנית והכא"מ יהיה זהה(‪:‬‬
‫‪r r x L μ0 I ⎛ 1‬‬
‫⎞ ‪μ IL ⎛ ⎛ x‬‬
‫⎞ ‪1‬‬
‫⎞⎞ ‪⎛ d − x‬‬
‫⎜‪⎟⎟ dy dx = 0 ⎜⎜ ln⎜ ⎟ + ln‬‬
‫‪⎜⎜ −‬‬
‫∫ ∫ = ‪φ = ∫ B ⋅ ds‬‬
‫⎟⎟ ⎟‬
‫⎠ )‪2π ⎝ x (d − x‬‬
‫⎠ ‪2π ⎝ ⎝ a‬‬
‫⎠⎠ ‪⎝ d − a‬‬
‫‪a 0‬‬
‫כעת ניתן למצוא את הכא"מ‪:‬‬
‫‪dφ μ 0 ILv ⎛ 1‬‬
‫⎞‪1‬‬
‫⎜⎜‬
‫=‬
‫⎟⎟ ‪−‬‬
‫‪dt‬‬
‫⎠ ‪2π ⎝ (d − x) x‬‬
‫‪ε =−‬‬
‫ב‪ .‬שני הנגדים מחוברים במקביל‪ ,‬לכן ההתנגדות השקולה היא‪:‬‬
‫‪2‬‬
‫‪R‬‬
‫‪3‬‬
‫= ‪RT‬‬
‫⇒‬
‫‪1‬‬
‫‪1‬‬
‫‪1‬‬
‫‪= +‬‬
‫‪RT R 2 R‬‬
‫לכן הזרם על המוט הוא‪:‬‬
‫‪3μ 0 ILv ⎛ 1‬‬
‫⎞‪1‬‬
‫⎜⎜‬
‫⎟⎟ ‪−‬‬
‫⎠ ‪4πR ⎝ (d − x) x‬‬
‫‪d‬‬
‫‪d‬‬
‫כיוון הזרם על המוט בתחום < ‪ 0 < x‬הוא כלפי מעלה ובתחום ‪< x < d‬‬
‫‪2‬‬
‫‪2‬‬
‫=‬
‫‪ε‬‬
‫‪RT‬‬
‫= ‪IT‬‬
‫הזרם מהפך וכיוונו כלפי מטה‪.‬‬
‫ג‪ .‬הזרם על הנגדים‪:‬‬
‫‪μ 0 ILv ⎛ 1‬‬
‫⎞‪1‬‬
‫⎜⎜‬
‫⎟⎟ ‪−‬‬
‫‪R‬‬
‫⎠ ‪2πR ⎝ (d − x) x‬‬
‫‪μ ILv ⎛ 1‬‬
‫⎞‪1‬‬
‫‪ε‬‬
‫⎜⎜‬
‫= ‪I2‬‬
‫‪= 0‬‬
‫⎟⎟ ‪−‬‬
‫‪2R‬‬
‫⎠ ‪4πR ⎝ (d − x) x‬‬
‫=‬
‫כיווני הזרמים מתהפכים במהלך תנועת המוט בהתאם לכיוון הזרם על המוט‪.‬‬
‫ד‪ .‬הכח החיצוני הדרוש למשיכת המוט שווה לכח המגנטי הפועל על המוט וכיוונו הפוך‪:‬‬
‫‪ε‬‬
‫= ‪I1‬‬
‫‪2‬‬
‫כיוון הכח המגנטי הוא שמאלה לכן כיוון הכח החיצוני הוא ימינה‪.‬‬
‫שימו לב שכיוון הכח לא משתנה במהלך כל תנועת המוט!!!‬
‫‪3μ 2 I 2 L2 v ⎛ 1‬‬
‫⎞‪1‬‬
‫⎜⎜‬
‫= ‪F = I T LB‬‬
‫‪= 0 2‬‬
‫⎟⎟ ‪−‬‬
‫‪RT‬‬
‫⎠ ‪8π R ⎝ (d − x) x‬‬
‫‪ε‬‬
Page 1 of 1
Induction
Let
by
, where is the coordinate of the lower side. During time
the frame moves downward
. Then
. The induced emf is
. The induced current is
. The force which acts on the frame (only
lower and upper sides contribute) is
.
The equation of motion reads
and the solution is
where
and
.
file://C:\Documents and Settings\mario\Application Data\SSH\temp\e_48_1_018_s.ht... 18/05/2010
Page 1 of 1
Induction
The flux through the small circle is
, so that
. The magnetic field produced by the circle
on the axis is (CALCULATE !):
Thus,
where
. Maximum is at
(CHECK !).
file://C:\Documents and Settings\mario\Application Data\SSH\temp\e_48_1_024_s.ht... 18/05/2010
Toroid’s self inductance
Submitted by: I.D. 037706835
The problem:
A toroid shaped inductor with a rectangle profile has an inner radius r1 , an outer radius r2 and a
height a. The toroid has N windings and current I flows through it.
1. What is the magnetic field in the entire area?
a
2. Show that the toroid’s self inductance is L = µ0 N 2 2π
ln ( r2
r1 ).
3. After placing an infinite wire at the center of the toroid (On it’s symmetry axis),
Calculate the mutual inductance between the toroid and the wire in two ways:
• By issuing a current in the infinite wire and calculating the EMF induced in the toroid.
• By issuing a current in the toroid and calculating the EMF induced in the infinite wire.
The solution:
1. The magnetic field cannot be in the z direction because of the symmetry, and cannot be in the
~ ·B
~ = 0. So that the only direction could be ϕ̂.
radial direction since ∇
Using Ampere’s law, we can build numerous closed curves and get the circulation of the magnetic
field, but only a circulation inside the toroid will give us a current through the surface of the circulation. Hence, the only magnetic field in the area will be at direction ϕ̂, inside the toroid (where
r1 < r < r2 , − a2 < z < a2 ):
Inside the toroid:
B · 2πr = µ0 · N · I
~ = µ0 N I ϕ̂
=⇒ B
2πr
(1)
Outside the toroid:
~ =0
B · 2πr = µ0 · 0 =⇒ B
(2)
2. According to the definition of the intuctance Φ = L · I:
Z
Z r2
aN 2 µ0 I
aN 2 µ0 r2
~
ΦB = N · B · d~s =
· dr =
ln · I
2πr
2π
r1
r1
aN 2 µ0 r2
ln
2π
r1
L =
(3)
(4)
3.a. Calculating the flux of the magnetic field from the wire through the toroid, and extracting the
mutual inductance Lwt :
µ0 Iw
ϕ̂
2πr
Z
~ w · d~s = aN µ0 ln r2 · Iw
= N B
2π
r1
aN µ0 r2
=
ln
2π
r1
~ wire =
B
ΦB
Lwt
(5)
(6)
(7)
1
3.b. Using Faraday’s law, and circulating on an infinite rectangle, where one of the sides is on the
infinite wire, and the other three ones are far enough to assume that the electrical field in their area
vanishes, we get that the circulation of the electrical field on the rectangle is equal to the EMF in
the wire, hence:
I
Z
~ · d~l = − ∂
~ · d~a
=
E
B
(8)
∂t
Z r2
aµ0 N It
aµ0 N r2 ˙
∂
∂ aµ0 N It r2
=−
(9)
= −
dr = −
ln
ln · It
∂t r1
2πr
∂t
2π
r1
2π
r1
aN µ0 r2
Ltw =
(10)
ln
2π
r1
And we got Lwt = Ltw as expected.
Another method is using the differential version of Faraday’s law, taking under consideration that
along the wire (at the area of the toroid, since only there a magnetic flux changes) there is an
electrical field in the ẑ direction (cylindrical coordinates) which has no dependence on ϕ since the
problem is symmetrical:
~ = µ0 N I · ϕ̂
B
2πr
~
~ = − ∂ E · ϕ̂ = − ∂ B
∇×E
∂r
∂t
Z
∂
~ =
~ · dr
E
B
∂t
Z
Z a Z r2
2
aN µ0
r2
∂
µ0 N I
~
=
E · d~z =
drdz =
ln ( )I˙
∂t − a r1 2πr
2π
r1
~ = E(r,z) ẑ
E
(11)
(12)
(13)
(14)
2
Ltw =
aN µ0 r2
ln
2π
r1
(15)
2
RLC circuit
Submitted by: I.D. 066072570
The problem:
For the given RLC circle:
1. find the law of connecting inductances in series and in parallel in general.
What is the total induction of the given circle?
2. The switch is on the right hand side for 3τsec and after is moved to the left side.
• Find the resonance frequency of the system.
• find the current through the resistor as a function of time.
The solution:
a. inductance in series:
I1 = I2 ⇒ ε = (L1 + L2 )I˙1 = Lef f I˙ ⇒ L1 + L2 = Lef f
(1)
inductance in parallel:
I1 + I2 =
ε
ε
ε(L1 + L2 )
ε
L1 L2
1
1
1
+
=
⇒
=
= Lef f ⇒
=
+
L1 L2
L1 L2
I1 + I2
L1 + L2
Lef f
L1 L2
(2)
in our case:
Lef f =
(L1 + L2 ) · L3
L1 + L2 + L3
(3)
after 3τ (RC) we get:
−t
V = ε(1 − e Rc ) = ε(1 − e−3 ) ≈ 4.75v
(4)
and the equation of the circuit is:
q
q
+ IR + LI˙ = 0 ⇒ + Rq̇ + Lq̈ = 0
c
c
(5)
we shall start analysis with the simplest case- R=0, in this case we have:
q̈ = −
1
q
Lc
(6)
1
and the solution is:
q = Aeiω0 t
(7)
where
s
ω0 =
2
1
= 8 · 106 Hrz
Lef f c
3
(8)
substitute q = Aeiωt in the equation we get:
r
iR
Γ2
i
2
2
−ω +
ω + ω0 = 0 ⇒ ω = Γ ± ω02 −
L
2
4
(9)
where
Γ=
R
L
(10)
the solution is
i
h
0 i
q = Re Aei(ω + 2 Γ)t
(11)
where
r
0
ω =
ω02 −
Γ2
4
(12)
substitute
q(t = 0) = c · v(3τ )
(13)
we get
A = 4.75 · 10−9
(14)
to find the curent:
Γ
i
Γ
0 i
I = q̇ = RE i(ω 0 + Γ)Aei(ω + 2 Γ)t = −Ae− 2 t · ( cos(ω 0 t) + ω 0 sin(ω 0 t))
2
2
eventually substitute the given parameters we get:
2
2
2
I = −4.75 · 10−9 e−37509t 37509 · cos(8 · 106 t) + 8 · 106 sin(8 · 106 t)
3
3
3
Amper
(15)
(16)
The first plot presents a voltage on the resistor R and on the second one the current through it as
a function of time.
2
3