Download Document 8894664

Problem 4:
The Coil
Consider a coil made of thin copper wire (radius ~ 0.25 mm) and has about 600 turns of
average diameter 25 mm over a length of 25 mm. What approximately should the
resistance and inductance of the coil be? The resistivity of copper at room temperature is
around 20 nΩ-m. Note that your calculations can only be approximate because this is not
at all an ideal solenoid (where length >> diameter).
Problem 4 Solution:
The resistance (NOTE: I screwed up and meant radius was 0.25 mm, not diameter)
ρ L ρ ⋅ N π d ( 20 nΩ m ) ⋅ ( 600 )( 25 mm )
R=
=
≈
≈ 4.8 Ω
2
A
π a2
( 0.25 mm )
The inductance of a solenoid we calculated in class to be:
(
L = μ0 n π R l ≈ 4π ×10
2
2
−7
TmA
-1
)
2
2
⎛ 600 ⎞ ⎛ 25 mm ⎞
⎜
⎟ π⎜
⎟ ( 25 mm ) ≈ 9 mH
⎝ 25 mm ⎠ ⎝ 2 ⎠
Problem 2:
An inductor consists of two very thin conducting cylindrical
shells, one of radius a and one of radius b, both of length h.
Assume that the inner shell carries current I out of the page,
and that the outer shell carries current I into the page,
distributed uniformly around the circumference in both
cases. The z-axis is out of the page along the common axis
of the cylinders and the r-axis is the radial cylindrical axis
perpendicular to the z-axis.
a) Use Ampere’s Law to find the magnetic field between
the cylindrical shells. Indicate the direction of the magnetic
field on the sketch. What is the magnetic energy density as
a function of r for a < r < b?
b) Calculate the inductance of this long inductor recalling that U B =
1 2
LI and using
2
your results for the magnetic energy density in (a).
c) Calculate the inductance of this long inductor by using the formula
G G
Φ = LI =
∫ B ⋅ dA and your results for the magnetic field in (a). To do this you
open surface
must choose an appropriate open surface over which to evaluate the magnetic flux. Does
your result calculated in this way agree with your result in (b)?
Problem 2 Solutions:
(a)The enclosed current Ienc in the Ampere’s loop with radius r is given by
I enc
Applying Ampere’s law,
G
G
⎧0, r < a
⎪
= ⎨I , a < r < b
⎪0, r > b
⎩
v∫ B ⋅ d s = B(2π r) = μ I
0 enc
, we obtain
⎧0, r < a
G ⎪⎪ μ0 I
B=⎨
ϕ̂, a < r < b (counterclockwise in the figure)
⎪ 2π r
⎪⎩0, r >
b
The magnetic energy density for a < r < b is
2
μ0 I 2
B2
1 ⎛ μ0 I ⎞
uB =
=
=
⎜
⎟
2 μ0 2 μ0 ⎝ 2π r ⎠
8π 2 r 2
It is zero elsewhere.
(b) The volume element in this case is 2π rhdr . The magnetic energy is : ⎛ μ I2 ⎞
μ I 2h ⎛ b ⎞
U B = ∫ uB dVol = ∫ ⎜ 02 2 ⎟ 2π h rdr = 0
ln ⎜ ⎟
8π r ⎠
4π
⎝a⎠
V
a⎝
μ I 2l ⎛ b ⎞ 1
Since U B = 0 ln ⎜ ⎟ = LI 2 , the inductance is
4π
⎝a⎠ 2
b
L=
μ0 h ⎛ b ⎞
ln ⎜ ⎟
2π ⎝ a ⎠
(c)
The magnetic field is perpendicular to a rectangular
surface shown in the figure. The magnetic flux through a
thin strip of area dA = ldr is
⎛μ I⎞
μ Ih
dΦ B = BdA = ⎜ 0 ⎟ ( h dr ) = 0 dr
2π r
⎝ 2π r ⎠
Thus, the total magnetic flux is
ΦB = ∫ dΦB = ∫
b
a
μ0 Ih
μ Ih b dr μ0 Ih ⎛ b ⎞
ln ⎜ ⎟
dr = 0 ∫
=
2π r
2π a r
2π
⎝a⎠
Thus, the inductance is
L=
which agrees with that obtained in (b).
μ h ⎛b⎞
ΦB
= 0 ln ⎜ ⎟
2π ⎝ a ⎠
I
‫מעגל ‪ ,RL‬בחינה ‪2002‬‬
‫נתון המעגל שמופיע באיור‪ ,‬בתחילה המפסק ‪ S‬סגור מזה זמן רב‪ .‬בזמן ‪ t=0‬המפסק ‪ S‬נפתח‪.‬‬
‫‪ .1‬מהו הזרם על כל נגד כאשר המפסק סגור והמעגל נמצא בשיווי משקל?‬
‫‪ .2‬מהו הזרם על הנגדים ‪ R1, R2‬כפונקציה של הזמן לאחר פתיחת המפסק? תאר כמשוואה וכגרף‪.‬‬
‫‪ .3‬מהו הפרש הפוטנציאלים על הנגדים ‪ R1,R2‬כפונקציה של הזמן לאחר פתיחת המפסק?‬
‫‪ .1‬כאשר המפסק סגור והזרם התייצב‪ L ,‬מתפקד כקצר‪ ,‬הנגדים ‪ R1,R2‬מחוברים במקביל ולהם נגד שקול ‪ . RT‬הזרם‬
‫‪R1‬‬
‫‪ε‬‬
‫= ‪ I‬וכן ‪ I1 R1 = I 2 R2‬כאשר‬
‫במעגל הינו‬
‫‪R1 + R2‬‬
‫‪RT‬‬
‫‪ I 2 ) I 2 = I‬הינו הזרם דרך הנגד ‪ ,(R2‬המתח על הנגד ‪R2‬‬
‫הינו ‪ . ∆V2 = I 2 R2‬באופן דומה ניתן למצוא את הערכים עבור ‪ ,R1‬הזרם דרך ‪ R1‬הוא גם הזרם במשרן הפרש‬
‫המתחים על המשרן הינו ‪.0‬‬
‫‪ .2‬לאחר פתיחת המפסק‪ ,‬מקור המתח בבעיה משתנה ועתה הנגדים מחוברים בטור‪ ,‬נחשב נגד שקול חדש‬
‫‪ RTT‬בהתאם‪ .‬עבור מעגל ‪ RL‬ידוע פתרון המשוואה הדיפרנציאלית‪:‬‬
‫)‬
‫‪L‬‬
‫(‬
‫‪EXP − tR‬‬
‫‪ε‬‬
‫‪RTT‬‬
‫= ) ‪ , I (t‬זהו הזרם שיוצר המשרן וכן הזרם דרך שני הנגדים‪.‬‬
‫איור סכמטי לפריקת משרן מוצג משמאל‪.‬‬
‫‪ .3‬מפל המתחים על הנגד ‪ R2‬עתה הוא ‪ ∆V2 = I (t ) R2‬ובאופן דומה עבור ‪.R1‬‬
Problem 5:
The LR circuit shown in the figure contains a resistor R1 and an inductance L in series
with a battery of emf ε 0 . The switch S is initially closed. At t = 0, the switch S is opened,
so that an additional very large resistance R2 (with R2 R1 ) is now in series with the
other elements.
(a) If the switch has been closed for a long time before t = 0, what is the steady current
I 0 in the circuit?
(b) While this current I 0 is flowing, at time t = 0, the switch S is opened. Write the
differential equation for I (t ) that describes the behavior of the circuit at times t ≥ 0.
Solve this equation (by integration) for I (t ) under the approximation that ε 0 = 0 .
(Assume that the battery emf is negligible compared to the total emf around the circuit
for times just after the switch is opened.) Express your answer in terms of the initial
current I 0 , and R1 , R2 , and L.
(c) Using your results from (b), find the value of the total emf around the circuit (which
from Faraday's law is −LdI / dt ) just after the switch is opened. Is your assumption in (b)
that ε 0 could be ignored for times just after the switch is opened OK?
(d) What is the magnitude of the potential drop across the resistor R2 at times t > 0, just
after the switch is opened? Express your answers in terms of ε 0 , R1 , and R2 . How does
the potential drop across R2 just after t = 0 compare to the battery emf ε 0 , if
R2 = 100R1 ?
Problem 5 Solutions:
(a) There is no induced emf before t = 0. Also, no current is flowing on R2.Therefore,
I0 =
ε0
R1
(b) The differential equation is
ε 0 − I (t)(R1 + R2 ) = L
dI (t)
dt
Under the approximation that ε 0 = 0 , the equation is
− I (t)(R1 + R2 ) = L
dI (t)
dt
The solution with the initial condition I(0) = I0 is given by
I (t) = I 0 exp(−
(R1 + R2 )
t)
L
(c)
ε =−L
Since I 0 =
ε0
R1
dI (t)
dt
= I 0 (R1 + R2 )
t =0
,
ε=
ε0
⎛
R ⎞
(R1 + R 2 ) = ⎜⎜1 + 2 ⎟⎟ ε 0 >> ε 0
R1
R1 ⎠
⎝
(∵ R 2 >> R1 )
Thus, the assumption that ε 0 could be ignored for times just after the switch is open is
OK.
(d) The potential drop across R2 is given by
ΔV2 =
⎛ R2
R2
ε = ⎜⎜
R1 + R2
⎝ R1 + R2
⎞⎛
R ⎞
R
⎟⎟⎜⎜1 + 2 ⎟⎟ ε 0 = 2 ε 0
R1 ⎠
R1
⎠⎝
If R2 = 100R1 ,
ΔV 2 = 100 ε 0
This is why you have to open a switch in a circuit with a lot of energy
stored in the magnetic field very carefully, or you end up very dead!!