PHGN200: All Sections
Recitation 7
April 10, 2007
1. The cross-sections of two hollow infinitely long cylindrical conductors are shown in blue and red in Fig. 1. The
blue cylindrical conductor carries a constant current, I12 , out of the page, with uniform current density J12 .
The red cylindrical conductor carries a current into the page, with current density J34 = 3J212 r .
r4
r1
r2
r3
Figure 1: The cross-sections of two hollow infinitely long cylindrical conductors are shown in blue and red. The
blue cylinder has inner radius r1 and outer radius r2 . The red cylinder has inner radius r3 and outer radius r4 .
(a) Find the magnitude of the magnetic field at a distance r, r1 < r < r2 , from the axis.
Solution: We apply Ampere’s Law, given by
I
~ · d~` = µo Iencl. ,
B
at a distance r from the axis, see Fig. 2.
r4
r
r1
r2
r3
Figure 2: An Amperian loop of radius r is shown by the dashed circle.
(1)
PHGN200: All Sections
Recitation 7
April 10, 2007
~ is parallel to d~`; thus, (1) yields
By symmetry, B
I
Bd` = µo Iencl. .
(2)
By symmetry, the magnetic field B is constant on the Amperian loop; thus, (2) yields
I
B d` = µo Iencl.
B2πr = µo Iencl.
µo Iencl.
,
B=
2πr
(3)
where Iencl. refers to the current enclosed by the Amperian loop. Iencl. can be found by introducing
the current density and then integrating, i.e.,
Z
~
Iencl. = J~ · dA
Z
= J dA,
(What happened to the dot product?)
Z r
=
J12 (2πζ) dζ
r1
Z r
= J12
2πζ dζ
r1
= J12 π r2 − r1 2 ,
(4)
where J12 =
I12
.
π(r2 2 −r1 2 )
Substituting (4) into (3) yields
µo I12
B=
2πr
r 2 − r1 2
r2 2 − r1 2
.
(b) Find the magnitude of the magnetic field at a distance r, r3 < r < r4 , from the axis.
Solution: We apply Ampere’s Law, given by
I
~ · d~` = µo Iencl. ,
B
at a distance r from the axis, see Fig. 3.
Page 2
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PHGN200: All Sections
Recitation 7
April 10, 2007
r
r4
r1
r2
r3
Figure 3: An Amperian loop of radius r is shown by the dashed circle.
~ is parallel or anti-parallel to d~`, i.e., B
~ · d~` = +Bd` or B
~ · d~` = −Bd`; thus, we take
By symmetry, B
the absolute value of both sides of (5) to yield
I
Bd` = |µo Iencl. | .
(6)
By symmetry, the magnetic field B is constant on the Amperian loop; thus, (6) yields
I
B d` = |µo Iencl. |
µo Iencl. ,
B=
2πr (7)
where Iencl. refers to the current enclosed by the Amperian loop. Iencl. can be found via current
densities, i.e.,
Z
~
~
|Iencl. | = J · dA
Z
Z
Z
~
~
~
~
~
~
= J12 · dA12 + J23 · dA23 + J34 · dA34 Z
Z
Z
= J12 dA12 + 0 dA23 − J34 dA34 Z r2
Z r
3J
ζ
12
(2πζ) dζ = J12 (2πζ) dζ −
2
r1
r3
r3 − r33 = I12 1 − 2
,
(8)
r2 − r1 2 I12
.
π(r2 2 −r1 2 )
Substituting (8) into (7) yields
µo I12 r3 − r3 3 1− 2
.
B=
2πr r2 − r1 2 where we have used the fact that J12 =
Page 3
PHGN200: All Sections
Recitation 7
April 10, 2007
2. A blue wire is wound evenly on a torus of rectangular cross section, see Fig. 4. Find the self inductance of
this torus.
y
w
I
x
r1
r2
z
Figure 4: A blue wire is wound evenly on a torus of rectangular cross section, with inner radius r1 and outer
radius r2 . There are N turns of the blue wire in all.
Solution: The magnetic flux through one square loop is given by (see Recitation 6, problem 2)
µo N Iw
r2
Φone loop =
ln
.
2π
r1
The magnetic flux through N square loops is given by
ΦN
loops
µo N 2 Iw
=
ln
2π
r2
,
r1
thus, the self inductance of the torus is given by
µo N 2 w
ln
L=
2π
Page 4
r2
.
r1
PHGN200: All Sections
Recitation 7
April 10, 2007
3. A blue wire carrying current I is wound evenly on a torus of rectangular cross section. There are N turns of
the blue wire in all. A red wire is thrown over the torus and is connected to a resistor, R, see Fig. 5. Find
the mutual inductance of this arrangement.
y
R
w
I
x
r1
r2
z
Figure 5: A blue wire carrying current I is wound evenly on a torus of rectangular cross section, with inner radius
r1 and outer radius r2 . There are N turns of the blue wire in all. A red wire is thrown over the torus and is
connected to a resistor, R.
Solution: The magnetic flux through the area enclosed by the red wire is given by (see Recitation 6,
problem 2)
µo N Iw
r2
ΦB =
ln
.
2π
r1
Thus, the mutual inductance is given by
µo N w
M=
ln
2π
Page 5
r2
r1
.
(9)
PHGN200: All Sections
Recitation 7
April 10, 2007
4. Consider the circuit shown in Fig. 6.
V
I1
S1
S2
R1
R2
L
I2
Figure 6: An LR circuit is shown.
(a) At time t = 0, switch S1 is closed and switch S2 is left open. Find current I1 as a function of time.
Solution: Applying Kirchhoff’s loop rule going in the counterclockwise direction through V , S1 , R1 ,
L, and R2 yields
V − I1 R1 − L
dI1
− I1 R2 = 0
dt Z
Z
− dt =
L
dI1
(R1 + R2 ) I1 − V
L
ln [(R1 + R2 ) I1 − V ] , where C1 = constant of integration
−t + C1 =
R1 + R2
R1 +R2
V
, where C2 = constant.
I1 = C2 e− L t +
R1 + R2
Using the initial conditions, i.e., I1 = 0 at t = 0, yields
I1 =
R1 +R2
V
1 − e− L t .
R1 + R2
(10)
(b) Find current I1 after a very long time.
Solution: Taking the limit as t → ∞ of (10) yields
I1 =
V
R1 + R2
Page 6
at t = ∞.
(11)
PHGN200: All Sections
Recitation 7
April 10, 2007
(c) After the current in the circuit has reached its final, steady-state value with switch S1 closed and S2
open, switch S2 is closed, thus short-circuiting the inductor. (Switch S1 remains closed.) Find current
I1 .
Solution: Applying Kirchhoff’s loop rule going in the counterclockwise direction through V , S1 , and
S2 yields
V − I1 R1 = 0
V
.
I1 =
R1
(d) Find current I2 as a function of time t that has elapsed since S2 was closed.
Solution: Applying Kirchhoff’s loop rule going in the counterclockwise direction through L, R2 , and
S2 yields
dI2
=0
−I2 R2 − L
dt
Z
Z
R2
1
−
dt =
dI2
L
I2
R2
I2 = Ce− L t ,
where C is some constant. At time t = 0, I2 is given by (11) (why?), thus, (12) yields
I2 =
R2
V
e− L t .
R1 + R2
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