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.
2 2
r −r1
o I12
Ans: B = µ2πr
.
r2 2 −r1 2
(b) Find the magnitude of the magnetic field at a distance r, r3 < r < r4 , from the axis.
µo I12 r3 −r3 3 Ans: B = 2πr 1 − r2 2 −r1 2 .
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.
Ans:
L=
µo N 2 w
2π
ln
r2
r1
.
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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.
Ans:
M=
µo N w
2π
ln
r2
r1
.
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PHGN200: All Sections
Recitation 7
April 10, 2007
4. Consider the circuit shown in Fig. 6.
V
I1
S1
S2
R1
L
I2
R2
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.
R +R
V
− 1L 2 t
Ans: I1 = R1 +R2 1 − e
.
(b) Find current I1 after a very long time.
Ans: I1 = R1V+R2 at t = ∞.
(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 .
Ans: I1 = RV1 .
(d) Find current I2 as a function of time t that has elapsed since S2 was closed.
R
Ans: I2 = R1V+R2 e− L2 t .
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