Name:_______________________ ___
PHY2061
11-18-04
Exam 3 Solutions
Closed book exam. A calculator is allowed, as is one 8.5×11” sheet of paper with your own
written notes. Please show all work leading to your answer to receive full credit. Answers
should be calculated to 2 significant digits. Exam is worth 100 points, 25% of your total
grade.
1. Consider the toroid shown with an inner radius a = 5 cm and an outer radius
b = 6 cm. The total number of turns of wire wound around the toroid is N = 400.
The core of the toroid has a rectangular cross-section with a thickness h = 0.5 cm.
(a) [6 points] If the wire carries a current of 2A, what is the magnitude of the
magnetic field inside the toroid at a radius midway between the inner and
outer radii?
Use Ampere’s Law. Consider a loop at a fixed radius inside toroid.
v∫
C
B ⋅ ds = µ0ienc
2π rB = µ0 Ni
B=
µ0 Ni 2ki
=
N
r
2π r
a+b
= 0.055 m
2
B = 2.91 mT
r=
1. Continued
(b) [8 points] Calculate the inductance of the toroid.
NΦB
L=
i
b dr
b
Φ B = ∫ B ⋅ da = 2kiNh ∫
= 2kiNh ln
S
a r
a
b
L = 2kN 2 h ln = 2.92 × 10−5 H
a
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Name:_______________________ ___
PHY2061
11-18-04
2. [8 points] A long straight wire with a cylindrical cross-section has a radius of 0.3 mm
and carries a current of 100 mA uniformly spread across its cross-section. What is the
ratio of the magnetic field at a radius of 0.1 mm to that at the surface of the wire?
Use Ampere’s Law. Consider a loop centered on wire axis.
v∫
C
B ⋅ ds = µ0ienc
2π rs Bs = µ0i (at surface of wire, r = rs )
Bs =
µ0i
2π rs
Inside wire:
current density j =
i
π rs 2
i
i 2π r 2
′
′
r
dr
π
=
2
0 πr 2
π rs 2 2
s
r
ienc = ∫ j ⋅ da = ∫
0
r
µ0 i
µi
π r2 = 0 2 r
2
2π r π rs
2π rs
µ 0i
r
2π rs 2
r 1
= =
Ratio: R =
µ0i
rs 3
2π rs
B=
3. A square loop of wire with edge length 7 cm is located in the plane of the paper inside
a homogeneous magnetic field of 0.35 T pointing into the paper. It is connected in
series with a resistor of 210 Ω. The magnetic field is now increased at a constant rate
by a factor of 4.5 in 18 s.
(a) [6 points] Calculate the magnitude of the induced EMF in the loop (in V)
during that time.
ε =−
dΦB
dt
Φ B = ∫ B ⋅ da = BA
S
∆B = B2 − B1 = ( f − 1) B1 where f = 4.5
( f − 1) B
∆B
= −A
∆t
∆t
2 ( 4.5 − 1)( 0.35 T )
ε = − ( 0.07 m )
= −0.333 mV
18 s
ε = −A
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Name:_______________________ ___
PHY2061
11-18-04
(b) [6 points] Calculate the current induced in the loop in Amps during that time.
ε
−3.33 ×10−4 V
i= =
= 1.588µ A
R
210 Ω
4. A rectangular loop of wire is located in the plane of the paper inside a homogeneous
magnetic field of 6 mT pointing into the paper. One side length is fixed at 50 cm, the
other side grows with time according to the formula A = A 0 (1 + at ) , where
A 0 = 50 cm and a = 2.5 s -1 .
(a) [6 points] Calculate the magnitude of the induced EMF in the loop.
ε =−
dΦB
dt
Φ B = ∫ B ⋅ da = BA = BwA 0 (1 + at )
S
ε = − BwA 0 a = − ( 0.006 T )( 0.5 m )( 0.5 m )( 2.5 ) = 3.75 mV
(b) [6 points] If the resistance of the wire increases as R = R0 (1 + λt ) , where
R0 = 3Ω and λ = 0.5 s-1, what would the acceleration of the moving wire need
to be in order to maintain a constant current in the wire? Hint: Consider
adding a quadratic time dependence to the length of the growing side.
1
⎛
⎞
let A = A 0 ⎜1 + at + aλ t 2 ⎟
2
⎝
⎠
BwA 0 a (1 + λt )
BwA 0 a
ε
i= =−
=−
= constant!
R
R0 (1 + λ t )
R0
d 2A
acceleration α = 2 = A 0 aλ = ( 0.5 m ) ( 2.5 s-1 )( 0.5 s-1 ) = 0.625 m/s 2
dt
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Name:_______________________ ___
PHY2061
11-18-04
5. A solenoid 42 cm long has a circular cross-section of 8.5 cm2. There are 420 turns of
wire carrying a current of 3.25A.
(a) [6 points] Find the magnitude of the magnetic field inside the solenoid
assuming that it is essentially infinite in length (neglect end effects).
Use Ampere’s Law. Consider a loop cutting through the solenoid.
v∫
C
B ⋅ ds = µ0ienc
BA = µ0 nAi
B = µ0 ni
N
420
=
= 1000
A 0.42 m
⇒ B = 4π (10−7 ) (1000 )( 3.25 A ) = 4.084 mT
n=
(b) [6 points] Find the total energy stored in the magnetic field inside the volume
of the solenoid. Neglect end effects.
U=
B2
1
V = µ0 n 2i 2 AA
2µ0
2
U = 0.5 ( 4π × 10−7 )(103 ) ( 3.25 ) (.42 ) ( 8.5 cm 2 )
2
2
U = 2.37 mJ
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1 m2
104 cm 2
Name:_______________________ ___
PHY2061
11-18-04
6. An oscillating LC circuit consists of a 2 mH inductive coil and a 3 nF capacitor. The
capacitor has a voltage drop of 2.8 V when the current through the coil is 1.2 mA.
(a) [6 points] Find the period of oscillation.
T=
1 2π
=
= 2π LC = 1.54 × 10−5 s
f
ω
(b) [6 points] Find the maximum charge on the capacitor.
(c) [6 points] Find the maximum current through the coil.
The inductor and capacitor exhibit oscillation, with the charge on the capacitor
being exchanged for current through the inductor. To solve this, note that the
total energy is constant:
1 2 1 q2
Li +
q = C ∆V
2
2C
2
1
1
2
= ( 0.002 H ) (1.2 × 10−3 A ) + ( 3 × 10−9 F ) ( 2.8 V )
2
2
−8
= 1.32 ×10 J
U tot =
U tot
U tot
Now set either maximum inductor energy or capacitor energy to U tot :
U tot =
2
1 qmax
⇒ qmax = 2CU tot = 8.9 × 10−9 C
2 C
U tot =
2U tot
1 2
Limax ⇒ imax =
= 3.63 mA
2
L
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Name:_______________________ ___
PHY2061
11-18-04
7. A frog can be considered a magnetic dipole when immersed in strong magnetic field
(a property known as diamagnetism).
(a) [3 points] Does the frog have greater potential energy when its magnetic
dipole moment is aligned with the direction of the magnetic field or opposite
to it?
The potential energy of a magnetic dipole in a magnetic field is: U = −µ ⋅ B ,
So the frog has more potential energy when it is aligned opposite to the field.
(b) [3 points] If the field is strong enough, it is possible to levitate the frog.
Would this occur near the center of a solenoid where the field is uniform, or
near the end of a solenoid where the field is diverging?
The force on a magnetic dipole in a non-uniform magnetic field is:
∂B
, so the force can only balance gravity when the frog is in the nonFz = µ z
∂z
uniform field near the end of the solenoid.
R
ε(t)
C
L
8. Consider an RLC circuit connected to an alternating-current power supply. The
power supply delivers an electric potential of ε ( t ) = ε 0 sin ωt .
(a) [6 points] Write down a differential equation for the charge q ( t ) on the
capacitor.
Apply Kirchoff’s second law that the sum of the voltage drops must be zero:
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Name:_______________________ ___
PHY2061
11-18-04
∆VR + ∆VC + ∆VL − ε = 0
q
di
+ L = ε (t )
C
dt
2
d q
dq 1
L 2 + R + q = ε (t )
dt
dt C
iR +
(b) [6 points] Estimate the current in the circuit for the following three limiting
1
cases (give your reasons!): 1) ω → 0 , 2) ω → ∞ , and 3) ω ≈
LC
1
Note that the complex impedance of an inductor is iω L and of a capacitor is
iωC
1) As ω → 0 , the capacitor becomes an open circuit and the inductor
becomes a short circuit (infinite impedance and zero impedance,
respectively). So the current would be zero.
2) As ω → ∞ , the capacitor becomes a short circuit and the inductor
becomes an open circuit (zero impedance and infinite impedance,
respectively). So the current would be zero.
1
3) For ω ≈
, the current is maximized and matched to the resistor.
LC
The current becomes
ε
R
(c) [6 points] How does adding a transformer to the output of the power supply
affect the behavior of the circuit, if the number of secondary windings on the
transformer (which are connected to the circuit) is half of the number of
primary windings (which are connected to the power supply)?
NS
∆VP
NP
So the magnitude of the voltage (and current) will drop by a factor 2, but all
other features (frequency, phase) remain the same.
The transformer steps-down the AC voltage: ∆VS =
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