Name:_______________________ ___
PHY2061
11-9-06
Exam 2 Solutions
1. [8 points] In the circuit shown, the resistance R1 = 12Ω . The battery voltages are
identical: ε1 = ε 2 = ε 3 = 1 V . What is the current (in amps) flowing through the middle
branch from a to b?
i2
i1
i3
1
Applying the junction rule: i1 + i2 = i3
Applying the loop rule to the left loop (LL), right loop (RL), and the full loop (FL) gives:
LL ⇒ ε − i1 R − i3 R − ε − i1 R = 0 ⇒ 2i1 + i3 = 0
RL ⇒ ε − i2 R − i3 R − ε − i2 R = 0 ⇒ 2i2 + i3 = 0
FL ⇒ ε − i1 R + i2 R − ε + i2 R − i1 R = 0 ⇒ i2 = i1
Plugging the FL equation into the junction rule equation, and then into the RL equation
gives:
2i2 + 2i2 = 0
⇒ i2 = i1 = 0 ⇒ i3 = 0
Thus no current flows from top to bottom in the middle branch (or anywhere for that matter).
Since all batteries have a potential difference of 1 V, one can see that the loop rule works out
if all currents are zero.
Page 1 of 9
Name:_______________________ ___
PHY2061
11-9-06
2. [6 points] A capacitor of capacitance C = 2 ×10−10 F contains a charge q on one of its
plates (and –q on the other). It is connected in to a resistor of resistance R = 200Ω such that it
forms a closed circuit. How much time must elapse so that the charge on the capacitor is
reduced to only 10% of its starting value?
Applying Kirchoff’s loop rule, the equation to solve is:
dq q
R + =0
dt C
⇒ q ( t ) = q0 e− t / RC
Thus the elapsed time so that 10% of the charge remains is determined by:
0.1q0 = q0 e −t / RC
t
RC
t = ( RC ) ln10 = ( 200Ω ) ( 2 ×10−10 C ) ( 2.30 ) = 92ns
ln10 =
3. [6 points] A stationary flat conductor carries a constant current i >0 in the direction from
top to bottom ( −yˆ ) in the presence of a magnetic field B that points into plane the paper
( −zˆ ). If the electrons that make up the current are not allowed to leave the conductor, and the
magnitude of their drift velocity is vd , indicate the direction and determine the magnitude of
any electric field created inside the conductor.
i
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
e-, vd
B
FB, E
y
x
z
This is an example of a Hall probe. If the
current goes from top to bottom, the free
electrons travel from bottom to top. They will
feel a magnetic force according to:
FB = (−e) v × B which is in the +x direction. In
equilibrium, an electric field will be set up to
balance this force, such that
F = (−e) ( E + v × B ) = 0 , so the electric field will
point in the +x direction with magnitude vd B
Page 2 of 9
Name:_______________________ ___
PHY2061
11-9-06
4. [6 points] An electron is accelerated from rest between 2 electric plates with a potential
difference of ΔV = 5 ×105 V . What is the velocity of the electron upon reaching the far
plate? The mass of the electron is me = 9.11×10−31 kg .
The potential energy of the electron when in starts from rest is translated into kinetic energy
as it approaches the second plate:
U = eΔV = EK = ( γ − 1) mc 2 We need to use special relativity here because the energy will
be comparable to the rest mass energy.
eΔV
+1 = γ =
mc 2
⇒ v = c 1−
1
1−
v2
c2
1
⎛ eΔV
⎞
+ 1⎟
⎜
2
⎝ mc
⎠
2
= 0.86c = 2.6 × 108 m/s
5. [6 points] A long wire has an electric charge density of λ = +2.5μC / m , where
1μ C = 10−6 C , as measured in the rest frame of the wire. What is the magnitude of the electric
field in the rest frame of an electron traveling at a velocity of v = 2.8 ×108 m/s parallel to the
wire at a distance r = 1 cm from the wire?
In the rest frame of the wire, and by using Gauss’ Law, we can solve for the electric field:
λh
v∫ E ⋅ dA = 2π rhE = 0
ε0
S
⇒E=
λ0
2πε 0 r
In the rest frame of the electron, however, the wire appears length contracted (since the
electron’s motion is parallel to the wire). Thus the charge per unit observed length has
increased since charge is conserved but length is not:
L=
L0
γ
⇒λ =
q
q
=γ
= γλ0
L
L0
So the field is:
Page 3 of 9
Name:_______________________ ___
PHY2061
11-9-06
E =γ
γ=
λ0
2πε 0 r
1
v2
1− 2
c
E = ( 2.785 )
1
=
2.82
1− 2
3
= 2.785
2 ( 9 × 109 Nm 2 / C2 )( 2.5 × 10−6 C/m )
( 0.01 m )
= 1.25 × 107 N/C
6. [6 points] A long cylindrical wire of radius r = 3mm carries 100A of current. If the current
density is uniform throughout the cross-section of the wire, what is the magnitude of the
magnetic field at a radius of r = 1mm within the wire?
i 100 A
=
= 3.54 × 106 A/m 2
2
A πR
Use Ampere’s Law to find the magnetic field by choosing a path C at a radius r = 1mm:
The current density is j =
v∫
C
B ⋅ ds = 2π rB = μ0ienc
ienc = jπ r 2
B=
μ0 ( iπ r 2 )
( 2π r ) (π R
2
)
=
μ0i
r = 2.2 ×10−3 T
2
2π R
Page 4 of 9
Name:_______________________ ___
PHY2061
11-9-06
7. [8 points] An infinitely long insulated wire carrying a current I = 50 A is bent into a 270º
arc ( 3π / 2 radians) of radius R=2 cm. The current comes in from infinitely far away from
above, and exits to the left infinitely far away. Find the magnitude of the field B at the center
of the arc. The wire begins and ends its turn at the locations indicated.
I
y
I
x
z
R
R
I
There are 3 components to the field: 2 semi-infinite segments of wire, and the ¾ circle. The
magnitude of the field from a semi-infinite wire is half of that of an infinite wire (see
homework, or look at direct integration of the Biot-Savart law):
1 μ 0i
= 2.5 ×10−4 T in z direction.
Bsemi −∞ =
2 2π R
From the circular wire, the field is:
μi
3π
Barc = 0 Φ where Φ =
4π R
2
3μ0i
= 1.18 × 10−3 T in zˆ
Barc =
8R
The magnetic field from each section contributes in the same direction (z). Thus the total
magnitude is:
B = Barc + 2 Bsemi −∞ = 1.68 × 10−3 T
Page 5 of 9
Name:_______________________ ___
PHY2061
11-9-06
8. A stationary neutral atom resides at the center of a Cartesian coordinate system and has a
magnetic dipole moment of 2.1× 10−23 J/T aligned in the +yˆ direction.
(a) [6 points] What is the ratio of the magnitude of the magnetic field from the atomic dipole
at y = 100 nm to that at y = 25 nm? (The radius of the atom is about 0.1 nm, where 1 nm =
10-9 m)
The field of a dipole at distances large compared to its size falls off with the cube power of
distance:
B∝
μ
y3
So the ratio is:
B ( y = 100nm ) 1
1
= 3=
B ( y = 25nm ) 4
64
(b) [6 points] If a magnetic field is present everywhere with the form
B = B0 y yˆ , where B0 = 0.5 T/m , what is the magnitude of the acceleration of the dipole if
the mass of the atom is 10-25 kg.
Fy = μ y
dBy
dy
= μ y B0 = ( 2.1× 10−23 J/T ) ( 0.5 T/m ) = 1.05 × 10−23 N
F 1.05 × 10−23 N
= 105 m/s 2
a= =
−25
10 kg
m
Page 6 of 9
Name:_______________________ ___
PHY2061
11-9-06
9. [6 points] The magnetic field of a large solenoid is used to keep a proton in a perfect
circular orbit. The solenoid has 1000 windings per meter of length and has a radius of 1 m. If
the proton has a velocity of v = 1.5 ×106 m/s , what is the minimum current needed to keep
the proton orbiting within the confines of the solenoid in a plane perpendicular to the
solenoid axis? The proton mass is m p = 1.67 × 10−27 kg and its charge is q = 1.6 × 10−19 C .
r
p
B = μ0 ni
p = qBr = qr μ0 ni
1.67 × 10−27 kg )(1.5 × 106 m/s )
(
mv
⇒i=
=
= 12.5 A
qr μ0 n (1.6 ×10-19 C ) (1m ) ( 4π 10−7 ) (1000 )
10. [6 points] A rod of length L = 0.5 m and mass m = 0.5 kg carries a current I = 20 A in the
direction shown. The rod is aligned parallel to the z axis, and a uniform magnetic field is
present: B = − B0 yˆ , B0 = 0.5 T . The rod is suspended by two massless wires which bring the
current to and away from the rod. The acceleration due to gravity is 10 m/s2 in the −yˆ
direction. What is the angle θ that the suspension wires make with respect to the magnetic
field direction?
B
y
z
θ
x
FM
I
Fg
F = iL × B
tan θ =
FB iLB ( 20 )( 0.5 )( 0.5 )
=
=
= 1 ⇒ θ = 45D
Fg mg
( 0.5)(10 )
Page 7 of 9
Name:_______________________ ___
PHY2061
11-9-06
11. An oscillating LC circuit consists of a 0.002 H inductive coil and a 4 μF capacitor.
The capacitor has a voltage drop of 0.75 V when the current through the coil is
0.03A.
(a) [6 points] Find the maximum charge on the capacitor.
(b) [6 points] Find the maximum current through the coil.
The total energy in the circuit is:
1
q2 1 2 1
+ Li = CV 2 + Li 2
U=
2C 2
2
2
1
1
2
2
U = ( 4 × 10−6 ) ( 0.75 ) + ( 0.002 )( 0.03) = 2.025 × 10−6 J
2
2
This must equal the energy when all charge is on the capacitor, or all current in
the inductor:
q2 1 2 1
1
+ Li = CV 2 + Li 2
2C 2
2
2
2
q
U = 2.025 × 10−6 J = max
2C
⇒ qmax = 4.0μ C
U=
U = 2.025 × 10−6 J =
1 2
Limax
2
⇒ imax = 0.045 A
Page 8 of 9
Name:_______________________ ___
PHY2061
11-9-06
12. [6 points] A circular conducting loop of wire increases in radius with time according to
r = vt where v is a constant. It is immersed in a constant magnetic field B perpendicular to
the plane of the loop. What is the induced EMF in the loop?
dΦB d
dA
d
dr
= ( BA ) = B
= B ( π r 2 ) = π B 2r
dt
dt
dt
dt
dt
2
ε = 2π Bv t
ε=
13. A solenoid 21 cm long has a circular cross-section of 17 cm2. There are 210 turns of wire
carrying a current of 2.5A.
(a) [6 points] Find the magnitude of the magnetic field inside the solenoid assuming that
it is essentially infinite in length (neglect end effects).
The magnetic field of a solenoid, obtained via Ampere’s Law for example, is:
B = μ0 ni = 4π 10−7
210
2.5 = π × 10−3
0.21
(b) [6 points] Find the total energy stored in the magnetic field inside the volume of the
solenoid. Neglect end effects.
U=
B
2
2μ0
(π 10 )
V=
−3 2
8π 10−7
⎛
( 0.21) ⎜17
⎝
1 m2 ⎞
−3
⎟ = 1.4 × 10 T
104 cm 2 ⎠
Page 9 of 9