Problem 1:
An ideal battery is connected to a 200 kΩ and a 300 kΩ resistor in series.
A voltmeter is used to measure the voltage across the battery and the 200 kΩ
resistor.
Its readings are V_{battery} = 6.0 V, V_{200kΩ} = 2 V.
What is the value is the internal resistance of the voltmeter and what will the
voltmeter read if it is used to measure the voltage across the 300 kΩ resistor?
Solution:
 Concepts:
Resistors in series and parallel
 Reasoning:
The voltmeter as has a shunt resistance R. It reads the voltage across this
resistance. When placed in a circuit, it correctly reads the battery
voltage but not the voltages across the resistors of the undisturbed
circuit. The battery voltage is V = 6 V.
The voltages across R_{1} (200 kΩ) and R_{2} (300 kΩ)
(without the voltmeter in the circuit) are
V_{1} = 6V R_{1}/(R_{1} + R_{2}) = 2.4 V and
V_{2} = 6V R_{2}/(R_{1} + R_{2}) = 3.6 V.
 Details of the calculation:
With the voltmeter in the circuit we have:
2V = 6V [R_{1}R/(R_{1} + R)]/[(R_{1}R/(R_{1}
+ R)) + R_{2}] or 2V = 6V/[1 + R_{2}(R_{1} + R)/(R_{1}R)].
1 + R_{2}(R_{1} + R)/(R_{1}R)] = 3, R_{2}(R_{1}
+ R)/(R_{1}R) = 2, R_{2}/R + R_{2}/R_{1} =
2, 1/R = 2/R_{2}  1/R_{1}
R = 600 kΩ is the internal resistance of the voltmeter.
When placed across the 300 kΩ resistor the voltmeter reading is
V_{300kΩ} = 6V [R_{2}R/(R_{2} + R)]/[(R_{2}R/(R_{2}
+ R)) + R_{1}] = 3 V.
Problem 2:
A long straight wire is connected to an ideal battery. When the room
temperature is 20.0 ^{o}C, the established temperature of the wire is
22.0 ^{o}C.
The wire is disconnected and onethird of it is cut off. The
remaining piece of the wire is then connected to the same battery in the same
room.
What is the new established temperature of the wire?
Assume Newton's law of cooling applies, i.e. the rate at which a system transfers
heat to its environment is proportional to the temperature difference between
the system and its environment.
Solution:
 Concepts:
Energy dissipated by a resistance, Newton's law of cooling,
equilibrium
 Reasoning:
Let l be the length of the wire that is connected to a voltage source, V.
The wire has resistance that is proportional to its length, R = cl, c =
proportional constant.
The rate at which energy is dissipated by the wire is P = V^{2}/R = V^{2}/cl.
Assume Newton's law of cooling applies.
We therefore have dQ/dt = kl(T_{wire}  T_{room}), k =
proportional constant.
dQ/dt is proportional to (T_{wire}  T_{room}) (Newton's law of
cooling). dQ/dt must also be proportional to the surface are, which is
proportional to the length of the wire.
 Details of the calculation:
In equilibrium P = dQ/dt, V^{2} = ckl^{2}(T_{wire}  T_{room}).
Let T_{1} be T_{wire} for l = l_{1} and T_{2} be
T_{wire} for l = l_{2} = 2l_{1}/3.
Then (T_{1}  T_{room}) = (4/9) (T_{2}  T_{room}).
(T_{2}  T_{room}) = (9/4)*2 ^{o}C = 4.5 ^{o}C,
T_{2} = 24.5 ^{o}C.
Problem 3:
A
long cylindrical solenoid of radius R and length L >> R is tightly wound with a
single layer of wire. The number of turns per unit length is N/L.
The wire breaks when the tension in the wire is greater than T. Find the
maximum current the wire can carry before breaking.
Solution:
 Concepts:
Ampere's law, the Lorentz force
 Reasoning:
The problem has enough symmetry to let us calculate B from Ampere's law alone.
If the axis of the solenoid is the zaxis and the currents flow in the
φdirection, then B = B_{z}k, B_{z}
= μ_{0}IN/L inside the coil and B = 0 outside the coil.
The current in the wires may be treated as a surface current with surface
current density
k. The field very close to a section of surface with
current density
k
may be decomposed into the field B_{1} produced by the
section itself and the field B_{2} produced by all
other surrounding sections. The force per unit length on each wire can be
found by from F= IL ×
B_{2}.
 Details of the calculation:
Consider
a closed loop of length L enclosing N wires of the coil as shown. The current
flowing through the loop is NI, and very near the wires this current NI produces
the same
B_{1}
field as a surface current with density k = NI/L flowing in the same
direction in a flat surface. Ampere's law yields B_{1} = μ_{0}k/2
= μ_{0}In/2 pointing into the z direction on the inside and into the z
direction on the outside of the coil. So the field near the N wires that
is produced by currents not flowing in the vicinity of the loop is
B_{2}
= μ_{0}In/2 k. From
F = IL × B_{2}
we now find the force per unit length F/L = μ_{0}I^{2}n/2
pointing in the positive r direction.
Consider a wire loop of radius R acted on by a radial force. The magnitude of
force per unit length is constant. If the loop is cut into two equal
pieces, then the total force pulling the pieces apart is
F_{z} = 2*(I^{2}nμ_{0}/2)∫_{0}^{π/2}cosθ
Rdθ = I^{2}nμ_{0}R.
For the two pieces of the loop to not accelerate away from each other we need 2T
= F_{z}, where T is the tension in the wire. Therefore T = I^{2}nμ_{0}R/2.
The maximum current the wire can carry before breaking is I = (2TL/(Nμ_{0}R))^{½}.
Problem 4:
A hollow ball with a volume V is held in place in a tank under water by a
wire under a sloped plank as shown in the figure. The water density is ρ and average ball density is ρ/5.
The plank makes an angle α with the horizontal, with tan(α) = 1/3. What is the tension in the wire if
the whole system is accelerating horizontally with acceleration a = g/6.
Solution:

Concepts:
Freebody diagrams
 Reasoning:
Let the xaxis point towards the right and the yaxis point
up. For the net force F on the ball we have F_{y} = 0,
F_{x} = ma.
 Details of the calculation:
The buoyant force acting on the ball is B_{t} = ρVg
j  ρVa i.
F_{y} = mg + B_{y}  Ncosα = ρVg/5 + ρVg  Ncosα = 4ρVg/5  Ncosα = 0.
N = 4ρVg/(5cosα).
F_{x} = Nsinα + T + B_{x} = ma. T = 4ρVg/15  ρVg/30
+ ρVg/6 = 2ρVg/5.
Problem 5:
A glass is filled to height h_{0} with a
volume of water V_{0} with density ρ. A
straw with uniform cross sectional area A is used to drink
the water by creating a pressure at the top of the straw
(P_{top}) that is less than atmospheric pressure (P_{Atm}). The
top of the straw is a distance h_{top} above the bottom of the glass.
(a) Determine the pressure (P_{top}) at the top of the straw as a
function of the height (h) of the water in the glass so that the velocity of
water coming out of the top of the straw remains constant at a value of v_{top}.
(b) How much work is done by the person in order to drink all the water in the
glass with the constant velocity v_{top}?
(c) How much time does it take to drink all the water in the glass with the constant velocity
v_{top}?
Solution:
 Concepts:
Bernoulli's equation, work and energy
 Reasoning:
Bernoulli's equation implies energy conservation. We ignore viscosity.
 Details of the calculation:
Let the y axis point up and the bottom of
the glass be at y = 0.
(a) P_{atm} + ρgh + ½ρv^{2
}= P_{top} + ρgh_{top}
+ ½ρv_{top}^{2}.
Here h is the height of the water in the glass and v is the speed with which h
decreases.
Assume that v is negligible compared to v_{top}. Then
P_{top}
= P_{atm}  ρg(h_{top}  h)  ½ρv_{top}^{2}.
(b)
The water has to be lifted and given some kinetic energy.
The work done lifting the water is
W = ∫_{0}^{h0}A_{glass}ρdh g(h_{top}  h) = A_{glass}ρgh_{0}(h_{top}
 h_{0}/2) = V_{0}ρg(h_{top}  h_{0}/2), since A_{glass}h_{0}
= V_{0}.
W_{kin} = ½V_{0}ρv_{top}^{2}. W_{total}
= V_{0}ρ[g(h_{top}  h_{0}/2) + ½v_{top}^{2}].
(c)
The volume flow rate is Av_{top}. The time it takes to drink all
the water is t = V_{0}/(Av_{top}).