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Prof. Darin Acosta Prof. Paul Avery March 9, 2005 PHY2049, Spring 2005
Exam 2 Solutions
- Two identical light bulbs A and B are connected in series to a constant voltage source. A wire is then connected across B. What is the brightness of A relative to its former brightness?
A light bulb will have a certain resistance R. The light it emits will be proportional to the energy dissipated by this resistance: P = i R^2_. So when two identical light bulbs are connected in series to a constant voltage V, the current passing through the circuit is i_ = V / 2 Rand the power dissipated in light bulb A is: 2 2 (^1 4 )
V V
P R
R R
Now when a wire is connected across light bulb B, that means that the resistance is bypassed by a wire in parallel to the bulb (i.e. the bulb is shorted out by a jumper). The resistance of bulb B becomes zero, and the current in the circuit is i = V / R. This current is twice as high as before, but the power dissipated by light bulb A will be four times more: 2 2 4 1
V
P P
R
Answer: 4
- In the circuit shown, what is the current (in amps) flowing through the 18V battery?
Let i 1 be the current through the 18V battery, i 2 the current through the middle branch
with the 6V battery, and i 3 the current through the rightmost branch.
By Kirchoff’s junction rule: i 1 (^) = i 2 (^) + i 3_._
By Kirchoff’s voltage sum rule applied to the rightmost loop:
2 3 3 2
6V 12 24
i i i i
By Kirchoff’s voltage sum rule applied to the leftmost loop:
1 2 2 3 2 2 2 2 2 2
2
3 2 1 2 3
18V 6 6V 12
i i i i i i i i i i
i
i i i i i
Answer: 1.0 Amps through 18V battery
- An electron is accelerated from rest by a potential difference of 8.0 kV. It then enters a uniform magnetic field of magnitude 1.0 T with its velocity perpendicular to the direction of the field. Calculate the radius in meters of its path in the magnetic field.
Setting the centripetal acceleration, a = v^2^ / r, equal to the acceleration due to the magnetic force, FB / m = qvB / m, yields the following relation for the radius of curvature
for a charged particle moving in a uniform magnetic field: mv r qB
To get the velocity, one must realize that the kinetic energy of the electron after accelerating from rest across an electric potential equals the initial potential energy:
( )( )(^ )
2
31 19 4 19
2 2 9.11 10^ kg^ 1.6^10 8000 V 3.0 10 m 1.6 10
E (^) K mv eV
mv meV
meV r qB
− − − −
× ×
⇒ = = = ×
×
Answer: 3.0 × 10 −^4 m
- A solenoid is 3.0 cm long and has a radius of 0.50 cm. It is wrapped with 500 turns of wire carrying a current of 2.0A. What is the magnetic field at the center of the solenoid?
One can apply Ampere’s Law to determine that for a very long solenoid, the magnetic field inside is given by:
B = μ 0 ni, where n is the number of windings per unit length and i is the current in the
wire. This formula will hold approximately for the center of a solenoid of finite length. Thus,
( ) (^ )
7 2 0
4 10 2 4.2 10 T
B = μ ni = π × −^ ^ = × −
Answer: 4.2 × 10 −^2 T
- A capacitor with an initial potential difference of 100V is discharged through a resistor when a switch between them is closed at t =0. At t =5 s, the potential difference across the capacitor is 50V. What is the time constant of the circuit?
A capacitor in series with a resistor satisfies the following differential equation by applying Kirchoff’s voltage sum rule:
0
dq q R dt C
The solution is: q t ( ) = q e 0 −^ t^ / RC
So the voltage on the capacitor is C ( ) 0^ t^ /^ RC^ 0 t^ / RC
q V t e V e C
= −^ = −
Plugging in for what we know:
0 5/ 0
0 100 V
5 50 V
ln 2
5 7.2 s ln 2
C RC C
RC
V V
V t V e
RC
τ RC
−
Answer: 7.2 s
- In the figure shown, two long straight wires with separation d carry currents i 1 and i 2 =2 i 1 out of the page. At what point on the x -axis shown is the net magnetic field due to the currents equal to zero?
The magnetic field from each wire will be in the same direction for x < 0 and x > d. Only for 0 < x < d are the fields lines opposite in direction. (Use the right-hand rule to get the direction). Thus, the net magnetic field is:
0 1 2
1 2 1 1
i i x d x i d x i x i d x i x d d x x
Answer: d /
- A horizontal power line carries a current of 7000A from south to north. Earth's magnetic field, with a magnitude of 60μT, is directed toward the north with a dip angle 60° downward relative to the horizontal. Find the magnitude of the magnetic force acting on a 100 m length of power line.
The magnetic field and the power line both point north, but the magnetic field points into the Earth at an angle of 60°.
The force on a current carrying wire is given by: F = i L × B
So the force points west with magnitude:
F = iLB sin θ = ( 7000 )( 100 )( 60 × 10 −^6 )( sin 60 )=36.4 N
Answer: 36.4 N
S N
i
B
