Physics 1214 - General Physics II
Midterm
Solutions
1. Point particles q1 , q2 , and q3 with charge 3.00 nC are situated at each of three corners of a square whose side
length is 0.2 meters. What are the magnitude and direction of the resultant force on a point particle q4 of charge
−1.00 µC if it is placed at the center of a square. Also, compute the potential energy of the system of four particles.
q1
q2
F1,4
F2,4
q4
F3,4
q3
qq ′
We use the formula F = k 2 to determine the forces between two charged particles. Since we can add forces
r
vectorially, we have
→
−
→
−
→
−
→
−
F net = F 1,4 + F 3,4 + F 2,4
However, since q4 has a negative charge and is equidistance from two particles, q2 and q3 which have the same charge,
→
−
→
−
→
−
→
−
the forces F 3,4 and F 2,4 are equal in magnitude and opposite in direction. Thus F 3,4 = − F 2,4 and as a result,
→
−
→
−
F net = F 1,4
We are already given the charges on q1 and q4 so we just need to find r. To find r, we simply form a right triangle
from the midpoint of the left side of the box to the point particle q4 . Then the distance from q1 to q4 will be the
hypotenuse of the triangle whose other two side lengths are 0.1 meters each. Thus
p
√
√
r = 0.12 + 0.12 = 2 · 0.12 = 2 · 0.1 m
We now have
−9
C 1.00 × 10−6 C
−
→
→
−
2 3.00 × 10
9
2
F net = F 1,4 = 8.99 × 10 N · m /C
= 1.35 × 10−3 N
√
2
2 · 0.1m
1
2
′
To compute the potential energy of the system, we use the formula U = k qqr . We have to compute the energy
between each pair of particles. This means
Utot = U1,2 + U1,3 + U1,4 + U2,3 + U2,4 + U3,4
q1 q3
q1 q4
q2 q3
q2 q4
q3 q4
q1 q2
+k
+k
+k
+k
+k
=k
r1,2
r1,3
r1,4
r2,3
r2,4
r3,4
√
√
Here, r1,2 = r1,3 = 0.2 m, r1,4 = r2,4 = r3,4 = 2 · 0.1 m, and r2,3 = 2 · 0.2 m. Plugging these distance values
in, the charges, and the value of k, into the formula for Utot gives
q2 q3
q2 q4
q3 q4
q1 q3
q1 q4
q1 q2
+√
+√
+√
+
+√
Utot = k
0.2 m 0.2 m
2 · 0.1 m
2 · 0.2 m
2 · 0.1 m
2 · 0.1 m
"
−9
−9
3.00 × 10 C 3.00 × 10 C
3.00 × 10−9 C 3.00 × 10−9 C
= 8.99 × 109 N · m2 /C2
+
0.2 m
0.2 m
3.00 × 10−9 C 3.00 × 10−9 C
3.00 × 10−9 C 1.00 × 10−6 C
√
√
+
−
2 · 0.1 m
2 · 0.2 m
#
3.00 × 10−9 C 1.00 × 10−6 C
3.00 × 10−9 C 1.00 × 10−6 C
√
√
−
−
2 · 0.1 m
2 · 0.1 m
= −5.71 × 10−4 N.
3
2. What must the emf E be in order for the current through the 7.00 Ω resistor to be 1.80 A? You may assume that
each emf source has negligible resistance.
+
24.0 V −
I
+
E
1.8 A
−
7.00 Ω
3.00 Ω
2.00 Ω
We will first consider a clockwise outer loop around the 24 V battery, 7 Ω resistor and 3 Ω resistor. The loop rule
here gives
24.0 V − (1.80 A)(7.00 Ω) − I(3.00 Ω) = 0
Solving for I gives I = 3.80 A.
We will now try to solve for E using the loop around the E battery, 7 Ω resistor and 2 Ω resistor. Here we have
Solving for E gives E = 8.6 V.
E − (1.80 A)(7.00 Ω) + (2.00 A)(2.00 Ω) = 0
4
3. A thin 50.0 cm-long metal bar with mass 0.750 kg rests on, but is not attached to two metal supports in a uniform
0.450 T magnetic field. A battery and a 25 Ω resistor in series are connected to the supports. What is the largest
terminal voltage the battery can have without breaking the circuit at the supports?
V
R
−
→
FB
X
X
X
X
X
X
X
X
−
→
B
I
X
X
X
X
w = mg
Max current will be when FB is equal to weight w = mg. Notice that this forces current to go from left to right
across the bar. So we will first calculate FB , which is the force on the entire metal bar with an induced current in a
magnetic field. The formula is
FB = ILB sin(φ)
◦
Here φ = 90 , B = 0.450 T and L = 0.5 m. We do not know I, and we are looking for a voltage V , so setting
I = V /R and FB = mg gives us the equation
V
mg = LB
R
Solving for V gives
(0.75 kg)(9.8m/s2 )(25.0Ω)
mgR
=
= 817 V
V =
LB
(0.500 m)(0.450 T)
5
4. A rectangular circuit is moved at a constant velocity of 3.0 m/s into, through, and then out of a uniform 1.25 T
magnetic field as shown below. The magnetic field region is considerably wider than 50.0 cm. Find the magnitude
and direction (clockwise or counterclockwise) of the current induced in the circuit as it is (a) going into the magnetic
field, (b) within the magnetic field, but still moving, and (c) moving out of the field.
3.0 m/s
−
→
B
75.0 cm
12.5 Ω
50.0 cm
(a) As the loop enters the field, the flux goes from zero to increasing out of the page. The induced magnetic field
must counteract this, and points into the page and the current travels clockwise through the circuit. The magnitude
of the induced emf is E = BLv. In terms of the current I, we also have E = IR. Setting these equal gives IR = BLv.
Solving for I gives
(1.25 T)(0.750 m)(3.0 m/s)
BLv
=
= 0.225 A
I=
R
12.5 Ω
(b) Once the loop is entirely inside the magnetic field, there is no change in flux. If there is no flux, there is no
induced magnetic field and therefore no current, so I = 0.
(c) As the loop leaves the field, the flux points out of the page but is decreasing. To balance this, the induced
magnetic field points out of the page and the current travels counterclockwise through the circuit. Like in part (a),
I = 0.225 A.
6
5. Consider a series R − L − C circuit with R = 200 Ω, L = 0.900 H and C = 2.00 µF at an angular frequency of
1000 rad/s. (a) Compute the impedance of the circuit. (b) What is the phase angle φ of the source voltage with
respect to the current? (c) Construct a phasor diagram for this series R − L − C circuit.
p
(a) To compute the impedance Z, we use Z = R2 + (XL − XC )2 . We are given the resistance R = 200 Ω, but
we need XL and XC . We use the formulas XL = ωL and XC = 1/ωC to get
1
1
=
= 500 Ω
XL = ωL = (1000 rad/s)(0.900 H) = 900 Ω, XC =
ωC
(1000 rad/s)(2.00 × 10−6 F)
Thus,
p
p
Z = R2 + (XL − XC )2 = (200 Ω)2 + (900 Ω − 500 Ω)2 = 447 Ω
(b) To compute the phase angle φ we use
900 Ω − 500 Ω
XL − XC
=
= 2.00
tan(φ) =
R
200 Ω
which gives tan(φ) = 63.4◦.
(c) First, we know that the voltage V leads the current since φ > 0, so VL > VC . This gives us the following
diagram:
VL
V
VL − VC
I
φ
VR
VC