1.
Applying Kirchhoff’s Rules, (a) find the current through the 12 Omega resistor. (b) Find the potential difference between a and b.
i
1
= i
2
+ i
3
6 i
1
+ 4 i
2
− 8 − 18 = 0
12 i
3
+ 8 − 4 i
2
= 0
Now, it’s just algebra. Plugging (1) into (2).
10 i
2
+ 6 i
3
= 26
Solve for i
3
, i
3
=
26 − 100 i
2
6
Plug into (3) and solve for i
2
,
2(26) + 8 − 20 i
2
− 4 i
2
= 0 so,
Use i
2 to get i
3
V ab is the voltage across the 12Ω Resistor, so i
3
=
26 − 10(5
6
/ 2)
= 1 / i
1
= i
2
+ i
3
= 16 / 6 A
6 A
V ab
= (12) i
3
= 2 V i
2
= 5 / 2 A
2.
A 5 µF capacitor is charged in a circuit with a 100V battery and 25 k Ω R . After the capacitor is fully charged, the battery is disconnected and the capacitor is allowed to discharge through the resistor.
(a) What is the maximum charge the capacitor can obtain?
(b) How long will it take for the capacitor to reach 1/2 of its maximum possible voltage?
(c) How much heat is lost in the resistor while the capacitor is discharging?
Discharging, q = C (1 − e − t
RC
)
V q max
= C = 5 µF (100 V ) = 50 mC max
= 100 V so, V
1 / 2 max
= 50 V
V
1 / 2 max
= V max
(1 − e − t
RC
) ln (1 / 2) =
− t
RC giving t = 0 .
0866 s q = q
0 e − t
RC
E = i = dq dt
=
− q
0
RC e − t
RC
Z
P = i 2 R = inf
0
P dt = q 2
0
RC 2
( C ) 2
RC 2 e −
2 t
RC
RC
2
R e − u |
0
E =
− 2 C
2
= 0 .
025 J
1

3.
Calculate the maximum kinetic energy of a proton accelerated in a cyclotron of radius 0.5 m and a magnetic field of 0.35 T.
4.
v =
RqB m
K = mv 2
2
= m (
RqB m
) 2
2
=
( Rqb ) 2
2 m
= 2 .
35 ∗ 10 −
13 J
Explain how the Hall effect works. How were the scientists able to determine that charge carriers were negative?
The Hall Effect was used to determine that charge carriers were negative. Send a current in the negative y direction down a thick conducting strip inside a B field that is directed perpendicular to the current and into the page. First, assuming, charge carriers were positive, the velocity of the charges would be in the direction of the current, then using F = q ( v × B ), the charges carriers would collect on the right (positive x) side of the metal strip. This would result in a positive voltage collective on the positive x side of the strip. If the charge carriers were negative, the velocity of the carriers would be in the opposite direction of the current. Again the force would push the carriers to the positive x side of the strip causing a negative charge buildup there and a negative voltage. The scientists saw a negative voltage most of the time causing them to believe that the charge carriers were negative, electrons.
5.
A conducting bar of mass = 100g hangs from two identical springs. Initially, the springs stretch 2cm due to the weight of the bar. A current of 2 A is run through the bar and a magnetic field is applied perpendicular to the current. If the springs stretch another 3cm, what is the magnitude and direction of the applied B field?
First, find the spring constant
F spring
+ F g
= 0 = 2 kx − mg = 0
So, k = mg
2 x
=
( .
1 kg )(9 .
8 m/s 2 )
2( .
02 m )
= 24 .
5
Now, we add the B field and we stretch more. Again sum the forces
F
B
+ F g
+ F spring
= 0
Since the B field is perpendicular ilB + mg − 2 kx = 0
(2 A )( l )( B ) + ( .
1 kg )(9 .
8 m/s 2 ) − 2(24 .
5)( .
05 m ) = 0
B = 18 .
4( l ) T out of the page
6.
Two long parallel conductors carrying current, I
1
= 3 A and I
2
= 2 A into the page. Determine the magnetic field at point P.
2

Using superposition and Ampere’s Law
B
1
=
µ
0
I
1
2 π ( .
05 m )
( − cos (67 .
3)ˆ − sin (67 .
y )
B
2
=
µ
0
I
2
2 π ( .
12)
( − cos (22 .
6)ˆ + sin (22 .
y
B = B
1
+ B
2
= − 7 .
6 ∗ 10 −
6 ˆ − 9 .
8 ∗ 10 −
6 ˆ
7.
A coaxial cable has an inner conducting cylinder of radius, a, surrounded by a thin conducting outer shell. Each conductor carries a current, I, as shown. Find the magnetic field in the regions:
(a) r < a
(b) a < r < b
(c) r > b
Use Ampere’s law, I
B · ds = µ
0 i enc
For a > r , we are not enclosing all the current, so we need to find a current density first, and use that to get the enclosed current.
B (2 πr ) = µ
0 i enc i enc
= j ( πr 2 ) where j = i
πa 2 so,
B =
µ
0 ir
2 πa 2
For a < r < b , we are enclosing all the current, so i enc
= i
B ((2 πr ) = µ
0 i enc
= µ
0 i
B =
µ
0 i
2 πr
For r > b , i enc
= + i − i = 0
So,
B = 0
3