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b.
Practice Problems
Magnetic field lines are (toward or away from) the N
geographic pole of the earth.
The earth's north geographic pole is nearest the
earth's (north or south) magnetic pole.
1.
a.
Charge
+
v
up
B
south
FB
east
b.
–
south
down
west
c.
+
east
down
north
d.
–
east
south
up
e.
+
west
south
up
f.
+
west
down
south
g.
–
north
east
up
h.
+
down
south
west
2.
FB = qvB = (1.6 x 10-19 C)(3 x 106 m/s)(1.30 T)
FB = 6 x 10-13 N south
3.
4.
FB = ILB
FB = (2500 A)(0.05 m)(1.2 T) = 150 N south
a.
FB = Fc  qvB = mv2/r  v = qBr/m
v = (2 x 1.6 x 10-19)(2)(0.01 m)/6.6 x 10-27) = 9.7 x 105 m/s
b.
K = ½mv2
K = ½(6.6 x 10-27 kg)(9.7 x 105 m/s)2 = 3.1 x 10-15 J
c.
p = mv
p = (6.6 x 10-27 kg)(9.7 x 105 m/s) = 6.4 x 10-21 kg•m/s
5.
a.
current
east
side
above
B
south
b.
down
north
east
11. a.
qV = ½mv2  V = ½mv2/q
V = ½(9.11 x 10-31)(2 x 107)2/-1.6 x 10-19 C = -1100 V
b. (up or down).
c.
FB = qvB
FB = (1.6 x 10-19 C)(2 x 107 m/s)(6 x 10-4 T) = 1.9 x 10-15 N
d.
FB = Fc = mv2/r  r = mv2/FB
r = (9.11 x 10-31 kg)(2 x 107 m/s)2/(1.9 x 10-15 N) = 0.19 m
e.
FB = Fe = qE
E = FB/q = 1.9 x 10-15/1.6 x 10-19 = 1.2 x 104 N/C north
12. a.
down
b.
Fe = FB  qE = qvB
v = E/B = 1000 N/C/0.5 T = 2000 m/s
13.
B = k'I/r = (2 x 10-7 T•m/A)(100 A)/1.0 m
B = 2 x 10-5 T (up)
14.
B = oI(N/L) = (4 x 10-7 T•m/A)(6.0 A)(10,000/0.50 m)
B = 0.15 T (up)
15. a.
FB = Fc  qvB = mv2/r
(1.60 x 10-19)(1 x 10-4)r = (9.11 x 10-31)(1 x 106)  r = .057 m
b.
counterclockwise
16. a.
c.
south
south
none
d.
north
east
down
B = k’I/r = (2 x 10-7)(5 A)/0.5 m = 2 x 10-6 T (up)
e.
west
below
south
b.
f.
east
below
north
FB = ILB = (2 A)(0.1 m)(2 x 10-6 T) = 4 x 10-7 N, (north)
g.
north
below
west
h.
up
north
west
i.
north
down
west
j.
down
west
north
k.
east/west
west
none
l.
west
south
up
6.
B = k'I/r = (2 x 10-7 T•m/A)(15 A)/0.05 m
B = 6 x 10-5 T (up)
7.
down
8.
c.
B = oI(N/L)
B = (4 x 10-7 T•m/A)(2.0 A)(400/0.10 m) = 0.010 T
9. a.
B = k'I/r = k'IA/r
direction is into the page
b.
FB = ILB = IBL(k'IA/r) = k'IAIBL/r
direction is away from wire A
10. a. Magnetic field lines are (toward or away from) the N
pole of a permanent magnet.
17. a. (1)
B between the clockwise current loops
FB on the electron beam
E that would keep the electron beam undeflected
(2)
F B = Fe
qvB = qE  v = E/B
b. (1) (a, b or c)
(2)
FB = Fc  qevB = mev2/r
me/qe = Br/v (v = E/B when Fe + FB = 0) = B2r/E
18. a.
B
B = A x B = (0.01 m2)(3 T) = 0.03 Wb
E
E = B/t = 0.03 Wb/0.1 s = 0.3 V
I
I = E/R = 0.3 V/10  = 0.03 A
b.
clockwise
19. a.
B
E
B = B x A = (0.50 T)(1 m2) = 0.50 Wb
E = B/t = 0.50 Wb/1 s = 0.50 V
down
south
south
I
I = E/R = 0.50 V/5  = 0.10 A
D
b.
clockwise
20. a.
clockwise
south
in the rod
c.
I = E/R = 40 V/2  = 20 A
d.
FB = ILB = (20 A)(2 m)(2 T) = 80 N
A
P = Fv = (80 N)(10 m/s) = 800 W
D
P = IV = (20 A)(40 V) = 800 W
g.
Energy/Power is conserved. The mechanical power is
converted into electrical power.
21.
I = E/R = vLB/R = (3 m/s)(1 m)(0.2 T)/1  = 0.6 A
FB
FB = ILB = (0.6 A)(1 m)(0.2 T) = 0.12 N
P
P = Fv = (0.12 N)(3 m/s) = 0.36 W
22. a.
E = /t = (B x A)/t = B(r2)/t = (5 T)()(0.1 m)2/0.1 s
E = 1.57 V
b.
I = E/R = 1.57 V/10  = 0.157 A, clockwise
E = B/t = 0.05 Wb/0.01 s = 5 V
12.
B
I = E/R = 5 V/10  = 0.5 A
13.
D
14.
Right hand, thumb (B) up, don't flip (decreasing flux),
fingers (I) wrap counterclockwise.
Right hand, thumb (B) down, don't flip, fingers (I)
wrap clockwise. I = E/R = vLB/R = (5)(0.2)(2)/(20) = 0.1 A
15.
D
f.
B = A x B = (0.01 m2)(5 T) = 0.05 Wb
 rotating to || = decreasing flux  0.05 Wb to 0 Wb.
11.
C
e.
I
E = (B x A)/t: A and B must remain unchanged with
respect to each other  D
10.
E = BLv = (2 T)(2 m)(10 m/s) = 40 V
b.
in the U-shaped rail and rod
9.
E = (B x A)/t: A and B must remain unchanged with
respect to each other  D
16.
A
There is no work done because the magnetic force is
perpendicular to the direction of motion (W = F||d).
17.
C
Fc = FB  mv2/r = qvB 
v = qrB/m = (10-19)(10-1)(10-1)/(10-27) = 10-21/10-27 = 106
18.
C
E = B/t = (A x B)/t: Even though A changes, the
amount of B enclosed by A doesn't change  E ' = E
19.
A
E = vLB  E  v with a positive slope.
20.
C
Only loop 1 has B (B = k'I/r). B is down on side of
loop, thumb (B) down, don't flip, fingers (I) clockwise.
21.
D
When currents are in the same direction, the force is
attracting  toward X and away from Z.
22.
Practice Multiple Choice
1.
A
2.
D
3.
C
Left hand, thumb (v) south, fingers (B) down, palm
(FB) west  FB is west.
C
23.
C
Right hand, thumb (I) east, hand points (side) north,
curl fingers 90o (B) up  B is up.
24.
Right hand, thumb (I) north, fingers (B) up, palm (FB)
east  FB is east.
25.
C
B
4.
C
5.
A
6.
C
7.
D
v = 2r/T
eBr/m = 2r/T  T = 2m/eB
Thumb (B) up, don't flip, fingers (I) counterclockwise.
Fingers (I) counterclockwise, thumb (B), points up.
In copper sheet, thumb (I) north, fingers (B) down,
palm (FB) west  P2 is at a higher V.
F B = Fe
qvB = qE  v = E/B = 6.0 N/C/2.0 T = 3 m/s
Right hand, thumb (I) left, fingers (B) up, palm (FB)
away from center  loop will expand.
FB = k'I1I2L/r: k', L and r don't change only currents
are doubled  FB  I1I2  (2)(2) = 4 x original F.
FB = k'I1I2L/r: k', L and r don't change only I changed.
If force is quadrupled then the product (I1)(I2) = 4.
8.
D
FB = FC  qvB = mv2/r  v = rqB/m  eBr/m. fingers
(B) down, palm (FB) toward center  counterclockwise
Left hand, thump (v) east, fingers (B) down, palm (FB)
south. The path is clockwise with constant radius.
Practice Free Response
1.
a.
B = B x A = (0.030 T)(0.20 m)2 = 1.2 x 10-3 Wb
b.
 = BA = (0.20 T – 0.030 T)(0.20 m)2 = 6.8 x 10-3 Wb
E = /t = 6.8 x 10-3 Wb/0.50 s = 0.0136 V
c.
(1)
I = E/R = 0.0136 V/0.60  = 0.0227 A
(2)
Right hand, thumb (B) down, flip hand (increasing
flux), fingers (I) wrap counterclockwise.
2.
3.
a.
Right hand, thumb (I) "north", palm (FB) "east", 
fingers (B) "up"
b.
F = ILB = ma
a = ILB/m = IDB/M
c.
v2 = vo2 + 2ad
v2 = 0 + 2(IDB/M)(L – 0)  v = (2IDBL/M)½
d.
v = (2IDBL/M)½
v = [(2)(200 A)(0.1 m)(5 T)(10 m)/(0.5 kg)] ½ = 63 m/s
e.
P = FBvav = (IDB)½(2IDBL/M)½
P = (I3D3B3L/2M)½
P = [(200 A)3(0.10 m)3(5 T)3(10 m)/2(0.5 kg)]½ = 3160 W
a.
The direction of the magnetic field is Into the page.
b.
The direction of the magnetic force on the particle is at
right angles to the velocity  no work is done by the
magnetic field and the speed is constant.
c.
FB = FC  qvB = mv2/r
(3.20 x 10-19 C)(0.090 T) = (1.45 x 10-25 kg)v/(1.75/2 m)
v = 1.74 x 105 m/s
d.
qV = ½mv2
(3.2 x 10-19 C)V = ½(1.45 x 10-25 kg)(1.74 x 105 m/s)2
V = 6860 V