Download Ch. 29/30 Practice Test — Solution

Ch. 29/30 Practice Test — Solution
Multiple Choice
1. (b) Torques from the magnetic force on the top and bottom wire cancel and any force on the
wire on the left contributes no torque, so we only need to consider the wire on the right. Since
everything is nice and perpendicular,
τ = rF = rIlB = (0.3 m)(2 A)(0.3 m)(0.05 T) = 0.009 N · m.
2. (b) Ampère-Maxwell equation:
3. (b) Fc = FB , so
mv 2
r
H
B · ds = µ0 0 dE
dt .
= qvB, which shows that B and v are inversely proportional.
4. (e) Magnetic field inside an ideal solenoid is uniform.
5. (e) The magnetic field does no work because the force is always perpendicular to the velocity.
6. (b)
~ = (15 A)(2.0î) m×(30î−40ĵ) mT = [30î×(−40ĵ)] mN = (−1200k̂) mN = (−1.20k̂) N
F~B = I~l×B
Alternatively, by the right hand rule, F~B points down. Its magnitude equals the area of the
~ which has base length |I~l| = (15 A)(2.0 m) and height
parallelogram created by I~l and B,
~
40 mT (the y-component of B).
7. (d) In the velocity selector, we set FE = FB,1 , or qE = qvB1 ⇒ v =
E
B1 .
In the magnetic field,
Fc = FB,2
mv 2
= qvB2
r
mv
mE
(9.11 × 10−31 kg)(4.0 × 103 V)
B2 =
=
=
qr
qrB1
(1.602 × 10−19 C)(0.004 m)(2.0 × 10−3 T)
= 0.00284 T
8. (d) Using B =
µ0 I
2πr ,
B=
µ0 I
µ0 I
4 µ0 I
4
+
= ·
= B0 .
2πa 2π(3a)
3 2πa
3
1
9. (b) The magnetic field inside the solenoid is perpendicular to the surface, so
Z
~ · dA = B(πr2 ) = µ0 N I · πr2
ΦB = B
l
−7
(4π × 10 )(8000)(5.0 A)
· π(0.020 m)2 = 1.58 × 10−5 Wb
=
4
Free Response
10. a. By conservation of energy,
∆U = Kf
1
q∆V = mv 2
2
1
eE = mv 2
2
mv 2
E=
.
2e
b. F~B points up initially, so the path is a half circle upward.
~ and ~v are perpendicular,
c. Within the magnetif field, Fc = FB . Since B
mv 2
mv
mv
= qvB ⇒ r =
=
.
r
qB
eB
~
d. ii. By Newton’s second law, F~e must oppose F~B , so from F~e points down. As F~e = q E
~
and q < 0, E points up.
P
i. By Newton’s second law,
Fy = 0 for the electron, so
FB = Fe
qvB = qE
E = vB.
11. a. In each case, take a circle of radius r sharing the axis of the cable as the closed path over
~ is parallel to ds at every point on this path.
which to apply Ampère’s law. Note that B
i. Since the current is uniformly distributed,
I
B · ds = µ0 Iin
B(2πr) = µ0 I ·
B=
2
r2
a2
µ0 Ir
.
2πa2
ii.
I
B · ds = µ0 Iin
b.
c.
d.
e.
B(2πr) = µ0 I
µ0 I
B=
.
2πr
Proceed as in part a. Since Iin = I − I = 0, B = 0.
B should increase linearly up to r ≤ a, be inversely proportional for a ≤ r ≤ b, and
decrease (concave up) until it reaches 0 at r = c.
~ are perpendicular, FB = qvB. Substituting B found in part a.i, FB =
i. Since ~v and B
qvµ0 I
2πr .
ii. Toward the center.
The answer would not change. Since the new set-up still maintains a radial symmetry, the
~
external current does not affect B.
12. a. Both toward the center (test charge is positive). Magnitude is smaller at P2 .
b. In each case, take as gaussian surface a cylinder of radius r and length l with the same
axis as the nonconductor. Since the electric field is radially outward, only the side of
~ is perpendicular to the side. By
the cylinder contributes to the electric flux. Also, E
Gauss’s law,
i.
I
~ · dA = qin
E
0
ρV
E(2πrl) =
0
ρ(πR2 l)
E=
(2πrl)0
ρR2
=
.
20 r
ii. The only difference from part i is qin , which is proportional to the base area of the
gaussian surface, so
ρR2 r2
ρr
E=
· 2 =
.
20 r R
20
c. (Use the right hand rule.)
d. In Ampère’s law, integrate over a circle of radius r sharing the axis of the conductor. Since
~ is parallel to ds everywhere on this path and the current is uniform,
B
I
~ · ds = µ0 Iin
B
B(2πr) = µ0 I ·
B=
3
r2
R2
µ0 Ir
.
2πR2