Ch20P 1,9,11,17,19,31,47,57,59,65,69,73,79,83,89,95
ΦM = B⊥A. In this case B⊥ = 1.2 mT, and A = 25cm2, so
ΦM = B⊥A = (1.2 × 10−3 T)(25 × 10−4 m2) = 3.0 × 10−6 T・m2 = 3.0 μWb.
1 The magnetic flux is given by Eq. (20.1):
B-field the magnetic flux through the loop is ΦMi = BA⊥,
where A⊥ is the encompassed area of the loop perpendicular to the B-field. In a time interval Δt
the B-field through the loop is gone so ΦMf = 0. The induced emf, which is equal in
magnitude to the rate of change of the magnetic flux in the loop, is then [see Eq. (20.3)]
0 BA
NBA
mf
mi
m
V
N
N
N
t
t
t
t
9 Before the loop is yanked from the
Plug in N = 1, B = 0.40 T, A⊥ = 0.25m2, and Δt = 200 ms = 0.200 s to obtain E = +0.50 V.
(Here the plus sign in E means that the induced emf would generate a current which in turn
produces a B-field whose flux is positive through the loop, as is ΦMi.)
11 The induced emf due to a flux change in a time interval
Δt is given by E = −NΔΦM/Δt. Take the
absolute values to obtain its magnitude|E| = N|ΔΦM|/Δt.
Plug in E = 60V, ΔΦM = 30mWb, and N = 100, and solve for Δt:
t
N
m
V
(150)(30mWb)
60V
75ms
B-field in the solenoid varies as a function of the current I it carries: B = μ0nI. The
magnetic flux of this field through the coil of cross-sectional area A is ΦM = BA = μ0nAI. As
the current in the solenoid changes by ΔI during a time interval Δt the corresponding change in ΦM
is ΔΦM = μ0nAΔI, which causes an emf in the N-turn coil:
N o nA I
(240)(4 E 7T m / A)(1000 / m)(2 E 4m2 )(4.9 A)
m
V
N
t
t
5E 3s
59mV
Note that n = 10/cm = 10 × 102/m.
17 The
19 The magnetic flux provided by the
B-field of the electromagnet through the loop of area A
is ΦM = BAcos θ [see Eq. (20.1)], where θ (= 30◦) is the angle between _B and the normal
direction of the plane containing the loop. As B changes from Bi to Bf the corresponding change in
ΦM is ΔΦM = BfAcos θ−BiAcos θ = (Bf −Bi )Acos θ, which causes an induced emf given by E =
−NΔΦM/Δt. Plug in N = 1, Bi = 0, Bf = 0.500 T, A = 4.0 cm×4.0 cm, θ = 30◦, and Δt = 200 ms to obtain
(0.5T 0)(.04m)2 cos(30o )
3.5mV
t
t
0.2s
Here the minus sign indicates that the induced emf tends to run a current whose magnetic field
produces a flux opposite in direction to the original flux, which is downward. Thus the
induced current must be counterclockwise viewed from the top so as to produce an upward
flux.
V
N
m
N
(Bf
Bi ) A cos( )
1
31 The wire moves at
v = 0.60 m/s relative to the magnetic field lines. Thus Eq. (20.4) gives
E = vBl = (0.60 m/s)(0.200T)(1.00m) = 0.12V,
with the east end of the wire at the higher voltage (positive).
47 Apply Eq. (20.4) to find the emf due to each length:
E = vBl = (22 m/s)(0.35T)(0.10m) = 0.77V.
Since there are 2 lengths per turn and a total of 25 turns the total emf is 2×25×0.77V =
39V.
57 (a) According to Eq. (19.1)
μ = 2πBr/I, where B is in T, r is in m, and I is in A. So the SI
unit of μ is T・m/A.
(b) The magnetic flux has the unit of magnetic field (B, in T) times area (A, in m2 ), i.e.,
T・m2.
(c) Solve for L from Eq. (20.8): L = NΦM/I. The unit of ΦM is T・m2, that of I is A, while
N is unitless. Thus H, the unit of L, is equivalent to T・m2/A.
(d) Solve for μ0 from Eq. (20.9): μ0 = Ll/N2A. Here L is in H, l is in m, A is in m2, while
N is unitless. Thus the unit of μ0 is H・m/m2 = H/m. [Note that this is consistent with the
expression T・m/A found in part (a) above, given that 1H = 1T・m2/A.]
L is defined in Eq. (20.8): NΦM = LI, where the magnetic flux ΦM in the
coil is caused by the current I running in the N-turn coil itself. Thus
59 The self-inductance
L
65
N
LI
m
69
L
(2.5H )(1.80 A)
1.8mWb
V
I/ t
10V
2.0 A / s
N
m
I
500(2mWb)
3.8 A
0.26 H
2.5 103
5.0 H
Note that we took the absolute value here since we know that L > 0. (In fact the signs of E
and ΔI/Δt are always opposite to each other so as to keep L positive.)
ΔI = −5.0A−(+5.0A) = −10 A, which took place in a time interval
Δt, generating an emf of E = −L(ΔI/Δt) [see Eq. (20.10)]. Plug in L = 1.5 H, E = 100 V,
and solve for Δt:
73 The change in current is
t
L I
V
(1.5H )( 10 A)
100V
0.15s
79 (a) from the problem statement
5τ ≈ 2.5 s, so τ ≈ 0.50 s.
(b) For an R-L circuit τ = L/R.
(c)
L
1.0 H
2.0
0.5s
(d) The current varies as a function of time as I = Im(1 − e−t/τ ), where Im = V/R is the
steady-state current when t _ τ . Plug in V = 12V, R = 2.0Ω, and t/τ = 10 to obtain
V
12V
I
1 e t/
1 e 10 6.0 A
R
2
Note that the factor e−10 is negligible in comparison with 1, meaning that the current has
practically reached its steady-state value at t = 10τ .
(e) Zero, as the current through the inductor has reached a constant value: VL ∝ ΔI/Δt = 0.
R
83 Initially there is no current:
I = 0. Thus Eq. (20.11) reads V = L(ΔI/Δt), which gives the
voltage of the battery to be
I
V
t
where we used Eq. (20.10). Solve for ΔI/Δt:
I V V
20V
V
t
L
L
L
10mH
20.0V
2.00 103 A / s
89 (a) Similar to the previous problem, the milliammeter is in series with an inductor so the
current I is the same for both. Before S1 is closed the current I through the inductor is zero.
An instant after it is closed I should remain zero, as it cannot change abruptly in an inductor.
(b) After a long time the current I must have reached Im, its steady-state value, whereupon
ΔI/Δt = 0. So the voltage difference across the inductor, being proportional to ΔI/Δt, drops
to zero. The load of the circuit is the 10-Ω resistor, so now I = Im = E/R = 10V/10Ω = 1.0 A.
L 2mH
0.2ms
(c)
R 10
(d) Before S1 is open and S2 is closed the current reading is 1.0 A, as discussed in part (b).
Just after that, the current should remain 1.0 A, in the same direction (positive), since no
abrupt change in I is possible in an inductor.
(e) The current varies as a function of time as I = Ime−t/τ , where τ = R/L is the time
constant. We wish to solve for solve for t/τ . First rewrite it as I/Im = e−t/τ , then take the
natural logarithm of both sides: ln(I/Im) = lne−t/τ = −t/τ. Thus
I
0.14 A
n
Im
1.0 A
(f) Since τ = 0.20 ms, t = 2.0 τ = 2.0(0.20 ms) = 0.40 ms.
t
n
95 The energy stored in an inductor=
1 2
LI
2
1.97
1
(40mH )(2.0 A) 2
2
80mJ