國立臺灣海洋大學資工系 1A 普通物理 97(2)作業 8 解答 (9800525/0601)
此資料專為教學用請勿流傳-楊志信
Ex.8-1, Prob.30-11 (HRW8e) Sol.
(a) Let ℓ be the length of a side of the square circuit.
Then the magnetic flux through the circuit is ΦB =
ℓ2B/2, and the induced emf is
A 2 dB
d
.
εind = − Φ B = −
dt
2 dt
Now B = 0.042 – 0.870 t and dB/dt = –0.870 T/s.
2.00 2
Thus, εind = −
( −0.870) = 1.74 (V).
2
The magnetic field is out of the page and decreasing
so the induced emf is counterclockwise around the
circuit, in the same direction as the emf of the
battery. The total emf is
ε + εi = 20.0 V + 1.74 V = 21.7 V.
(b) The current is in the sense of the total emf
(counterclockwise).
Ex.8-3A, Prob.30-92 (HRW8e) Sol.
(a) The self-inductance per meter is
L/ℓ = μ0 n2 A = (4π×10−7)(1×104)π(0.016)2
= 0.101 ≈ 0.10 (H/m).
(b) The induced emf per meter is
ε/ℓ = (L/ℓ)(di/dt) = (0.101)(13) = 1.31 ≈ 1.3 (V/m).
Ex.8-3B, Prob.30-98 (HRW8e) Sol.
(a) We use UB = Li2/2, to solve for the selfinductance:
L = 2UB/ i2
= 2(25.0×10−3)/(60.0×10−3)2 = 13.88 ≈ 13.9 (H).
(b) Since UB ∝ i2, for UB to increase by a factor of 4,
i must increase by a factor of 2. Therefore, i should
be increased to 2(60.0 mA) = 120 mA.
Ex.8-2, Prob.30-85 (HRW8e) Sol.
In Fig.30-77, the battery is ideal, ε = 10 V, R1 = 5.0
Ω, R2 = 10 Ω, and L = 5.0 H. Switch S is closed at
time t = 0. Just afterwards, what are (a) i1, (b) i2, (c)
the current iS through the switch, (d) the potential
difference V2 across resistor 2, (e) the potential difference VL across
the inductor, and (f) the
rate of change di2/dt? A
long time later, what are
(g) i1, (h) i2, (i) iS, (j) V2,
(k) VL, and (1) di2/dt?
Sol. (a) When switch S is just closed,
V1 = ε and i1 = ε/R1 = 10 V/5.0 Ω = 2.0 A.
(b) Since now VL = ε, we have i2 = 0.
(c) iS = i1 + i2 = 2.0 A + 0 = 2.0 A.
(d) Since VL = ε, V2 = ε –VL = 0.
(e) VL = ε = 10 V.
(f) di2/dt = VL /L = 10 V/ 5.0 H = 2.0 A/s.
(g) After a long time, we still have V1 = ε, so i1 = 2.0 A.
(h) Since now VL = 0, i2 = ε/R2 = 10 V/10 Ω = 1.0 A.
(i) is = i1 + i2 = 2.0 A + 1.0 A = 3.0 A.
(j) Since VL = 0, V2 = ε – VL = ε = 10 V.
(k) VL = 0.
(l) di2/dt = VL /L = 0 .