UWIS, Physik II, Lösung Serie 5

Thomas Kuster

26. November 2004

1 Metallflugzeug

s_{AB} = 20m Spannweite (von A nach B)
v = 50ms^{−1} Geschwindigkeit
B = 43μT = 4.3 · 10^{−5}T Magnetfeld senkrecht zur Flugrichtung
U_{AB} = V induzierte Spannung von A nach B
F_{L} = qvB sin φ  (1)
F_{E} = Eq  (2)

(3)

0 = F_{L} + F_{E} {(1) & (2) \atop =} qvB \underbrace{sin φ}_{v⊥B⇒1} +Eq ⇒ vB + E = 0 ⇒ E = −vB (4)

U_{AB} = Es_{AB} {(4) \atop =} −vBs_{AB} = −50ms^{−1} 4.3 · 10^{−5}T 20m =

= −0.043m^{2}s^{−1}Vsm^{−2} ⇒ |U_{AB}| = 0.043V

2 Rechteckige Spule

a = 0.4m Seitenlänge der Spule
b = 0.3m Seitenlänge der Spule
A⃗ = (\begin{matrix}0.4 \\ 0 \\ 0 \end{matrix}) m × (\begin{matrix}0 \\ 0.3 \\ 0 \end{matrix}) m = (\begin{matrix}0 \\ 0 \\ 0.12 \end{matrix}) m^{2} Flächennormale
N = 12 Windungen der Spule
I = 3A Strom in der Spule
B⃗ = (\begin{matrix}0.3 \\ 0 \\ 0.4 \end{matrix}) T homogenes Magnetfeld

m⃗ = NIA⃗
M⃗ = m⃗ × B⃗

2.1 Magnetisches Moment der Spule (m⃗)

(5) ⇒ m⃗ = NIA⃗ = 12 · 3A (\begin{matrix}0 \\ 0 \\ 0.12 \end{matrix}) m^{2} = (\begin{matrix}0 \\ 0 \\ 4.32 \end{matrix}) Am^{2}

2.2 Drehmoment der Spule (M⃗)

(5) & (5) ⇒ M⃗ = m⃗ × B⃗ = (\begin{matrix}0 \\ 0 \\ 4.32 \end{matrix}) Am^{2} × (\begin{matrix}0.3 \\ 0 \\ 0.4 \end{matrix}) T =

(\begin{matrix}0 \\ 1.296 \\ 0 \end{matrix}) Cs^{−1} m^{2} JC^{−1}sm^{−2} = (\begin{matrix}0 \\ 1.296 \\ 0 \end{matrix}) J = (\begin{matrix}0 \\ 1.296 \\ 0 \end{matrix}) Nm

3 Ferromagnetismus

\frac{n}{l} = 50Windungen cm^{−1} = 5000Windungen m^{−1}
I = 2A Strom in der Spule
B = 1.72T Magnetfeld total
μ_{0} = 4π · 10^{−7}VsA^{−1}m^{−1}

3.1 Angelegtes Feld (B_{0})

B_{0} ≈ \frac{μ_{0}nI}{l} = 4π · 10^{−7}VsA^{−1}m^{−1}5000m^{−1}2A = 0.0126T

3.2 Magnetisierung (M)

B = B_{0} + μ_{0}M ⇒ M = \frac{B − B_{0}}{μ_{0}} = \frac{1.72T − 0.0126T}{4π · 10^{−7}VsA^{−1}m^{−1}} =

1358732Vsm^{−2}AmV^{−1}s^{−1} ≈ 1.359 · 10^{6}Am^{−1}

3.3 Permeabilität (μ)

B = μB_{0} ⇒ μ = \frac{B}{B_{0}} = \frac{1.72T}{0.0126T} ≈ 137

UWIS, Physik II, Lösung Serie 5 3

4 Induktion

B = 0.6T Magnetfeld
v = 8ms^{−1} Geschwindigkeit des Metallstab
l = 0.15m Metallschienenabstand
R = 25Ω Widerstand

4.1 Spannung (U)

U = −\frac{BΔA}{Δt} = −B \frac{lΔs}{Δt} = −Blv = −0.6T · 0.15m · 8ms^{−1} =

−0.72Vsm^{−2}m^{2}s^{−1} = −0.72V

4.2 Strom (I)

U = RI ⇒ I = \frac{U}{R} = \frac{−0.72V}{25Ω} = 0.0288A = 28.8mA

4.3 Kraft (F_{B})

F = IlB \underbrace{sin φ}_{=1} = 0.0288A · 0.15m · 0.6T = 0.002592AmT =

0.002592N = 2.592mN

4.4 Leistung (P)

P = UI = 0.72V · 0.0288A = 0.020736JC^{−1}Cs^{−1} = 20.7mW