Physics Problems & Solutions: #11

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Nuclear & Particle Physics - Stern-Gerlach Experiment

A beam of neutral hydrogen atoms in their ground state is moving into the plane of this page and passes through a region of a strong inhomogeneous magnetic field that is directed upward in the plane of the page. After the beam passes through this field, a detector would find that it has been

A. deflected upward
B. deflected to the right
C. undeviated
D. split vertically into two beams
E. split horizontally into three beams

(GR9677 #11)

Solution:

The Stern–Gerlach experiment → spin discovery

A beam of neutral atom passes through inhomogeneous magnetic field will split vertically into 2 beams representing spin-up and spin-down electrons.

Answer: D

Electromagnetism - Capacitor

The capacitor in the circuit shown above is initially charged. After closing the switch, how much time elapses until one-half of the capacitor’s initial stored energy is dissipated?

A. RC
B. RC/2
C. RC/4
D. RC ln 2
E. ½ RC ln 2

(GR9277 #11)

Solution:

U_t= ½U₀ , t = ?
Energy of capacitor: U = ½CV²

U_t = ½U₀→ ½CV_t² = ½ × ½CV₀2
V_t² = ½ V₀2

Potential difference of capacitor: V_t= V₀e^−t/RC
V_t²= V₀2e^−2t/RC
½ V₀2= V₀2e^−2t/RC→ ½ = e^−2t/RC
ln ½ = − 2t/RC → ln 2 = 2t/RC
t = ½ RC ln 2

Answer: E

Electromagnetism - Vector Identity

Which of the following equation is a consequence of the equation $\nabla \times \bold H = \dot{\bold D}+\bold J$ ?

A. $\nabla \cdot (\dot{\bold D}+\bold J )=0$
B. $\nabla \times (\dot{\bold D}+\bold J )=0$
C. $\nabla(\dot{\bold D}+\bold J )=0$
D. $\dot{\bold D}+\bold J =0$
E. $\dot{\bold D}\cdot\bold J =0$

(GR8677 #11)

Solution:

Vector identity: ∇ · (∇ × A) = 0

$\nabla \cdot (\nabla \times \bold H) =\nabla \cdot (\dot {\bold D} + \bold J)=0$

Answer: A

Optics - Converge Lens

An object is located 40 centimeters from the first of two thin converging lenses of focal lengths 20 cm and 10 cm, respectively, as shown in figure above. The lenses are separated by 30 cm. The final image formed by the two lenses is located

A. 5.0 cm to the right of the second lens
B. 13.3 cm to the right of the second lens
C. infinitely to the right of the second lens
D. 13.3 cm to the left of the second lens
E. 100 cm to left right of the second lens

(GR0177 #11)

Solution:

¹/_S + ¹/_S'= ¹/_f

For the first lens:

¹/_S₁ + ¹/_S₁'= ¹/_f₁

¹/₄₀ + ¹/_S₁' = ¹/₂₀

S₁' = 40 cm

The image is located 40 cm behind (to the right of) the first lens, which is 10 cm behind the second lens. Therefore, S₂ = −10.

For the second lens:

¹/_S₂ + ¹/_S₂'= ¹/_f₂

¹/₍₋₁₀₎ + ¹/_S₂' = ¹/₁₀

S₂' = 5 cm

The final image is located 5 cm to the right of the second lens.

Answer: A