Electromagnetism - Faraday’s law

A circular wire loop of radius R rotates with an angular speed ω in a uniform magnetic field B, as shown in the figure. If the emf ɛ induced in the loop is ɛ₀ sin ωt, then the angular speed of the loop is

A. ɛ₀ R/B
B. 2πɛ₀/R
C. ɛ₀/BπR²
D. ɛ₀²/BR²
E. tan⁻¹(ɛ₀/Bc)

(GR9677 #46)

Solution:

Faraday's Law: ɛ = − dΦ/dt
with  Φ = NBA  

Given:
N = 1
B uniform (constant)
ɛ =  ɛ₀ sin ωt

→ ɛ = − B dA/dt =  ɛ₀ sin ωt

− B dA =  ɛ₀ sin ωt dt 

 B ∫dA =  ɛ₀ ∫ sin ωt dt 

− BπR² = − ^ɛ₀/_ω  cos ωt

ω = ɛ₀ cos ωt / BπR²

at t = 0,
ω = ɛ₀ / BπR²

Answer: C

Alternative solution:
A = πR²  cos ωt
ɛ = − dΦ/dt = − B dA/dt 
ɛ = − BπR²  d(cos ωt)/dt 
ɛ₀ sin ωt = BπR²ω  sin ωt 
ɛ₀= BπR²ω  

ω = ɛ₀ / (BπR²)

Thermal Physic - Critical Isotherm

Questions 46-47.

Isotherms and coexistence curves are shown in the PV diagram above for a liquid-gas system. The dashed lines are the boundaries of the labeled regions. Which numbered curve is the critical isotherm? 

A. 1
B. 2
C. 3
D. 4
E. 5

(GR9277 #46)

Solution:

Critical isotherm (critical temperature) → dP/dV = 0

• the derivative of the curve is zero
• the tangent to the curve results in a horizontal line
• the point where the vertical and horizontal dashed lines cross

Answer: B

Thermal Physics - Blackbody Radiation

A blackbody at temperature T₁ radiates energy at a power level of 10 milliwatts (mW). The same blackbody, when at a temperature 2T₁, radiates energy at a power level of 

A. 10 mW
B. 20 mW
C. 40 mW
D. 80 mW
E. 160 mW

(GR8677 #46)

Solution:

Blackbody radiation (Stefan-Boltzmann's law): u = σT⁴

Constant σ = u / T⁴

u₁T₁⁴ = u₂T₂⁴
10 / T₁⁴ = u₂ / (2T₁)⁴
u₂ = (10)(2T₁)⁴ / T₁⁴ = (10)(16) =160

Answer: E

Quantum Mechanics - Free Particle

A free particle with initial kinetic energy E and de Broglie wavelength λ enters a region in which it has potential energy V. What is the particle’s new de Broglie wavelength?

A. λ(1 + E/V)
B. λ(1 − V/E)
C. λ(1− E/V)^{− 1}
D. λ(1 + V/E)^½
E. λ(1 − V/E)^{− ½}

(GR0177 #46)

Solution:

The initial kinetic energy of free particle:


De Broglie wavelength:


Energy of the particle when it enters the region:


So, its wavelength becomes:




Answer: E