Electromagnetism - Faraday’s law

The circuit shown is in a uniform magnetic field that is into the page and is decreasing in magnitude at rate of 150 tesla/second. The ammeter reads

A. 0.15 A
B. 0.35 A
C. 0.50 A
D. 0.65 A
E. 0.80 A

(GR9677 #02)

Solution:

V − IR − ɛ = 0
I = (V − ɛ)/R

ɛ = − dΦ/dt = −AdB/dt

Given:
dB/dt = −150 t/s (minus sign because it’s decreasing)
A = (0.1 m)² = 0.01 m²
R = 10 Ω
V = 5 V

ɛ = − (0.01)(−150) = 1.5 V
I = (5 − 1.5)/10 = 3.5/10 = 0.35 A

Answer: B

Nuclear & Particle Physics - Bragg Diffraction

The longest wavelength X-ray that can undergo Bragg diffraction in a crystal for a given family of planes of spacing d is

A. d/4
B. d/2
C. d
D. 2d
E. 4d

(GR9277 #02)

Solution:

Bragg’s law: 2d sin θ = nλ

Maximum → sin θ = 1

and n = 1 (1st order)

λ = 2d

Answer: D

Classical Mechanics - Satellite

A Satellite orbits the Earth in a circular orbit. An Astronaut on board perturbs the orbit slightly by briefly firing a control jet aimed toward the Earth’s center. Afterward, which of the following is true of the satellite’s path?

A. It is an ellipse
B. It is a hyperbola
C. It is a circle with larger radius
D. It is a spiral with increasing radius
E. It exhibits many radial oscillations per revolution.

(GR8677 #02)

Solution:

Initially, the object orbits the Earth in a circular orbit.

Perturbs the orbit slightly means giving some extra momentum, so the orbit won't be circular any longer, and will be elliptic, not enough to be Parabolic or Hyperbolic.

Also, logically the astronaut will not want the satellite to have v = v_escape (parabolic) or v v_escape (hyperbolic).

Answer: A


Notes:

Types of Orbits	Eccentricity	Energy	Velocity
Circular	e = 0	E = Vmin
Elliptic	0 e 1	Vmin E 0	v vescape
Parabolic	e = 1	E = 0	v = vescape It will escape the gravitational pull of the planet. If v is increased it will become a hyperbolic orbit.
Hyperbolic	e 1	E 1	v vescape It escapes the gravitational pull of the planet and continues to travel infinitely until it is acted upon by another body with sufficient gravitational force.

The orbital eccentricity, e is the amount by which its orbit deviates from a perfect circle.

Classical Mechanics - Circular Motion

The coefficient of static friction between a small coin and the surface of a turntable is 0.30. The turntable rotates at 33.3 revolutions per minute. What is the maximum distance from the center of the turntable at which the coin will not slide?

A. 0.024 m
B. 0.048 m
C. 0.121 m
D. 0.242 m
E. 0.484 m

(GR0177 #02)

Solution:

The coin will not slide if  F_centripetal = F_friction
mω²r = μ_smg
r = μ_sg/ω² =  μ_sg/(4π²f²)

Given:
μ_s = 0.30
f = 33.3 revolutions per minute = 33.3/60 = (100/3)(1/60) = 5/9
and take π² = (3.14)² ≈ 10

Radius:
r = (0.3)(10)/[4(10)(25/81)] = (3)(81)/[4(10)(25)] = 243/1000 = 0.243

Answer: D


Notes:


F_centripetal = mv²/r = mω²r
F_friction = μ_sN = μ_smg