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

Electromagnetism - Kirchoff's Law

One end of Nichrome wire of length 2L and cross-sectional area A is attached to an end of another Nichrome wire of length L and cross-sectional area 2A. If the free end of the longer wire is at an electric potential of 8.0 volts, and the free end of the shorter wire is at an electric potential of 1.0 volt, the potential at the junction of the two wires is most nearly equal to

A. 2.4 V
B. 3.3 V
C. 4.5 V
D. 5.7 V
E. 6.6 V

(GR0177 #85)

Solution:

Ohm’s Law: V = IR

Kirchoff's Law: I_in = I_out = I

I_in = (V₁ − V)/R₁
I_out = (V − V₂)/R₂

Resistance: R = ρL/A

Both are Nichrome wires, ρ₁ = ρ₂ = ρ

R₁ = ρ(2L/A) = 2R
R₂ = ρ(L/2A) = 0.5R

I_in = I_out
(8 − V)/2R = (V − 1)/0.5R
4 − ½V = 2V − 2
6 = ⁵⁄₂V 
V  =  2.4 V

Answer: A

Electromagnetism - Faraday’s law

A coil of 15 turns, each of radius 1 centimeter, is rotating at a constant angular velocity ω = 300 radians per second in an uniform magnetic field of 0.5 Tesla, as shown in figure. Assume at time t = 0 that the normal n̂ to the coil plane is along the y-direction and that the self-inductance of the coil can be neglected. If the coil resistance is 9 ohms, what will be the magnitude of the induced current in milliamperes?

A. 225π sin ωt
B. 250π sin ωt
C. 0.08π cos ωt
D. 1.7π cos ωt
E. 25π cos ωt

(GR0177 #86)

Solution:

Ohm’s Law: ɛ = V = IR
→  I = ɛ / R

Faraday's Law: ɛ = − dΦ_B / dt

Magnetic Flux:  dΦ_B =  NB_⊥dA

At time t = 0, the normal to the coil plane is along the y-direction.

It means: n̂ ∥ ŷ → B ∥ A → Φ_B = 0

→ Φ_B =  NBA sin ωt, since Φ_B (t = 0) = 0

Φ_B = − NB₀ πr² sin ωt
ɛ = − dΦ_B / dt = NB₀ πr² ω cos ωt

I =  ɛ / R = (NB₀ωr²/ R) π cos ωt

N  = 15 turns
B₀ = 0.5 Tesla
ω = 300 rad/s
r  = 1 cm =  10⁻²  m
R = 9 ohms

I = (15 ×  0.5 × 300 × 10⁻⁴ / 9) π cos ωt
= (25 × 10⁻³) π sin ωt Ampere
= 25 π sin ωt milliAmpere

Answer: E