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

See Question #32. The voltage across resistor R₄ is

A. 0.4 V
B. 0.6 V
C. 1.2 V
D. 1.5 V
E. 3.0 V

(GR9277 #33)

Solution:

Parallel: V_t = V₁ = V₂ = V₃ = ...

Series: V_t = V₁ + V₂ + V₃ + ...

V₄ = V₃ = V₃₄
V₃₄ + V₅ = V₂ 
V₂ + V₁ = V_t 

Or, using Kirchoff's Law:
Loop 1: V_t  = V₁ + V₂  
Loop 2: V_t  = V₁ + V₃₄ + V₅

From Question #32:
R₃₄ = 20
R₃₄₅ = 50
R₂₃₄₅ = 25

R_t = R₁ + R₂₃₄₅ = 50 + 25 = 75
I_t = V_t / R_t = 3/75 = I₁
V₁ =  I₁ R₁ = (3/75) × 50 = 2

Thus, V₃₄ + V₅ = V_t −V₁ = 3 − 2 = 1

V₃₄ must be less than 1
Only (A) and (B) could be TRUE

Since R₄ = R₅ but R₄ is in parallel with R₃ → V₄ must be less than V₅ 

Answer: A

Complete Calculation:

I₁ =  I₂ + I₃₄₅

R₃₄₅ = 50 = R₂
I₃₄₅ = I₂ =  I₁ / 2 = (3/75) / 2 = 3/150

R₃₄ and R₅ are in series
I₃₄₅ = I₃₄ = I₅
V₄ = V₃₄ = I₃₄ R₃₄ = (3/150) × 20 = 60/150 = 0.4

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