PHYS 1120 — EXAM 2 REVIEW

Electromagnetism • Forces, Induction, Circuits • 16 Problems + Quiz + Book Problems

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Q1Force Direction on Proton (RHR 2)
CONCEPT
A positive charge moving through a magnetic field experiences a force: F = qvB sinθ
METHOD
Use Right-Hand Rule #2: Point fingers in direction of v, curl toward B. Thumb = F direction (for + charge).
(a) v → (right), B ↑ (up) ⇒ F out of page (⊙)
(b) v ↑ (up), B ⊗ (into page) ⇒ F right (→)
(c) v ⊙ (out of page), B → (right) ⇒ F up (↑)
MNEMONIC: "FBI" — F = Thumb, B = Curl-to, I(v) = Fingers point
For NEGATIVE charges (electrons), flip the force direction!
(a) v right, B up v B F (out) (b) v up, B into page v B F (c) v out of page, B right v (out) B F
Q2Find B from Known v and F (RHR 2 Reverse)
CONCEPT
Given v and F directions, find B by reversing RHR 2. Point fingers along v, thumb along F — palm pushes toward B.
Ex 1: v → (right), F ↑ (up) ⇒ B out of page (⊙)
Ex 2: v → (right), F ⊗ (into page) ⇒ B down (↓)
KEY
For electrons: apply RHR as if positive, then flip B direction.
F = qvB ⇒ B = F/(qv)
Common mistake: forgetting to flip for negative charges!
Ex 1: v right, F up → B = ? v F B (out) Ex 2: v right, F into page → B = ? v F(in) B (down)
Q3Forces Between Parallel Wires
RULE
Same direction currents → ATTRACT
Opposite direction currents → REPEL
F/ℓ = μ₀I₁I₂ / (2πR₁₂)
WORKED EXAMPLE
I₁ = I₂ = 25 A, R = 35 cm = 0.35 m, ℓ = 15 cm = 0.15 m
F/ℓ = (4π×10⁻⁷)(25)(25) / (2π)(0.35) = 3.57×10⁻⁴ N/m
F = (3.57×10⁻⁴)(0.15) ≈ 5.4×10⁻⁴ N
MNEMONIC: "Same = Snuggle, Opposite = Oh no!" (attract vs repel)
Same Direction = ATTRACT I₁ I₂ ATTRACT Opposite = REPEL REPEL
Q4+ Charge in B Out of Page → Clockwise Circle
CONCEPT
A positive charge moving in a uniform B field follows a circular path. F is always perpendicular to v, acting as centripetal force.
R = mv / (qB)
DIRECTION
+ charge enters B⊙ (out of page) moving right → RHR gives F downward → curves down → Clockwise circle
ANSWER
Clockwise circular path
Negative charge in same setup → COUNTERCLOCKWISE
······ ····· ····· ····· ······ B out + v CW
Q5Variable Resistor → Induced Current in Loop A
SETUP
Circuit with variable resistor (slider). Nearby loop A detects induced current.
CHAIN OF REASONING
Slider moves LEFT → R decreases → I increases (V=IR, V fixed)
→ B increases (B ∝ I) → Φ through loop A increases
→ Lenz's Law: induced current opposes increase
→ Induced B opposes original B
ANSWER
(b) Counterclockwise in loop A
MNEMONIC: Lenz = "Nature is Lazy" — opposes all change
Circuit R (slider) R↓ I↑ B↑ Loop A CCW
Q6KE of Particle in Circular Path
KEY PRINCIPLE
Magnetic force is always perpendicular to velocity. Therefore:
W = F · d · cos(90°) = 0
Zero work done → No change in kinetic energy → Speed stays constant (only direction changes)
ANSWER
(c) Remains constant — F ⊥ v means W = 0 means ΔKE = 0
B field changes DIRECTION of velocity, never SPEED!
+ v F F ⊥ v → W=0 → ΔKE=0 |v| = constant
Q7Steady 1.5A in Solenoid → Loop Current
KEY PRINCIPLE
Faraday's Law: EMF is induced only when flux is changing.
ε = −dΦ/dt
Steady (constant) current → Constant B → Constant Φ → dΦ/dt = 0 → ε = 0 → No induced current
ANSWER
ZERO — a steady current produces no change in flux
This is the #1 Faraday trap! "Steady" = "constant" = NO induction. Only CHANGING currents induce.
Solenoid (1.5A steady) B (constant) Loop I = 0 dΦ/dt = 0 → ε = 0
Q8Bar Magnet Through Loop
PHASE 1: APPROACHING (N-end down)
Φ increasing → Lenz opposes increase → induced B opposes magnet's B → CCW current (from above) → Loop REPELS magnet
PHASE 2: LEAVING (N-end still down)
Φ decreasing → Lenz opposes decrease → induced B supports magnet's B → CW current (from above) → Loop ATTRACTS magnet
ANSWER
Current REVERSES direction as magnet passes through. Approaching: CCW. Leaving: CW.
MNEMONIC: "Coming in? Push away. Going out? Pull back." — Lenz always fights change.
Phase 1: Approaching N S v CCW REPELS Phase 2: Leaving N S v CW ATTRACTS
Q94× Speed → 4R Radius
FORMULA
R = mv / (qB)
R is directly proportional to v (m, q, B held constant).
If v → 4v, then R → 4R
ANSWER
New radius = 4R (radius scales linearly with speed)
Don't confuse with KE (KE ∝ v²) — radius is LINEAR in v!
R Original speed v 4R Speed 4v 4×
Q10Solenoid + Bar Magnet → Attract (Unlike Poles)
METHOD
1. Use RHR 1 to find solenoid polarity: curl fingers in direction of current, thumb points to N pole.
2. Compare solenoid pole facing magnet to magnet pole facing solenoid.
3. Unlike poles → ATTRACT. Like poles → REPEL.
ANSWER
Unlike poles face each other → ATTRACT (pull toward each other)
MNEMONIC: RHR 1 — "Curl with Current, Thumb = North"
S N Solenoid S N Bar Magnet ATTRACT N faces S = unlike poles
Q11Electron Beam Between Magnets
SETUP
Electron beam travels horizontally between magnets. B field points downward.
METHOD
1. Pretend it's a POSITIVE charge → apply RHR 2
2. v horizontal, B down → RHR gives force in one direction
3. FLIP for electron (negative charge) → force in opposite direction → UPWARD
ANSWER
(c) Accelerated upward — "accelerated" here means direction changes, not speed
"Accelerated" in physics = any change in velocity (including direction). B field can accelerate without changing speed!
N S B e− beam F (up)
Q12Switch Opens → Voltmeter Reading Increases
CONCEPT
When switch S opens, a parallel branch is removed from the circuit.
CHAIN
Remove parallel branch → Total R increases → Current from battery may decrease → But the voltage redistribution across remaining components changes → Voltmeter reading INCREASES
ANSWER
Voltmeter reading INCREASES when switch opens
WHY
Opening the switch removes a current path. The remaining resistor now gets a larger share of the battery voltage (less voltage dropped elsewhere).
V S (open) V ↑ Remove branch → R↑ → V reading ↑
Q13R and 4R in SERIES, P(4R)=40W → P(R)=?
KEY PRINCIPLE
In SERIES, current I is the same through all resistors.
P = I²R ⇒ P ∝ R (same I)
LOGIC
P(4R) = I²(4R) = 40W
P(R) = I²(R) = 40W ÷ 4 = 10W
ANSWER
P(R) = 10W — In series, power is proportional to resistance
Series: P ∝ R (bigger R = MORE power)
Don't confuse with parallel!
R P=? 4R P=40W I (same) Series: P ∝ R P(R) = 40/4 = 10W
Q14R and 5R in PARALLEL, P(R)=50W → P(5R)=?
KEY PRINCIPLE
In PARALLEL, voltage V is the same across all resistors.
P = V²/R ⇒ P ∝ 1/R (same V)
LOGIC
P(R) = V²/R = 50W
P(5R) = V²/(5R) = 50W ÷ 5 = 10W
ANSWER
P(5R) = 10W — In parallel, power is INVERSELY proportional to resistance
Parallel: P ∝ 1/R (bigger R = LESS power)
OPPOSITE of series!
R P=50W 5R P=? V same Parallel: P ∝ 1/R → P(5R) = 50/5 = 10W
Q15Alpha Particle in B Field (INTERACTIVE)
GIVEN
Alpha particle: q = 2e = 3.2×10−19 C, m = 6.64×10−27 kg
R = mv/(qB), Diameter = 2R, a = qvB/m
R = 0.670 mm | Diameter = 1.340 mm
a = 1.89×109 m/s²
Alpha has charge 2e (not 1e)! Forgetting = double the radius.
······· ··· ·· ·· ··· ······· B out α R D = 1.34 mm
Q16Rod on Rails (INTERACTIVE)
GIVEN
Rod of length L = 0.50 m slides on rails, external resistor R = 1.50 Ω
ε = BLv, I = ε/R, F = BIL
ε = 3.000 V | I = 2.000 A | F = 0.800 N
VIFf CHAIN
v (speed) → ε (EMF) → I (current) → F (force on rod)
Each depends on the previous!
Force on the rod OPPOSES motion (Lenz's Law applied mechanically)
R Rod (L) v B (in) ε = BLv = 3.0 V I
REFSeries vs Parallel Power Comparison
PropertySERIESPARALLEL
What's shared?Current I (same through all)Voltage V (same across all)
Power formulaP = I²RP = V²/R
Power vs RP ∝ R (more R = more P)P ∝ 1/R (more R = less P)
Bigger resistor gets...MORE powerLESS power
ExampleR & 4R: P(4R) = 4×P(R)R & 5R: P(R) = 5×P(5R)
Common mistakeUsing V²/R (wrong! V differs)Using I²R (wrong! I differs)
FormulaVariablesUsed In
F = qvB sinθForce on moving chargeQ1, Q2, Q11
R = mv/(qB)Circular orbit radiusQ4, Q9, Q15
F/ℓ = μ₀I₁I₂/(2πd)Force between parallel wiresQ3
ε = −dΦ/dtFaraday's Law (EMF)Q5, Q7, Q8
ε = BLvMotional EMFQ16
B = μ₀nISolenoid B fieldQ10
Question 1 of 16
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