Exam 2 — Complete Solutions with Variations

Some textbooks teach a "left-hand rule" for negative charges. Avoid it — it creates confusion when you're also using RHR #1 for field direction (which uses the right hand regardless of charge). The universal strategy: always use the right hand, then flip the answer at the end if your particle is negative.

This rule also applies to current in a wire. If you were told "electron current going up" (which is unusual notation), flip it to "conventional current going down" and then use the standard RHR. Conventional current is always the direction positive charges would move, which is opposite to actual electron motion in a metal.

Given v (or I) and F, the direction of B is uniquely determined only up to the constraint that B must be perpendicular to F. In 3D, this still leaves a whole plane of possible B directions. What the reverse-RHR gives you is the simplest B direction that produces the given F — but there could be other components of B parallel to v that add no force but contribute to total B. On an exam, the answer is usually just "B⊗" or "B⊙" without such subtleties.

This is the rule students keep forgetting because it's the opposite of charge behavior. Like charges repel, but like currents attract. The physical reason is the cross-product geometry of F = IL × B, which flips the sign relative to charge-charge Coulomb interaction.

A simple mnemonic: "wires want to merge when they flow together." Or: "currents are friendly — they attract friends (same direction) and repel enemies (opposite direction)." Whatever helps you remember — just don't confuse it with static charges.

In the same B field (into page) with the same initial velocity (right), a proton and an electron circle in opposite directions. The proton's force is up, so it curves up and circles counterclockwise. The electron's force is down, so it curves down and circles clockwise. This is the key mnemonic for every particle-in-field problem.

Whether the induced current goes clockwise or counterclockwise depends heavily on the orientation of loop A relative to the main circuit. The problem provides a picture on the exam showing exactly where A is. If A is on the opposite side of the main wire, the direction flips. Always draw the actual geometry before invoking Lenz — don't just memorize an answer.

The work-energy theorem says ΔKE = W_net. If W = 0, then ΔKE = 0, and the speed can't change. This is the universal answer for ANY question about a charged particle moving in a pure magnetic field — KE is always conserved.

What CAN change KE: only an electric field or a collision. If the problem adds an E field (as in a cyclotron) or mentions the particle hitting something, KE can change. But in a pure B field with no other influences, KE is rock-solid constant.

Students often see "solenoid + loop" and panic, thinking they need to compute B, find flux, and so on. They forget the #1 rule of induction: nothing is induced unless something is changing. Before doing any math, ask yourself: is anything in this picture actually changing? If not, the answer is zero, and you've saved yourself 10 minutes.

When would this problem have a nonzero answer? If the current was increasing or decreasing, or if the solenoid was moving toward or away from the loop, or if the loop was expanding or contracting, or rotating. Any "change" produces an induced EMF. No change = no induction.

When current flows from one end to another of a resistor, the end where current enters is at higher potential (because V = IR, and the voltage drops in the direction of current flow). So "a" being higher means the induced current flows into a from outside the resistor, and out of b.

Another way to think about it: the moving rod acts like a battery. The Lorentz force qv × B pushes positive charges in a specific direction inside the rod, and they pile up at one end. That end is the "+" terminal of the equivalent battery.

The current direction flips at the exact moment the magnet is halfway through the loop. At that instant, flux is at its maximum and momentarily constant — zero EMF for one instant. Just before that, flux is rising; just after, flux is falling. So the induced current must reverse sign to keep opposing the change.

This is exactly why the loop decelerates the magnet both coming in and going out. In both phases, the induced current creates a force on the magnet that opposes the motion — Lenz's Law enforcing energy conservation. If you drop a very strong magnet through a long copper tube, you can actually see it fall much slower than gravity would allow, because the continuously reversing induced currents keep slowing it down.

Here's the subtle but important twist: while R doubles when v doubles, the orbit period stays the same! T = 2πR/v = 2π(2R)/(2v) = 2πR/v. The doubled radius exactly cancels the doubled speed.

This is the cyclotron resonance principle: no matter how fast a charged particle is going, it always takes the same time to complete one orbit. This is why cyclotrons work — a fixed-frequency AC electric field can resonantly accelerate particles that start slow and end fast, without needing to retune.

Cyclotron frequency: f = qB/(2πm). Only depends on charge, mass, and field. Not on energy or radius.

A current-carrying solenoid behaves externally exactly like a bar magnet. The field lines come out one end (labeled N) and enter the other (labeled S), exactly like a permanent magnet's. The only difference is that you can turn it on and off with the switch.

To find which end of a solenoid is N: point the fingers of your right hand in the direction of current flow around the coils. Your thumb points toward the N end. Alternatively, look at the end face: if the current appears to flow counterclockwise, that end is N; if clockwise, it's S.

The last multiple-choice option on the original exam reads "electrons are NOT accelerated because magnetic fields don't change speed." This is a trap — it's true that magnetic fields don't change speed (Problem 6), but they DO change velocity (direction). Acceleration in physics means change in velocity, not just change in speed. So there IS acceleration — the electrons curve as they pass through.

Don't confuse "speed" with "velocity" on physics exams. Uniform circular motion has constant speed but nonzero acceleration (centripetal). The same is true for curved paths in magnetic fields.

On multiple-choice exams, circuit problems often have an easy answer if you think about topology before doing math. Ask:

1. What does the switch do? (Complete a new loop? Short out an element?)
2. Does that change the total resistance? (Adding parallel always decreases; removing parallel always increases.)
3. How does that affect the current? (Lower R → more I; higher R → less I.)
4. How does that affect the voltage across the element of interest? (V = IR, so if I changes and R is fixed, V changes proportionally.)

This chain of reasoning — topology → total R → current → V = IR — handles most "what happens when switch X is closed?" questions in under 30 seconds, no numbers required.

In parallel, power goes inversely with resistance. Smaller resistor hogs more current, dissipates more power. Bigger resistor acts as a "gentler" path with less current and less heat.

This is opposite to series (see next problem), where bigger R dissipates MORE power. The distinction is the key: in series, share I; in parallel, share V. Everything else flows from that one fact.

Compare these two problems carefully — they're asking the same question (how does power scale with R?) but the answer depends entirely on whether the resistors are in series or parallel:

• Problem 14 (parallel): R → 10 W, 2R → 5 W. Doubling R halves the power.
• Problem 15 (series): R → 30 W, 3R → 90 W. Tripling R triples the power.

Students who memorize one rule and apply it everywhere get these wrong. The correct mental model: ask which quantity is shared, then use the power formula that matches.

Memorize this matrix:

• Series: same I → use P = I²R → bigger R wins
• Parallel: same V → use P = V²/R → smaller R wins

Students constantly conflate "angle from the plane" with "angle from the normal." The formula Φ = BA cos φ uses the angle from the normal. To convert:

• If a problem says "B is 60° above the plane of the loop" → angle from normal is 90° − 60° = 30°
• If it says "B makes 20° with the normal" → you use 20° directly
• If it says "B is in the plane" → angle from normal is 90° → Φ = 0
• If it says "B is perpendicular to the plane" → angle from normal is 0° → Φ = BA (max)

Always draw the normal vector first and clearly label the angle you're measuring. If you can't see what φ is, you can't solve the problem.

Mechanical power supplied by whoever's pushing the rod: P_mech = F·v = (0.20)(7.5) = 1.5 W. Electrical power dissipated in the resistor: P_elec = εI = (1.5)(1.0) = 1.5 W. They match exactly.

This is Lenz's Law as energy conservation: the mechanical work done against the drag force is exactly converted into electrical energy, which is then dissipated as heat in the resistor. Nothing is lost, nothing is created. Energy in = energy out.

If you ever get a motional EMF problem where P_mech ≠ P_elec, you've made a mistake. The equality is not a coincidence; it's a conservation law. Use it as a sanity check on every problem.

The zero-field point must be on the side of the weaker current. Intuitively: you need to be "closer to the weak wire" to make its field as strong as the distant strong wire's field. In mathematical terms: the field decays as 1/r, so to balance a 3× stronger current, you must be 3× farther from the strong wire than from the weak wire.

Why not between the wires? With antiparallel currents, the B fields between the wires both point the same way (say, both "into the page" or both "out of the page" depending on your viewing direction). Adding two same-direction fields can never give zero. So the cancellation must happen outside.

What if the currents were PARALLEL (same direction)? Then the fields BETWEEN the wires point opposite ways, and the zero-field point would be between them. Using the same algebra: for parallel currents 25 A and 75 A, 40 cm apart, the null is where 25/x = 75/(0.40 − x) → x = 10 cm from the 25 A wire, or 30 cm from the 75 A wire.

Summary rule: zero-field point is always on the "closer to the weaker current" side. For antiparallel: OUTSIDE, past the weaker wire. For parallel: BETWEEN, closer to the weaker wire.

If you boil all of Exam 2 down to one sentence: moving charges create magnetic fields, magnetic fields exert forces on moving charges, and changing fields create EMFs. That's it. Every problem is an application of one of these three truths.

The right-hand rule is just nature's way of encoding the geometry. Lenz's Law is just energy conservation. Faraday's Law is just "change creates voltage." Kirchhoff's rules are just "charge and energy are conserved." Everything else is arithmetic.

Good luck on the exam. Be confident, show your work, and trust the formulas. You've got this.