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  2. MCAT Chemical and Physical Foundations of Biological Systems
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MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 4c Magnetism Charged Particle Motion

Study 4c Magnetism Charged Particle Motion in MCAT Chemical and Physical Foundations of Biological Systems with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on 4c Magnetism Charged Particle Motion, giving you a quick way to review the definitions, rules, and examples that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 4c Magnetism Charged Particle Motion

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QUESTION

Find the required speed for no deflection if E=300 N/CE=300\,\text{N/C}E=300N/C and B=0.10 TB=0.10\,\text{T}B=0.10T in a velocity selector.

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ANSWER

v=3.0×103 m/sv = 3.0\times 10^3\,\text{m/s}v=3.0×103m/s. Speed for no deflection satisfies v=E/Bv=E/Bv=E/B, resulting in 3.0×1033.0\times 10^33.0×103 m/s from given field strengths.

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Flashcard 1: Find the required speed for no deflection if E=300 N/CE=300\,\text{N/C}E=300N/C and B=0.10 TB=0.10\,\text{T}B=0.10T in a velocity selector.

Answer: v=3.0×103 m/sv = 3.0\times 10^3\,\text{m/s}v=3.0×103m/s. Speed for no deflection satisfies v=E/Bv=E/Bv=E/B, resulting in 3.0×1033.0\times 10^33.0×103 m/s from given field strengths.

Flashcard 2: State the cyclotron angular frequency for a nonrelativistic charge in a uniform magnetic field.

Answer: ω=∣q∣Bm\omega = \frac{|q|B}{m}ω=m∣q∣B​. Cyclotron frequency emerges from equating magnetic force to centripetal force, independent of velocity for nonrelativistic cases.

Flashcard 3: What is the direction of the magnetic field around a long straight wire with current (right-hand rule)?

Answer: Circles around wire; thumb III, fingers give B⃗\vec{B}B. The right-hand rule for a straight wire follows from Ampere's law, with magnetic field lines encircling the current.

Flashcard 4: State Lenz's law in one sentence about the direction of induced current.

Answer: Induced current opposes the change in ΦB\Phi_BΦB​. Lenz's law, a consequence of energy conservation, dictates that induced current creates a field opposing the flux change.

Flashcard 5: State Faraday's law for the induced emf in a loop in terms of changing magnetic flux.

Answer: E=−dΦBdt\mathcal{E} = -\frac{d\Phi_B}{dt}E=−dtdΦB​​. Faraday's law states that induced emf equals the negative rate of change of magnetic flux through the loop.

Flashcard 6: State the magnetic field magnitude inside an ideal long solenoid with turn density nnn and current III.

Answer: B=μ0nIB = \mu_0 n IB=μ0​nI. Inside a solenoid, the uniform field results from the additive contributions of tightly wound current loops, proportional to turn density and current.

Flashcard 7: State the magnitude of the magnetic field at distance rrr from a long straight wire carrying current III.

Answer: B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}B=2πrμ0​I​. The field strength decreases inversely with distance, derived from Ampere's circuital law for an infinite straight wire.

Flashcard 8: What is the magnetic flux through a flat surface of area AAA in a uniform field BBB?

Answer: ΦB=BAcos⁡θ\Phi_B = BA\cos\thetaΦB​=BAcosθ. Magnetic flux quantifies the component of the magnetic field passing through the surface, given by the dot product of field and area vectors.

Flashcard 9: Identify the relationship between F⃗\vec{F}F and v⃗\vec{v}v for magnetic forces on a moving charge.

Answer: F⃗⊥v⃗\vec{F}\perp\vec{v}F⊥v (magnetic force does no work). The magnetic force is always perpendicular to velocity due to the cross product, ensuring zero work done as F⃗⋅v⃗=0\vec{F}\cdot\vec{v}=0F⋅v=0.

Flashcard 10: What is the vector form of the magnetic force on a current-carrying wire segment?

Answer: F⃗=I L⃗×B⃗\vec{F} = I\,\vec{L}\times\vec{B}F=IL×B. The vector force on a wire segment results from the cross product of the length vector in the direction of current and the magnetic field.

Flashcard 11: State the formula for the magnetic force magnitude on a straight wire of length LLL carrying current III.

Answer: F=ILBsin⁡θF = ILB\sin\thetaF=ILBsinθ. The force on a current-carrying wire integrates the Lorentz force over charge carriers, depending on current, length, field, and angle.

Flashcard 12: State the SI unit of magnetic field BBB in base units.

Answer: 1 T=NA⋅m1\,\text{T} = \frac{\text{N}}{\text{A}\cdot\text{m}}1T=A⋅mN​. The tesla is defined from the force on a current-carrying wire, expressed in base SI units as newtons per ampere-meter.

Flashcard 13: How does the magnetic force direction change for a negative charge compared with a positive charge?

Answer: It is opposite to v⃗×B⃗\vec{v}\times\vec{B}v×B. A negative charge reverses the force direction due to the sign of qqq in the Lorentz force equation.

Flashcard 14: What is the direction of the magnetic force on a positive charge relative to v⃗\vec{v}v and B⃗\vec{B}B?

Answer: Along v⃗×B⃗\vec{v}\times\vec{B}v×B (right-hand rule). The direction follows from the cross product in the Lorentz force, determined by the right-hand rule for positive charges.

Flashcard 15: Identify the condition on the velocity direction that makes the magnetic force on a charge equal to zero.

Answer: θ=0∘\theta = 0^\circθ=0∘ or 180∘180^\circ180∘ (,v⃗∥B⃗\vec{v}\parallel\vec{B}v∥B,). Magnetic force vanishes when velocity is parallel or antiparallel to the magnetic field because sin⁡θ=0\sin\theta=0sinθ=0 at these angles.

Flashcard 16: What is the vector form of the magnetic force on a moving charge (Lorentz magnetic force)?

Answer: F⃗=q v⃗×B⃗\vec{F} = q\,\vec{v}\times\vec{B}F=qv×B. The vector form of the Lorentz magnetic force arises from the cross product of velocity and magnetic field vectors, scaled by the charge.

Flashcard 17: State the formula for the magnetic force magnitude on a charge moving in a magnetic field.

Answer: F=∣q∣vBsin⁡θF = |q|vB\sin\thetaF=∣q∣vBsinθ. The magnitude of the magnetic force on a moving charge is derived from the Lorentz force law, incorporating the charge, velocity, magnetic field strength, and the sine of the angle between velocity and field vectors.

Flashcard 18: State the cyclotron period for a nonrelativistic charge in a uniform magnetic field.

Answer: T=2πm∣q∣BT = \frac{2\pi m}{|q|B}T=∣q∣B2πm​. The period is the time for one circular orbit, derived from angular frequency as T=2π/ωT=2\pi/\omegaT=2π/ω.

Flashcard 19: What is the pitch of the helical path for a charge with velocity component v∥v_{\parallel}v∥​ along B⃗\vec{B}B?

Answer: pitch=v∥T=2πmv∥∣q∣B\text{pitch} = v_{\parallel}T = \frac{2\pi m v_{\parallel}}{|q|B}pitch=v∥​T=∣q∣B2πmv∥​​. The pitch represents axial advance per cyclotron period, combining parallel velocity with the orbital period in the magnetic field.

Flashcard 20: State the expression for the radius of circular motion of a charge moving perpendicular to uniform BBB.

Answer: r=mv∣q∣Br = \frac{mv}{|q|B}r=∣q∣Bmv​. The radius balances centripetal force with magnetic force for perpendicular motion, yielding r=mv∣q∣Br = \frac{mv}{|q|B}r=∣q∣Bmv​.

Flashcard 21: Identify the direction of motion for a positive charge entering uniform B⃗\vec{B}B into the page with v⃗\vec{v}v to the right.

Answer: Upward (force toward top of page). For positive charge with velocity rightward and field into the page, the Lorentz force directs upward via the right-hand rule, causing initial upward deflection.

Flashcard 22: Identify the direction of motion for an electron entering uniform B⃗\vec{B}B out of the page with v⃗\vec{v}v upward.

Answer: Deflects to the left. An electron's negative charge reverses the force direction, leading to leftward deflection when velocity is upward and field is out of the page.

Flashcard 23: Identify the speed selector condition for undeflected motion through crossed fields E⃗⊥B⃗\vec{E}\perp\vec{B}E⊥B.

Answer: v=EBv = \frac{E}{B}v=BE​. In crossed fields, undeflected motion occurs when electric and magnetic forces balance, so qE=qvBqE = qvBqE=qvB.