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    Magnetic Field

    Flashcards for topic Magnetic Field

    Intermediate26 cardsphysics

    Preview Cards

    Card 1

    Front

    Define magnetic field (B) mathematically and explain how to determine its direction

    Back

    Magnetic field is defined by the equation: F⃗=qv⃗×B⃗\vec{F} = q\vec{v} \times \vec{B}F=qv×B

    Where:

    • F⃗\vec{F}F is the magnetic force on a charged particle
    • qqq is the charge
    • v⃗\vec{v}v is the velocity

    Direction determination:

    1. There is one line through any point where a charged particle experiences no force when moving along it
    2. This line defines the direction of the magnetic field
    3. The force is perpendicular to both velocity and magnetic field (right-hand rule)

    Unit: Tesla (T) = newton/ampere·meter = weber/meter²

    Card 2

    Front

    How does the motion of a charged particle differ in a uniform magnetic field when its velocity is (a) perpendicular vs (b) at an angle to the field?

    Back

    (a) When velocity is perpendicular to field:

    • Particle moves in a circular path
    • Radius: r=mvqBr = \frac{mv}{qB}r=qBmv​
    • Force provides centripetal acceleration
    • Speed remains constant (force perpendicular to velocity)

    (b) When velocity is at an angle θ to field:

    • Motion follows a helical path
    • Velocity component parallel to field (v∥v_\parallelv∥​) remains unchanged
    • Velocity component perpendicular to field (v⊥v_\perpv⊥​) creates circular motion
    • Resulting path is a helix with radius r=mv⊥qBr = \frac{mv_\perp}{qB}r=qBmv⊥​​
    • The particle advances along field lines while circling around them
    Card 3

    Front

    Explain cyclotron frequency and derive its formula for a charged particle in a uniform magnetic field

    Back

    Cyclotron frequency is the frequency of revolution of a charged particle moving in a circular path in a uniform magnetic field.

    Derivation:

    1. For circular motion: qvB=mv2rqvB = \frac{mv^2}{r}qvB=rmv2​ (magnetic force = centripetal force)
    2. Time period: T=2πrvT = \frac{2\pi r}{v}T=v2πr​
    3. Substituting r=mvqBr = \frac{mv}{qB}r=qBmv​ into time period: T=2πmqBT = \frac{2\pi m}{qB}T=qB2πm​
    4. Frequency: ν=1T=qB2πm\nu = \frac{1}{T} = \frac{qB}{2\pi m}ν=T1​=2πmqB​

    Key insight: Cyclotron frequency depends only on q/mq/mq/m ratio and field strength, not on velocity or radius of motion.

    Card 4

    Front

    Compare and explain the relationship between electric and magnetic fields from the perspective of reference frames

    Back

    Reference frame transformation reveals the fundamental connection between electric and magnetic fields:

    1. In a stationary frame (S):

      • A current-carrying wire produces magnetic field only
      • A moving charge experiences magnetic force perpendicular to velocity
    2. In a moving frame (S') moving with the charge:

      • The charge appears stationary
      • No magnetic force can act on a stationary charge
      • Must experience electric force instead
      • The wire's charges appear to have different densities due to length contraction

    Key insights:

    • Electric and magnetic fields are aspects of a single electromagnetic field
    • Whether observed as electric or magnetic depends on observer's reference frame
    • Special relativity unifies these as a single entity
    • A purely magnetic field in one frame appears as combination of electric and magnetic fields in another
    Card 5

    Front

    Explain the Lorentz force law and how it combines electric and magnetic forces on a charged particle

    Back

    The Lorentz force law combines electric and magnetic forces on a charged particle: F⃗=qE⃗+qv⃗×B⃗\vec{F} = q\vec{E} + q\vec{v} \times \vec{B}F=qE+qv×B

    Components:

    • Electric force (qE⃗q\vec{E}qE):

      • Parallel to electric field for positive charges
      • Opposite to electric field for negative charges
      • Independent of particle velocity
      • Can change particle's kinetic energy
    • Magnetic force (qv⃗×B⃗q\vec{v} \times \vec{B}qv×B):

      • Perpendicular to both velocity and magnetic field
      • Zero when velocity parallel to field
      • Cannot change particle's kinetic energy (no work done)
      • Magnitude: FB=qvBsin⁡θF_B = qvB\sin\thetaFB​=qvBsinθ

    Applications:

    • Particle accelerators
    • Mass spectrometers
    • Hall effect devices
    • Cyclotrons and synchrotrons
    Card 6

    Front

    What is the force exerted on a wire carrying current i perpendicular to a uniform magnetic field B when the wire can slide along parallel conducting rails?

    Back

    The force on the wire is given by:

    • F = i × L × B
    • Direction: perpendicular to both the current direction and magnetic field
    • In a sliding wire setup with parallel rails, this force causes the wire to move along the rails
    • Magnitude depends on:
      • Current strength (i)
      • Length of wire between rails (L)
      • Magnetic field strength (B)

    Example: A 10 cm wire carrying 2A current perpendicular to a 0.5T magnetic field experiences a force of 0.1N along the rails.

    Card image
    Card 7

    Front

    What happens to a charged particle's kinetic energy when it moves in a uniform magnetic field, and why?

    Back

    The kinetic energy of a charged particle in a uniform magnetic field:

    • Remains constant throughout its motion
    • Is not changed by the magnetic force

    This occurs because:

    • The magnetic force (F⃗=qv⃗×B⃗\vec{F} = q\vec{v} \times \vec{B}F=qv×B) is always perpendicular to the velocity
    • Work done by the force is zero: W=F⃗⋅d⃗=0W = \vec{F} \cdot \vec{d} = 0W=F⋅d=0 (since force and displacement are perpendicular)
    • The force changes only the direction, not the magnitude of velocity

    Consequences:

    • The speed of the particle remains constant
    • The radius of the circular path is constant
    • For non-perpendicular motion, the particle follows a helical path with constant pitch

    Example: An electron moving through a bubble chamber in a magnetic field travels in a perfect circle or helix with unchanging speed, allowing scientists to calculate its momentum from the radius of curvature.

    Card image
    Card 8

    Front

    What is the direction of magnetic force on a positive charge moving perpendicular to a magnetic field?

    Back

    The magnetic force on a positive charge moving perpendicular to a magnetic field:

    • Is perpendicular to both the velocity and the magnetic field vectors
    • Follows the right-hand rule: when you point your fingers in the direction of velocity and curl them toward the magnetic field, your thumb points in the direction of force
    • Has magnitude F = qvB (where q is charge, v is speed, B is magnetic field strength)
    • Example: A proton moving west in a vertically upward magnetic field experiences a force directed north

    Note: For negative charges, the force direction is opposite to what the right-hand rule predicts.

    Card image
    Card 9

    Front

    Given a conducting wire of mass m that can slide on parallel rails in a uniform magnetic field, what minimum coefficient of static friction μ is required to prevent the wire from sliding?

    Back

    The minimum coefficient of static friction needed is:

    • μ = (iLB)/(mg)

    Where:

    • i = current in the wire
    • L = separation between rails
    • B = magnetic field strength
    • m = mass of the wire
    • g = acceleration due to gravity

    This represents the point where the magnetic force (iLB) equals the maximum static friction force (μmg). If μ is less than this value, the wire will accelerate along the rails.

    Card image
    Card 10

    Front

    How would you describe the motion of a sliding wire on parallel rails in a magnetic field when the static friction coefficient is insufficient to prevent movement?

    Back

    When static friction is insufficient to prevent movement:

    1. The wire accelerates along the rails with:

      • a = (iLB - μmg)/m
      • If μ = 0 (frictionless), then a = iLB/m
    2. Motion characteristics:

      • Direction: perpendicular to both current and magnetic field
      • Constant acceleration if current remains constant
      • Kinetic friction opposes motion but doesn't prevent it
    3. Energy transformation:

      • Electrical energy converts to mechanical energy
      • Acts as a simple motor converting electrical energy to linear motion
      • Increasing current increases acceleration
    Card image

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