Magnetic Field Due to A Current

Flashcards for topic Magnetic Field Due to A Current

Intermediate47 cardsphysics

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Card 1

Front

Derive the expression for the magnetic field at a distance dd from a long, straight current-carrying wire using Ampere's Law.

Back

Using Ampere's Law:

  1. Draw a circular path of radius dd around the wire
  2. The magnetic field B\vec{B} is tangential to this path with constant magnitude
  3. The circulation is: Bdl=Bdl=B(2πd)\oint \vec{B} \cdot \vec{dl} = B \oint dl = B(2\pi d)
  4. The current enclosed by the path is ii
  5. By Ampere's Law: Bdl=μ0i\oint \vec{B} \cdot \vec{dl} = \mu_0 i
  6. Therefore: B(2πd)=μ0iB(2\pi d) = \mu_0 i
  7. Solving for BB: B=μ0i2πdB = \frac{\mu_0 i}{2\pi d}

The direction follows from the right-hand rule: grip the wire with your right hand with thumb pointing along current direction; your fingers curl in the direction of the magnetic field.

Card 2

Front

Given a solenoid with nn turns per unit length carrying current ii, derive the expression for the magnetic field inside the solenoid using Ampere's Law.

Back

B=μ0niB = \mu_0 n i

Derivation using Ampere's Law:

  1. Draw a rectangular path PQRS with:

    • Side PQ parallel to the solenoid axis inside the solenoid
    • Side RS outside the solenoid
    • Sides QR and SP perpendicular to the solenoid axis
  2. Calculate circulation:

    • Along PQ: PQBdl=Bl\int_P^Q \vec{B} \cdot \vec{dl} = Bl (where ll is length of PQ)
    • Along QR, RS, SP: Bdl=0\int \vec{B} \cdot \vec{dl} = 0 (B is either zero outside or perpendicular to dl\vec{dl})
    • Total circulation: Bdl=Bl\oint \vec{B} \cdot \vec{dl} = Bl
  3. Current enclosed by path:

    • Number of turns enclosed = nlnl
    • Total current = nlinli
  4. Apply Ampere's Law: Bl=μ0nliBl = \mu_0 nli

  5. Solving for B: B=μ0niB = \mu_0 ni

Note: This is valid for infinite solenoids. For practical applications, the approximation works well when the solenoid length is at least 5-6 times its diameter.

Card 3

Front

When applying the Biot-Savart Law to various current configurations, how do you determine whether the magnetic field at a point goes into the plane or comes out of the plane?

Back

To determine if the magnetic field goes into or out of a plane:

  1. Identify the direction of current (i) and the position vector (r) from the current element to the point

  2. Apply the right-hand rule to the cross product (dl × r):

    • Place your right hand along the current direction dl
    • Curl your fingers toward the position vector r
    • Your thumb points in the direction of the magnetic field
  3. Conventional notation:

    • ⊗ (encircled cross) indicates field going into the plane
    • ⊙ (encircled dot) indicates field coming out of the plane

Example: For a straight current-carrying wire in a plane with current flowing upward, points to the left of the wire experience a magnetic field going into the plane, while points to the right experience a field coming out of the plane.

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Card 4

Front

If two parallel wires carry currents in the same direction, what forces do they experience and why? How does this relate to the magnetic field pattern around each wire?

Back

When two parallel wires carry currents in the same direction:

  1. They attract each other with a force per unit length given by: FL=μ0I1I22πd\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} Where d is the separation distance between wires.

  2. The physical explanation:

    • Each wire produces circular magnetic field lines
    • Wire 1 experiences a force due to being in Wire 2's magnetic field
    • Wire 2 experiences a force due to being in Wire 1's magnetic field
    • When currents flow in the same direction, each wire sits in a region where the other's field lines are oriented to produce an attractive force
  3. This principle forms the basis for defining the ampere: two parallel wires 1m apart carrying 1A each experience a force of 2 × 10⁻⁷ N per meter.

If the currents flow in opposite directions, the wires would repel each other with the same magnitude of force.

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Card 5

Front

How can you apply the right-hand rule to determine the magnetic dipole moment of a current-carrying loop, and what is its mathematical representation?

Back

To determine the magnetic dipole moment of a current-carrying loop:

  1. Curl your right-hand fingers in the direction of the current flow
  2. Your extended thumb points in the direction of the magnetic dipole moment vector

Mathematical representation:

  • The magnetic dipole moment vector is given by: μ=IA\vec{\mu} = I\vec{A}
  • Where II is the current and A\vec{A} is the area vector perpendicular to the loop

Properties:

  • The magnetic dipole moment has units of A⋅m²
  • Its direction is perpendicular to the plane of the loop
  • The strength is proportional to both current and loop area
  • For multiple turns N: μ=NIA\vec{\mu} = NI\vec{A}

Example: A 5 cm radius loop carrying 2A current in counterclockwise direction (when viewed from above) has a magnetic dipole moment of 2×π×(0.05)2=0.01572 \times \pi \times (0.05)^2 = 0.0157 A⋅m² pointing upward.

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Card 6

Front

When a circular current loop is placed in an external magnetic field, what determines the torque experienced by the loop, and how is this related to the right-hand rule?

Back

The torque experienced by a current loop in an external magnetic field:

  • Is given by: τ=μ×B\vec{\tau} = \vec{\mu} \times \vec{B} where μ\vec{\mu} is the magnetic dipole moment
  • Has magnitude: τ=μBsinθ\tau = \mu B \sin\theta where θ\theta is the angle between μ\vec{\mu} and B\vec{B}
  • Is maximum when the dipole moment is perpendicular to the field (θ=90°\theta = 90°)
  • Is zero when the dipole moment is parallel or antiparallel to the field (θ=0°\theta = 0° or 180°180°)

Right-hand rule connection:

  • The dipole moment direction (from right-hand thumb rule) determines how the loop orients in the field
  • The loop experiences torque that tends to align its dipole moment parallel to the external field
  • The torque direction follows the right-hand rule for cross products

Example: A circular coil with its dipole moment at 30° to a 0.5T magnetic field will experience a torque that rotates it toward alignment with the field lines.

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Card 7

Front

What is the pattern of magnetic field lines produced by a circular current loop, and how does this pattern relate to a bar magnet?

Back

The magnetic field lines of a circular current loop:

  • Form closed loops that emerge from one face, curve around the outside, and enter through the opposite face
  • Are densely packed and roughly parallel inside the loop (stronger field)
  • Become less dense as distance from the loop increases (weaker field)
  • Are symmetrical about the axis of the loop
  • On the axis, point in one direction through the loop and in the opposite direction outside the loop

This pattern is identical to that of a short bar magnet or magnetic dipole, with one face acting as a north pole and the opposite face as a south pole.

Example: A compass needle placed at various positions around the loop will align tangentially to these field lines, always pointing from north to south.

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Card 8

Front

What specific notation convention is used to represent current direction perpendicular to a plane, and how does it relate to vector field visualization?

Back

The notation convention for current direction perpendicular to a plane:

  1. Current coming OUT of the plane: Represented by a dot (⊙)

    • Mnemonic: Visualize the tip of an arrow pointing toward you
    • Represents positive z-direction in standard right-handed coordinates
  2. Current going INTO the plane: Represented by a cross (⊗)

    • Mnemonic: Visualize the fletching (feathers) of an arrow moving away from you
    • Represents negative z-direction in standard right-handed coordinates

This notation is used consistently in vector field visualization for any vector pointing perpendicular to the plane (including magnetic fields, electric fields, angular momentum, etc.).

The convention complements the representation of vectors lying within the plane, which are shown as arrows pointing in their respective directions, forming a complete system for visualizing 3D vector fields on a 2D surface.

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Card 9

Front

How does Ampere's Law apply to finding the magnetic field inside a solenoid, and what path should be chosen to simplify the calculation?

Back

To apply Ampere's Law to a solenoid:

  1. Choose a rectangular path with:

    • One side parallel to the solenoid axis inside the solenoid
    • One side outside the solenoid
    • Two sides perpendicular to the axis
  2. Calculate circulation: Bdl=μ0Ienclosed\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}

  3. For this path:

    • Only the side inside contributes to the integral (B·dl = Bl)
    • Other sides contribute zero (B perpendicular to dl or B = 0)
  4. The enclosed current is: Ienclosed=nliI_{enclosed} = nli (where n = turns per unit length, l = length of path, i = current)

  5. Thus: Bl=μ0nliBl = \mu_0 nli, giving: B=μ0niB = \mu_0 ni

This uniform field exists only inside a long solenoid and points along the axis direction.

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Card 10

Front

How does the right-hand rule apply to determining magnetic field direction in circular current loops and solenoids, and what are the field characteristics in these configurations?

Back

Right-Hand Rule for Magnetic Fields

Single Circular Loop

  • Right-hand thumb rule:
    • Curl fingers of right hand in direction of current flow
    • Extended thumb points in direction of magnetic field through loop center
  • Field direction indicators:
    • Counterclockwise current → field points toward viewer
    • Clockwise current → field points away from viewer
  • Field pattern:
    • Lines emerge from one face (north pole)
    • Loop around and enter opposite face (south pole)
    • Strongest at center, weaker with distance

Solenoid (Multiple Loops)

  • Field characteristics:
    • Adjacent loops with same current direction create reinforcing internal fields
    • External fields from adjacent turns tend to cancel
    • Field strength increases with more loops (proportional to number of turns)
  • Field properties inside long solenoid:
    • Nearly uniform field strength (B = μ₀ni)
    • Parallel field lines
    • Field strength depends on current (i), turn density (n)
  • Overall pattern:
    • Resembles bar magnet with north and south poles at ends
    • Almost zero field outside a tightly wound solenoid
    • Practical application: MRI machines utilize this principle for precise imaging
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