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    Permanent Magnets

    Flashcards for topic Permanent Magnets

    Intermediate85 cardsphysics

    Preview Cards

    Card 1

    Front

    How does a compass needle behave when subjected to two perpendicular magnetic fields, and what determines its final position?

    Back

    When a compass needle is subjected to two perpendicular magnetic fields:

    1. The needle aligns itself with the resultant magnetic field vector (Br)
    2. The resultant field is the vector sum of:
      • The horizontal component of Earth's magnetic field (BH)
      • The external perpendicular magnetic field (B)

    The final position is determined by:

    • The needle experiences forces mBr on its north pole and -mBr on its south pole
    • These forces produce zero net torque only when the needle aligns parallel to Br
    • The equilibrium angle θ between the needle and the original direction is given by: tan⁡θ=BBH\tan\theta = \frac{B}{B_H}tanθ=BH​B​

    This principle is utilized in tangent galvanometers and other magnetic measuring instruments.

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    Card 2

    Front

    In a scenario where a magnetic needle deflects from its north-south alignment, calculate the strength of an unknown perpendicular magnetic field given:

    • The horizontal component of Earth's magnetic field is 3.0 × 10^-5 T
    • The needle deflects 30° from its original position

    Back

    Using the tangent law of perpendicular magnetic fields:

    Step 1: Identify the formula relating the fields and deflection angle: B=BHtan⁡θB = B_H \tan\thetaB=BH​tanθ

    Step 2: Substitute the known values:

    • BH = 3.0 × 10^-5 T
    • θ = 30°

    Step 3: Calculate the unknown field: B=(3.0×10−5 T)×tan⁡(30°)B = (3.0 × 10^{-5} \text{ T}) \times \tan(30°)B=(3.0×10−5 T)×tan(30°) B=(3.0×10−5 T)×(0.577)B = (3.0 × 10^{-5} \text{ T}) \times (0.577)B=(3.0×10−5 T)×(0.577) B=1.73×10−5 TB = 1.73 × 10^{-5} \text{ T}B=1.73×10−5 T

    The perpendicular magnetic field strength is 1.73 × 10^-5 T.

    Note: This principle is fundamental to field measurement techniques and explains why a compass needle deviates in the presence of external magnetic sources.

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    Card 3

    Front

    Why does a moving-coil galvanometer include a soft-iron core, and how does this design choice affect its sensitivity?

    Back

    A soft-iron core is included in a moving-coil galvanometer to:

    1. Concentrate and intensify the magnetic field in the gap between the pole pieces
    2. Provide a low-reluctance path for magnetic flux, increasing field strength without requiring a larger permanent magnet
    3. Create a more uniform radial field distribution around the coil

    This design significantly enhances sensitivity because:

    • The sensitivity equation is: sensitivity = nAB/k
    • Increasing B directly improves sensitivity
    • The stronger field produces greater torque for a given current
    • Lower currents produce measurable deflections, making the instrument capable of detecting very small currents

    The soft-iron core thus transforms what would be a moderately sensitive instrument into a high-precision measuring device suitable for detecting microampere-level currents.

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    Card 4

    Front

    What is the Tan-A position in a deflection magnetometer, and how is the bar magnet oriented relative to the compass needle?

    Back

    The Tan-A position in a deflection magnetometer is characterized by: • The arms of the magnetometer are aligned along the magnetic east-west direction • The bar magnet is placed parallel to the arm's length (east-west orientation) • The compass needle is in "end-on" position relative to the bar magnet • The distance 'd' is measured from the center of the compass to the center of the magnet • This arrangement allows measurement of M/BH (ratio of magnetic moment to horizontal component of Earth's field) • The magnetic field due to the bar magnet is perpendicular to Earth's horizontal field in this position

    Formula used: MBH=4πμ0(d2−l2)22dtan⁡θ\frac{M}{B_H} = \frac{4\pi}{\mu_0}\frac{(d^2-l^2)^2}{2d}\tan\thetaBH​M​=μ0​4π​2d(d2−l2)2​tanθ, where θ is the deflection angle

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    Card 5

    Front

    What experimental techniques eliminate reading errors in compass-based measurements of magnetic deflection angles?

    Back

    Techniques to eliminate reading errors in magnetic deflection angle measurements:

    1. Symmetrical reading technique:

      • Read both ends of the compass pointer
      • Calculate the average of the two readings
      • Eliminates errors from pivot point eccentricity
    2. Mirror alignment method:

      • Place a plane mirror beneath the compass scale
      • Position your eye so the pointer aligns with its reflection
      • Ensures perpendicular viewing angle to eliminate parallax
    3. Rotation compensation:

      • Take readings, then rotate the compass 180°
      • Repeat measurements in the new orientation
      • Average all readings to cancel systematic errors
    4. Multiple measurement protocol:

      • Vary the distance between magnet and compass
      • Plot results and identify outliers
      • Use statistical methods to determine best-fit values

    Example application: When determining M/BH ratio (magnetic moment/horizontal field) using Tan-A position of Gauss, these techniques can improve measurement precision from ±5% to better than ±1%.

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    Card 6

    Front

    How does the relationship between cotθ (the cotangent of deflection angle) and (d²-l²)²/2d verify the inverse square law for magnetic fields in Tan-A position experiments?

    Back

    The linear relationship between cotθ and (d²-l²)²/2d directly verifies the inverse square law because:

    • For Tan-A position: cotθ = (4π/μ₀)(BH/M)[(d²-l²)²/2d]
    • When plotted, this produces a straight line through the origin
    • This confirms that magnetic field B decreases with the square of distance (1/r²)
    • The linearity demonstrates that the equation B = (μ₀/4π)(2M/r³) is valid
    • The straight-line fit with zero y-intercept validates that no additional terms affect the relationship

    This relationship is derived from the basic principle that magnetic field due to a magnetic pole is inversely proportional to the square of the distance from the pole.

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    Card 7

    Front

    What is the mathematical difference between Tan-A and Tan-B position experiments when verifying the inverse square law for magnetic fields, and why do both approaches yield valid results?

    Back

    Tan-A and Tan-B positions differ in their mathematical relationships:

    • Tan-A position (end-on):

      • cotθ = (4π/μ₀)(BH/M)[(d²-l²)²/2d]
      • Plots cotθ vs. (d²-l²)²/2d
    • Tan-B position (broadside-on):

      • cotθ = (4π/μ₀)(BH/M)(d²+l²)³/²
      • Plots cotθ vs. (d²+l²)³/²

    Both yield valid results because:

    1. Both derive from the same fundamental inverse square law
    2. Both produce linear graphs through the origin
    3. Each represents a different geometric configuration of the same physical system
    4. The mathematical differences account for the different field orientations
    5. The consistent linearity in both cases provides stronger validation than either alone

    This demonstrates the robustness of the inverse square law regardless of measurement configuration.

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    Card 8

    Front

    What is the mathematical expression for the restoring torque on a magnetic needle oscillating in a uniform magnetic field, and how does this relate to simple harmonic motion?

    Back

    The restoring torque on an oscillating magnetic needle in a uniform magnetic field is:

    Γ=−MBHsin⁡θ\Gamma = -MBH\sin\thetaΓ=−MBHsinθ

    Where:

    • MMM = magnetic moment of the needle
    • BHBHBH = horizontal component of the magnetic field
    • θ\thetaθ = angular displacement from equilibrium

    For small oscillations, sin⁡θ≈θ\sin\theta \approx \thetasinθ≈θ, so: Γ=−MBHθ\Gamma = -MBH\thetaΓ=−MBHθ

    This creates simple harmonic motion because:

    • The torque is proportional to displacement
    • The torque is directed opposite to displacement
    • The angular acceleration α=Γ/I=−(MBH/I)θ=−ω2θ\alpha = \Gamma/I = -(MBH/I)\theta = -\omega^2\thetaα=Γ/I=−(MBH/I)θ=−ω2θ

    This gives a time period: T=2πIMBHT = 2\pi\sqrt{\frac{I}{MBH}}T=2πMBHI​​

    Application: This principle is used in oscillation magnetometers to determine MBHMBHMBH.

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    Card 9

    Front

    How does the behavior of field lines differ between an electric dipole and a magnetic dipole when intersecting a closed surface that encloses only one pole?

    Back

    When a closed surface encloses only one pole of a dipole:

    For magnetic dipoles:

    • Field lines entering the surface must equal those exiting
    • The net magnetic flux remains zero (∮B⃗·dS⃗=0)
    • Field lines from the enclosed pole must curve back through the surface to reach the opposite pole
    • This reflects the absence of magnetic monopoles

    For electric dipoles:

    • Field lines can terminate on or originate from charges
    • The net electric flux equals the enclosed charge divided by ε₀ (∮E⃗·dS⃗=qenclosed/ε₀)
    • If only a positive charge is enclosed, there is a net outward flux
    • If only a negative charge is enclosed, there is a net inward flux

    This fundamental difference demonstrates why isolated electric charges can exist while magnetic monopoles cannot.

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    Card 10

    Front

    Compare and contrast Tan-A vs Tan-B positions in a deflection magnetometer, explaining how each configuration allows the calculation of MBH\frac{M}{B_H}BH​M​ and the physical principles behind them.

    Back

    Deflection Magnetometer Configurations

    Tan-A Position (End-On)

    • Geometric Arrangement:
      • Arms aligned east-west
      • Bar magnet placed parallel to arm (end-on to compass)
      • Creates field perpendicular to Earth's horizontal field (BHB_HBH​)
    • Field Equation: B=μ04π2Md(d2−l2)2B = \frac{\mu_0}{4\pi}\frac{2Md}{(d^2-l^2)^2}B=4πμ0​​(d2−l2)22Md​
    • Deflection Equation: MBH=4πμ0(d2−l2)22dtan⁡θ\frac{M}{B_H} = \frac{4\pi}{\mu_0}\frac{(d^2-l^2)^2}{2d}\tan\thetaBH​M​=μ0​4π​2d(d2−l2)2​tanθ
    • Distance Dependence: Fourth power (d2−l2)2(d^2-l^2)^2(d2−l2)2 in denominator
    • Field Strength: Stronger field effect

    Tan-B Position (Broadside-On)

    • Geometric Arrangement:
      • Arms aligned north-south
      • Bar magnet placed perpendicular to arm (broadside-on to compass)
      • Also creates field perpendicular to Earth's horizontal field
    • Field Equation: B=μ04πM(d2+l2)3/2B = \frac{\mu_0}{4\pi}\frac{M}{(d^2+l^2)^{3/2}}B=4πμ0​​(d2+l2)3/2M​
    • Deflection Equation: MBH=4πμ0(d2+l2)3/2tan⁡θ\frac{M}{B_H} = \frac{4\pi}{\mu_0}(d^2+l^2)^{3/2}\tan\thetaBH​M​=μ0​4π​(d2+l2)3/2tanθ
    • Distance Dependence: 3/2 power (d2+l2)3/2(d^2+l^2)^{3/2}(d2+l2)3/2
    • Field Strength: Generally weaker field effect

    Common Underlying Principle

    Both configurations rely on the tangent law of perpendicular fields:

    • The compass needle aligns with the resultant of two perpendicular magnetic fields
    • The tangent of deflection angle equals the ratio of perpendicular fields: tan⁡θ=BBH\tan\theta = \frac{B}{B_H}tanθ=BH​B​
    • This creates a "magnetic force balance" allowing precise measurement of MBH\frac{M}{B_H}BH​M​

    Note: Both methods measure the same physical quantity but through different geometric arrangements, providing verification through multiple approaches.

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