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Magnetism Part II

Magnetism Part II. Field and Flux. Origins of Magnetic Fields. Using Biot-Savart Law to calculate the magnetic field produced at some point in space by small current elements. Using Ampere’s Law to calculate the magnetic field of a highly symmetric configuration carrying a steady current.

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Magnetism Part II

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  1. Magnetism Part II Field and Flux

  2. Origins of Magnetic Fields • Using Biot-Savart Law to calculate the magnetic field produced at some point in space by small current elements. • Using Ampere’s Law to calculate the magnetic field of a highly symmetric configuration carrying a steady current

  3. Biot-Savart Law • The vector dB is perpendicular both to ds and to the unit vector r directed from ds to P. • The magnitude of dB is inversely proportional to r2, where r is the distance from ds to P • The magnitude of dB is proportional to the current and to the magnitude ds of the length element ds. • The magnitude of dB is proportional to sin θ, where θ is the angle between the vectors ds and r.

  4. dB = (μo/4π)[(Ids x r)/r2] μo = 4π x 10-7 Tm/A B = (μoI/4π) ∫(ds x r)/r2 The integral is taken over the entire current distribution The magnetic field determined in these calculations is the field created by a current-carrying conductor This can be used for moving charges in space Biot-Savart Law

  5. Problem • Consider a thin, straight wire carrying a constant current I and placed along the x axis as shown. Determine the magnitude and direction of the magnetic field at point P due to this current. P a I

  6. P r a θ x I

  7. Solution • ds x r = k(ds x r) = k (dx sin θ) • k being the unit vector pointing out of the page • dB = dB k = (μoI/4π) [(dx sin θ)/r2]k • sin θ = a/r so r = a/ sin θ = a csc θ • x = -a cot θ • dx = a csc2 θ • dB = (μoI/4π) (a csc2 θ sin θ d θ)/(a2 csc2 θ) • B =(μoI/4πa) ∫sin θ d θ = (μoI/4πa) cos θ from θ = 0 to θ= π • B = (μoI/2πa)

  8. Problem • Calculate the magnetic field at point O for the current-carrying wire segment shown. The wire consists of two straight portions and a circular arc of radius R, which subtends and angle θ. The arrow heads on the wire indicate the direction of the current.

  9. Solution A’ • The magnetic field at O due to the current in the straight segments is zero because ds is parallel to r along these paths • In the semicircle ds is perpendicular to r so ds x r is ds A O C C’

  10. Solution Continued • dB = (μoI/4π) ( ds/R2) • B = (μoI/4π)/R2 ∫ds • B = [(μoI/4π)s]/R2 • s = rθ • B = (μoI/4πR) θ

  11. HW Problem • Consider a circular loop of radius R located on the yz plane and carrying steady current I. Calculate the magnetic field at an axial point P a distance x from the center of the loop. y z P I

  12. HW Cont’d • Ch. 30 prob. # 8,16 and 20

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