magnetism part ii field and flux. origins of magnetic fields using biot-savart law to calculate the...
TRANSCRIPT
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
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.
Biot-Savart Law
• 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
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.
I
P
a
I
a
P
x
rθ
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)
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.
Solution
• 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
OA
A’
C
C’
Solution Continued
• dB = (μoI/4π) ( ds/R2)
• B = (μoI/4π)/R2 ∫ds
• B = [(μoI/4π)s]/R2
• s = rθ
• B = (μoI/4πR) θ
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
PI
HW Cont’d
• Ch. 30 prob. # 8,16 and 20