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