the biot-savart law
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The Biot-Savart Law. AP Physics C Montwood High School R. Casao. Biot and Savart recognized that a conductor carrying a steady current produces a force on a magnet. - PowerPoint PPT PresentationTRANSCRIPT
The Biot-Savart Law
AP Physics C
Montwood High School
R. Casao
• Biot and Savart recognized that a conductor carrying a steady current produces a force on a magnet.
• Biot and Savart produced an equation that gives the magnetic field at some point in space in terms of the current that produces the field.
• Biot-Savart law says that if a wire carries a steady current I, the magnetic field dB at some point P associated with an element of conductor length ds has the following properties:– The vector dB is perpendicular to both ds (the
direction of the current I) and to the unit vector rhat directed from the element ds to the point P.
– The magnitude of dB is inversely proportional to r2, where r is the distance from the element ds to the point P.
– The magnitude of dB is proportional to the current I and to the length ds of the element.
– The magnitude of dB is proportional to sin , where is the angle between the vectors ds and rhat.
• Biot-Savart law:
2
o
rπ4r̂xdsIμ
dB
• o is a constant called the permeability of free space; o =4· x 10-7 Wb/A·m (T·m/A)
• Biot-Savart law gives the magnetic field at a point for only a small element of the conductor ds.
• To determine the total magnetic field B at some point due to a conductor of specified size, we must add up every contribution from all elements ds that make up the conductor (integrate)!
2o
rr̂xds
π4Iμ
dB
• The direction of the magnetic field due to a current carrying element is perpendicular to both the current element ds and the radius vector rhat.
• The right hand rule can be used to determine the direction of the magnetic field around the current carrying conductor: – Thumb of the right hand in the
direction of the current.– Fingers of the right hand curl around
the wire in the direction of the magnetic field at that point.
Magnetic Field of a Thin Straight Conductor
• Consider a thin, straight wire carrying a constant current I along the x axis. To determine the total magnetic field B at the point P at a distance a from the wire:
• Use the right hand rule to determine that the direction of the magnetic field produced by the conductor at point P is directed out of the page.
• This is also verified using the vector cross product (ds x rhat): fingers of right hand in direction of ds; point palm in direction of rhat (curl fingers from ds to rhat); thumb points in direction of magnetic field B.
• The cross product (ds x rhat) = ds·rhat·sin rhat is a unit vector and the magnitude of a unit vector = 1.
• (ds x rhat) = ds·rhat·sin ds·sin
• Each length of the conductor ds is also a small length along the x axis, dx.
• Each element of length ds is a distance r from P and a distance x from the midpoint of the conductor O. The angle will also change as r and x change.
• The values for r, x, and will change for each different element of length ds.
• Let ds = dx, then ds·sin becomes dx·sin • The contribution to the total magnetic field at point
P from each element of the conductor ds is:
2o
rθsindx
π4Iμ
dB
• The total magnetic field B at point P can be determined by integrating from one end of the conductor to the other end of the conductor.
• The distance a from the midpoint of the conductor O to the point P remains constant.
• Express r in terms of a and x.
• Express sin in terms of a and r.
21
22 xa
ara
θsin
21
22222 xarxar
• For an infinitely long wire:
• From the table of integrals:
23
22
o
23
22
o
21
2222o
2o
2o
xa
dxπ4aIμ
xa
dxaπ4Iμ
B
xa
axa
dxπ4Iμ
B
rθsindx
π4Iμ
rθsindx
π4Iμ
dB
2
12222
322 xaa
x
xa
dx
aπ2Iμ
B
aπ4Iμ2
11aπ4Iμ
aπ4Iμ
B
aπ4Iμ
B
aaaπ4Iμ
B
xa
xaπ4aIμ
xaa
xπ4aIμ
B
o
ooo
21
221
2
o
21
2221
22
o
21
222o
21
222
o
• For a conductor with a finite length:
• From the table of integrals:
x
x
x
x
x
x
x
x
x
x
x
x
23
22
o
23
22
o
21
2222o
2o
2o
xa
dxπ4aIμ
xa
dxaπ4Iμ
B
xa
axa
dxπ4Iμ
B
rθsindx
π4Iμ
rθsindx
π4Iμ
dB
2
12222
322 xaa
x
xa
dx
21
22
o
21
22
o
21
2221
22
o
21
2221
22
o
x
21
222o
x
21
222
o
xaaπ2
xIμB
xa
x2aπ4Iμ
B
xa
x
xa
xaπ4Iμ
B
xa
x
xa
xaπ4Iμ
B
xa
xaπ4aIμ
xaa
xπ4aIμ
B
xx
• When the angles are provided:– express r in terms of a and the angle :
– Because angles are involved, we need to change dx to d:
- Take the derivative of x:
θcscaθsin
ar
ra
θsin
θcotaθtana
xxa
θtan
dθθcscadx
dθθcscadx2
2
• To determine the magnitude of the magnetic field B, integrate:
12
oθ
θ
o
θ
θ
o
θ
θ 22
2
o
θ
θ 2
2
o
2
θ
θ
oθ
θ
θcosθcos-aπ4
Iμθcos-
aπ4
IμB
dθθsinaπ4
IμB
θcsca
dθθsinθcsca
π4
IμB
θcsca
θsindθθcsca
π4
IμB
r
θsindx
π4
IμdB
2
1
2
1
2
1
2
1
2
1
2
1
• The magnetic field lines are concentric circles that surround the wire in a plane perpendicular to the wire.
• The magnitude of B is constant on any circle of radius a.
• The magnitude of the magnetic field B is proportional to the current and decreases as the distance from the wire increases.
Magnetic Field of a Current Loop• To determine the magnetic field B at the
point O for the current loop shown:
• The magnetic field at point O due to the straight segments AA' and CC' is zero because ds is parallel to rhat along path AA' and ds is antiparallel to rhat along path CC'.
• For the curved portion of the conductor from A to C, divide this into small elements of length ds.
• Each element of length ds is the same distance R away from point O.
0180sin0;0sin
θsinr̂dsr̂xds
• Each element of length ds contributes equally to the total magnetic field B at point O.
• The direction of the magnetic field B at point O is down into the page.
• At every point from A to C, ds is perpendicular to rhat, therefore:
• Integrate from A to C:
ds90sin1dsθsinr̂dsr̂xds
C
A 2o
2
C
A
oC
A Rds
π4Iμ
rr̂xds
π4Iμ
dB
• Pull the constant R out in front of the integral and integrate from A to C:
• The distance s is the arc length from A to C; arc length s = R·. Revising the equation:
2o
C
A2o
Rπ4sIμ
dsRπ4Iμ
B
Rπ4θIμ
Rπ4θRIμ
Rπ4sIμ
B o2
o2
o
Magnetic Field on the Axis of a Circular Current Loop
• Consider a circular loop of wire of radius R in the yz plane and carrying a steady current I:
• To determine the magnetic field B at a point P on the axis a distance x from the center of the loop:– Divide the current loop into small elements
of length ds.– Each element of length ds is the same
distance r to point P on the x axis.– Each element of length ds contributes equally
to the total magnetic field B at point P.
2
o
rr̂xds
π4Iμ
dB
• Express r in terms of R and x:
• Each element of length ds is perpendicular to the unit vector rhat from ds to point P.
• Substituting into the integral equation:
222 xRr
ds90sin1dsθsinr̂dsr̂xds
22o
xRds
π4Iμ
dB
•Notice that the direction of the magnetic field contribution dB from element of length ds is at an angle with the x axis.
• At point P, the magnetic field contribution from each element of length ds can be resolved into an x component (dBx) and a y component (dBy).
• The dBy component for the magnetic field from an element of length ds on one side of the ring is equal in magnitude but opposite in direction to the dBy component for the magnetic field produced by the element of length ds on the opposite side of the ring (180º away). These dBy components cancel each other.
• The net magnetic field B at point P is the sum of the dBx components for the elements of length ds.
• The direction of the net magnetic field is along the x axis and directed away from the circular loop.
θcosxR
dsπ4Iμ
dB
θcosxR
dsπ4Iμ
dB
θcosdBdB
22o
22o
x
• Express R2 + x2 in terms of an angle :
• Substituting into the integral equation:
21
22 xR
RrR
θcos
21
2222o
22o
xR
RxR
dsπ4Iμ
dB
θcosxR
dsπ4Iμ
dB
• Pull the constants out in front of the integral:
• The sum of the elements of length ds around the closed current loop is the circumference of the current loop; s = 2··R
dsxRπ4
RIμB
xR
Rdsπ4Iμ
dB
23
22
o
23
22
o
• The net magnetic field B at point P is given by:
23
22
2o
23
22
2o
23
22
o
xR2
RIμB
xRπ4
RIμπ2B
Rπ2xRπ4
RIμB
• To determine the magnetic field strength B at the center of the current loop, set x = 0:
R2Iμ
B
R2RIμ
R2
RIμ
R2
RIμB
R2
RIμ
xR2
RIμB
o
3
2o
6
2o
32
2o
23
2
2o
23
22
2o
• For large distances along the x axis from the current loop, where x is very large in comparison to R:
3
2o
6
2o
32
2o
23
2
2o
23
22
2o
x2RIμ
B
x2
RIμ
x2
RIμB
x2
RIμ
xR2
RIμB
• The magnetic dipole of the loop is the product of the current I and the area A of the loop: = I··R2
23
22
o
23
22
2o
2
xRπ2
μμB
xR2
RIμB
πμ
RI