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