This video describes the Biot-Savart Law and explains each part of the equation. Jun 05, 2016 biot- savart law This law tells us about the magnetic field ( magnitude and direction ) produced by moving charges. The Biot-Savart law is a mathematical description of the magnetic field d B that arises from a current I flowing along an infinitesimal path element d l called the current element.

The application of the Biot-Savart law on the centerline of a current loop involves integrating the z-component. The symmetry is such that all the terms in this element are constant except the distance element dL, which when integrated just gives the circumference of the circle. The magnetic field is then.

Physics 2420: In–Class Problems Law of Biot–Savart

- A wire carrying a current I is shaped as shown below. Find the magnitude of the magnetic field at point P using the Law of Biot–Savart. The identity òdx[x
^{2}+ b^{2}]^{–3/2}= x/b^{2}[x^{2}+b^{2}]^{1/2}+ C may be of assistance. - A wire carrying a current I is shaped as shown below. The arcs are circular of radii a and b where a > b. The straight pieces are radial from the centre of the shape. Each large arc subtends an angle of 2π/3. Find the magnitude of the magnetic field at the centre of the shape using the Law of Biot–Savart. The relationship S = rθ may be of use.
- A loop of wire has the shape of two concentric semicircles connected by two radial segments. The loop carries current I as shown. Find the magnetic field at the point P using the Law of Biot–Savart.
- An arc of thin wire of radius R lies in the positive quadrant of the xy plane. It carries a current I clockwise. Determine the integral expression for the magnetic field acting at x = a. Don't bother evaluating the integral.
- A flat ribbon of length L and width W lies in the yz plane as shown below. It carries surface current
**K**=Cz. Determine the magnetic field*k***P**= (b, 0, 0). - A semi–circular ribbon has inner radius R and outer radius R + a. The ribbon carries a surface current
**K**= Csφ where s is radial distance and φ is the angular unit vector in polar coordinates. Determine the magnetic field acting at the centre of the curvature. - A flat disk of radius R lies in the xy plane as shown below. It carries surface charge density Σ . The disk is given angular velocity Ω φ where φ is the cylindrical rotational unit vector. Since the charge is moving, we now have a surface current density
= Σ sΩ φ where s is the distance from the centre of the disk. Determine the magnetic field acting at*K***P**= (0, 0, b). - A sphere of radius R is shown below. It carries volume charge density ρ . The sphere is given angular velocity Ω φ about the z axis where φ is the spherical rotational unit vector. Since the charge is moving, we now have a volume current density
= ρ r*J**sin*(θ)Ω φ where r is the radial distance from the centre of the sphere and angle θ is measured from the z–axis. Determine the magnetic field acting at**P**= (0, 0, b). - A cylinder has its central axis oriented in the z direction. Its base is in the xy plane. The base has radius R and infinite height. It carries a current density
**J**= Cs, where s is the cylindrical radial distance. Determine the magnetic field acting at a distance b from the centre of the base.*z* - Which of the following are possible magnetic fields?
- B
_{x}= 2xy, B_{y}= 2z+3, and B_{z}= 5–2yz. - B
_{x}= x^{2}+y^{2}, B_{y}= 3, and B_{z}= –xz. - B
_{x}= y^{3}–z^{3}, B_{y}= z^{3}–x^{3}, and B_{z}= x^{3}–y^{3}. - B
_{x}= ln(x), B_{y}= y/x, and B_{z}= 2y/x. - The diagrams below show a current–carrying loop of radius a located in the xz plane and a point P located in the yz plane. As shown in Figure 1, point P is a distance R from the centre of the loop at an angle α to the y axis. As shown in figure 2, point A is a point on the loop at an angle β from the z axis. Vector
**r**=**P**–**A**is the distance from points A to P. - Obtain expressions for
**P**,**A**,**r**, and din Cartesian (*l*) coordinates.*ijk* - Find d
in Cartesian coordinates.*l ´*r - Use the Law of Biot–Savart to obtain integral expressions for the Cartesian components of the magnetic field at point P.
- Assume R >> a and use the identity (1 – x)
^{z}= 1 + zx for small x to expand and simplify the integrands in part (c). Integrate to showwhere m = Iπ a

^{2}is the dipole moment of the loop. - Figure 3 shows the relationship between the
and*k*axes and the*j*and θ axes. Use this relationship to show that*r* - Expressing the vectors in Cartesian form.
- The cross product d
´*l***r**is - The Law of Biot-Savart states that
- These are messy integrals. However, an examination of the symmetry if the problem leads up to expect that B
_{x}must be zero as contributions to from opposite sides of the loop will cancel. Using MAPLE confirms that B_{x}= 0. - Examining Figure 3, we see that the magnetic field lying along the r axis is

We label the pieces of the wire A, B, and C. The field due to a segment of infinitesimal piece of wire is given by

d**B** = (μ_{0}/4π)I [d** l** ´ (

**r**/r

^{3})] , (1)

where **r** is the vector from the segment to the point of interest. For the radial pieces A and C, d** l** = d

**and thus d**

*r***´ (**

*l***r**/r) = 0. We need only consider the contribution from B.

Next we choose a set of axis and pick an arbitrary piece of segment B. We then must express d** l** and

**r**in Cartesian coordinates. Here we have d

**=**

*l***dxand**

*i***r**= –

**x –**

*i***h. As well, r = |**

*j***r**| = [x

^{2}+ h

^{2}]

^{½}.

Thus equation (1) becomes

d**B** = (μ_{0}/4π)I [** i** ´ (–

**x –**

*i***h)]dx/[x**

*j*^{2}+ h

^{2}]

^{3/2}.

Doing the cross product yields

d**B** = –** k** (μ

_{0}/4π)Ihdx/[x

^{2}+ h

^{2}]

^{3/2}.

This tell us that the magnetic field at point P is directed into the page which agrees with what we get from the Right–Hand Rule.

We integrate over the entire length of the segment to get **B**

| = ò |

= – _{0}/4π)I ò_{–a}^{L–a} h/[h^{2} + x^{2}]^{3/2} dx | |

(μk_{0}/4π)I {x / h[h^{2} + x^{2}]^{½} |_{–a}^{L–a}} | |

(μk_{0}/4π)(I/h){ (L–a)/ [h^{2} + (L–a)^{2}]^{½} + a/[h^{2} + a^{2}]^{½}} |

Note that in the limit L >> h, this reduces to the familiar result for a field due to a long wire

B_{wire} = (μ_{0}/4π)(I/h) .

The field due to a segment of infinitesimal piece of wire is given by

d**B** = (μ_{0}/4π)I [d** l** ´ (

**r**/r

^{3})] , (1)

where **r** is the vector from the segment to the point of interest. For the radial pieces in the diagram above, d** l** = d

**r**and thus d

**´**

*l***r**= 0. We need only consider the contribution from curved arcs.

The field due to each side curve is the same by symmetry and is directed into the paper at the origin using the right hand rule. The field due to the top and bottom curves is the same by symmetry and is also directed into the paper at the origin using the right hand rule.

In general we have arc segments to deal with. Symmetry suggests that we deal with the integrals involving circles and arcs in polar coordinates. Consider a current carrying arc such as is shown below where we have considered an arbitrary portion of the arc. We will end up integrating over θ from –α to +α . First however, we must express d** l** and

**r**in Cartesian coordinates. This is a little trickier to do here than in the previous question so let’s move d

**and**

*l***r**to the origin and do a little geometry and trigonometry. Remember vectors can be moved around as long as we do not change length or orientation.

Recall that the magnitude of d** l** is dl = | d

**| = Rdθ . Thus d**

*l***= Rdθ [**

*l*

*i**cos*(θ) –

*j**sin*(θ) ]. As well,

**r**= R [–

*i**sin*(θ) –

*j**cos*(θ)] since r = |

**r**| = R. Thus equation (1) becomes

d**B** = (μ_{0}/4π)(I/R^{3}){ Rdθ [ *i**cos*(θ) – *j**sin*(θ) ] ´ R [–*i** sin*(θ) – *j**cos*(θ)] } .

Working out the cross product yields

d**B** = –** k** (μ

_{0}/4π)(I/R)dθ [

*cos*

^{2}(θ) +

*sin*

^{2}(θ)] = –

**(μ**

*k*_{0}/4π)(I/R)dθ .

So the magnetic field at the origin of the arc is directed into the page which is what we already found from the Right–Hand Rule.

Next, we integrate over the entire angular extent of the segment to get **B**

| = ò |

= – _{0}/4π)(I/R) ò_{–α}^{α}dθ | |

(μk_{0}/4π)(I/R) {θ |_{–α}^{α}} | |

(μk_{0}/2π)(Iα/R) |

For the top and bottom coils, R = b, and α = 2π/3. For the side coils R = a and α = π/3. Thus the total field is

**B = **–

**2(μ**

*k*_{0}I/2π){(2π/3b) + (π/3a)} = –

**(μ**

*k*_{0}I/3)(2/b + 1/a) .

Note that in the limit a = b, this reduces to the familiar result for a field due to a circular wire

B = μ_{0}I/π a .

The field due to a segment of infinitesimal piece of wire is given by

d**B** = (μ_{0}/4π)I [d** l** ´ (

**r**/r

^{3})] , (1)

where **r** is the vector from the segment to the point of interest. For the radial pieces in the diagram above, d** l** = d

**r**and thus d

**´**

*l***r**= 0. We need only consider the contribution from curved arcs.

The field due to the large arc out of the paper at the point P using the right hand rule. The field due to the small arc is directed into the paper at the P. From question 3, we found that the field due to an arc was

B ** = **(μ

_{0}I/2π) (α/r) .

For both arcs, α = π/2. For the top arc r = 2R, and for the bottom r = R. Taking out of the page as positive

B = (μ_{0}I/4)(1/2R – 1/R) = –μ_{0}I/8R .

The net field is into the paper.

The field due to a segment of infinitesimal piece of wire is given by

d**B** = (μ_{0}/4π)I [d** l** ´

**r**] / r

^{3}, (1)

where **r** is the vector from the segment to the point of interest. Putting the origin at the centre of the radius of curvature and picking an arbitrary piece or the wire d** l**, we find

**S**=

**R**

*i**cos*(φ) +

**R**

*j**sin*(φ) and

**P**=

**a. Therefore**

*i***r**=

**P**–

**S**=

**[a – R**

*i**cos*(φ)] –

**R**

*j**sin*(φ). The magnitude of this vector is r = [a

^{2}– 2aR

*cos*(φ) + R

^{2}]

^{½}. Note that d

**= –φ Rdφ , since the current moves in the clockwise or negative direction. To do the cross product, we need to express d**

*l***in**

*l***notation. Placing d**

*ijk***at the origin as shown in the next diagram, doing some geometry, and determining the components of d**

*l***, we see that d**

*l***= Rdφ [**

*l*

*i**sin*(φ) –

*j**cos*(φ)].

Thus equation (1) for this problem is

d**B** = (μ_{0}/4π)I Rdφ [*i**sin*(φ)–*j**cos*(φ)]´ [** i** [a–R

*cos*(φ)]–

**R**

*j**sin*(φ)]/[a

^{2}–2aR

*cos*(φ)+R

^{2}]

^{3/2}.

Doing the cross product yields

And we have

d**B** = ** k** (μ

_{0}/4π)IRdφ [a

*cos*(φ) – R]/[ a

^{2}–2aR

*cos*(φ)+R

^{2}]

^{3/2}.

This tells us that the magnetic field at point P is directed out of the page which agrees with what we get from the Right–Hand Rule.

We integrate over the entire length of the segment from 0 to π/2 to get **B**

| = ò |

= – _{0}IR/4π) ò_{0}^{½π }dφ [acos(φ) – R]/[ a^{2}–2aRcos(φ)+R^{2}]^{3/2} |

The field due to a surface current is given by

**B** = (μ_{0}/4π) ò_{surface} dS [** K** ´

**r**] / r

^{3}, (1)

where **r** is the vector from the segment of the surface current to the point of interest. Using the given origin and picking an arbitrary piece of the wire dS, we find **S** = ** j** y +

**z and**

*k***P**=

**b. Therefore**

*i***r**=

**P**–

**S**=

**b –**

*i***y –**

*j***z. The magnitude of this vector is r = [b**

*k*^{2}+ y

^{2}+ z

^{2}]

^{½}. The surface element is dS = dydz where -½W £ y £ ½W and 0 £ z £ L. Thus equation (1) for this problem is

**B** = (μ_{0}/4π) ò_{½W}^{½W} dy ò_{0}^{L} dz {** k**Cz ´ [

**b –**

*i***y –**

*j***z]} / [b**

*k*^{2}+ y

^{2}+ z

^{2}]

^{3/2}

Doing the cross product yields

And we have

**B** = (μ_{0}/4π)C ò_{½W}^{½W} dy ò_{0}^{L} dz {** i** zy +

**zb} / [b**

*j*^{2}+ y

^{2}+ z

^{2}]

^{3/2}.

Using MAPLE to integrate the components of the magnetic field, we find B_{x} = 0 and

The field due to a surface current is given by

**B** = (μ_{0}/4π) ò_{surface} dS [** K** ´

**r**] / r

^{3}, (1)

where **r** is the vector from the segment of the surface current to the point of interest. The object has planar circular (polar coordinate) symmetry, so the variables of integration should be s and φ with the area element being dS = sdsdφ where R £ s £ R+a and 0 £ φ £ π . All directions should be given in standard ** ijk** notation for convenience in dealing with the cross product. We know that φ = –

*i**sin*(φ) +

*j**cos*(φ). The origin is at the centre of curvature. Picking an arbitrary piece of the object dS, we find

**S**=

**x +**

*i***y and**

*j***P**= 0. Therefore

**r**=

**P**–

**S**= –

**x –**

*i***y. The magnitude of this vector is r = [x**

*j*^{2}+ y

^{2}]

^{½}. We need to express x and y in terms of s and φ . We know x = s

*cos*(φ) and y = s

*sin*(φ), so

**r**= –

**s**

*i**cos*(φ) –

**s**

*j**sin*(φ) and r = s. Thus equation (1) for this problem is

**B** = (μ_{0}/4π) ò_{R}^{R+a} ds ò_{0}^{π} dφ s {Cs[–*i**sin*(φ) + *j**cos*(φ)] ´ [–** i** s

*cos*(φ) –

**s**

*j**sin*(φ)]} / s

^{3}

Doing the cross product yields

And we have

**B** = ** k** (μ

_{0}/4π) C ò

_{R}

^{R+a}ds ò

_{0}

^{π}dφ =

**(μ**

*k*_{0}/4) CR .

The field due to a surface current is given by

**B** = (μ_{0}/4π) ò_{surface} dS [** K** ´

**r**] / r

^{3}, (1)

where **r** is the vector from the segment of the surface current to the point of interest. The object has planar circular (polar coordinate) symmetry, so the variables of integration should be s and φ with the area element being dS = sdsdφ where 0 £ s £ R and 0 £ φ £ 2π . All directions should be given in standard ** ijk** notation for convenience in dealing with the cross product. We know that φ = –

*i**sin*(φ) +

*j**cos*(φ). The origin is at the centre of the disk. Picking an arbitrary piece of the disk dS, we find

**S**=

**x +**

*i***y and**

*j***P**=

**b. Therefore**

*k***r**=

**P**–

**S**= –

**x –**

*i***y +**

*j***b. The magnitude of this vector is r = [x**

*k*^{2}+ y

^{2}+ b

^{2}]

^{½}. We need to express x and y in terms of s and φ . We know x = s

*cos*(φ) and y = s

*sin*(φ), so

**r**= –

**s**

*i**cos*(φ) –

**s**

*j**sin*(φ) +

**b and r = [s**

*k*^{2}+ b

^{2}]

^{½}. Thus equation (1) for this problem is

**B** = (μ_{0}/4π)ò_{0}^{R} ds ò_{0}^{2π}dφ s{Σ sΩ[–*i**sin*(φ)+*j**cos*(φ)]´ [–** i**s

*cos*(φ)–

**s**

*j**sin*(φ)+

**b]}/[s**

*k*^{2}+b

^{2}]

^{3/2}

Doing the cross product yields

And we have

**B** = (μ_{0}/4π) Σ Ω ò_{0}^{R} ds ò_{0}^{2π }dφ s^{2} [** i **b

*cos*(φ) +

**b**

*j**sin*(φ) +

**s] / [s**

*k*^{2}+ b

^{2}]

^{3/2}

The ** i** and

**terms vanish upon integration over φ , leaving**

*j*The field due to a volume current is given by

**B** = (μ_{0}/4π) ò_{volume} dV [** J** ´

**g**] / g

^{3}, (1)

where **g** is the vector from the segment of the surface current to the point of interest. The object has spherical symmetry, so the variables of integration should be r, θ , and φ with the volume element being dV = r^{2}*sin*(θ)drdθ dφ where 0 £ r £ R, 0 £ θ £ π and 0 £ φ £ 2π . All directions should be given in standard ** ijk** notation for convenience in dealing with the cross product. We know that φ = –

*i**sin*(φ) +

*j**cos*(φ). The origin is at the centre of the sphere. Picking an arbitrary piece of the sphere dV, we find

**S**=

**x +**

*i***y +**

*j***z and**

*k***P**=

**b. Therefore**

*k***g**=

**P**–

**S**= –

**x –**

*i***y +**

*j***(b – z). The magnitude of this vector is g = [x**

*k*^{2}+ y

^{2}+ z

^{2}+ b

^{2}– 2bz]

^{½}. We need to express x, y and z in terms of r, θ , and φ . We know x = r

*sin*(θ)cos(φ), y = r

*sin*(θ)

*sin*(φ), and z = r

*cos*(θ) so

**r**= –

**r**

*i**sin*(θ)cos(φ) –

**r**

*j**sin*(θ)

*sin*(φ) +

**[b – r**

*k**cos*(θ)] and g = [r

^{2}+ b

^{2}– 2br

*cos*(θ)]

^{½}. Thus equation (1) for this problem is

**B** = (μ_{0}/4π) ò_{0}^{R} dr ò_{0}^{2π }dφ ò_{0}^{π} dθ r^{2}*sin*(θ) {ρ r*sin*(θ)Ω [–*i**sin*(φ)+*j**cos*(φ)]´ [– ** i** r

*sin*(θ)cos(φ) –

**r**

*j**sin*(θ)

*sin*(φ) +

**(b – r**

*k**cos*(θ))]}/[r

^{2}+ b

^{2}– 2br

*cos*(θ)]

^{3/2}

Doing the cross product yields

And we have

The ** i** and

**terms vanish upon integration over φ , leaving**

*j*B_{z} = (μ_{0}/2) ρ Ω ò_{0}^{R} dr ò_{0}^{π} dθ r^{4 }*sin*^{3}(θ) / [r^{2} + b^{2} – 2br*cos*(θ)]^{3/2} .

The integrand may be converted to a series by expanding about s = 0. Only the first term gives a non-zero result for the integration over θ . We are left with

B_{z} = 2μ_{0} ρ ΩR^{5 }/ 15b^{3} .

The field due to a volume current is given by

**B** = (μ_{0}/4π) ò_{volume} dV [** J** ´

**r**] / r

^{3}, (1)

where **r** is the vector from the segment of the surface current to the point of interest. The object has cylindrical symmetry, so the variables of integration should be s, φ , and z. All directions should be given in standard ** ijk** notation for convenience in dealing with the cross product. We take the origin is at the bottom of the cylinder and assume that the point of interest is on the y axis. Picking an arbitrary piece of the cylinder dV = sdsdφ dz, we find

**S**=

**x +**

*i***y +**

*j***z and**

*k***P**=

**b. Therefore**

*j***r**=

**P**–

**S**= –

**x +**

*i***(b – y) –**

*j***z. The magnitude of this vector is r = [x**

*k*^{2}+ y

^{2}+ z

^{2}+ b

^{2}– 2by]

^{½}. We need to express x, y and z in terms of s, φ , and z. We know x = s

*cos*(φ), y = s

*sin*(φ), and z = z so

**r**= –

**s**

*i**cos*(φ) +

**[b – s**

*j**sin*(φ)] –

**z and r = [z**

*k*^{2}+ s

^{2}+ b

^{2}– 2bs

*sin*(φ)]

^{½}. Thus equation (1) for this problem is

The cross product yields

Thus our equation becomes

The ** j** term vanishes leaving

The integral over z yields

The integrand may be converted to a series by expanding about s = 0. Only the first term gives a non-zero result for the integration over φ . We are left with

B_{z} = – μ_{0}CR^{3}/6b .

Noting that the current in this case I = ò_{0}^{R} dr ò_{0}^{2π }dφ sJ = 2π CR^{3}/3 , the above answer reduces to B_{z} = – μ_{0}I / 4π b. The same result as a long thin wire.

Real magnetic fields have zero divergence, Ñ·**B** = 0

(a) |
| = ¶ B |

= ¶ (2xy)/¶ x + ¶ (2z + 3)/¶ y + ¶ (5 – 2yz)/¶ z | ||

= 2y + 0 – 2y | ||

= 0 |

So this may be a magnetic field.

(b) |
| = ¶ B |

= ¶ (x | ||

= 2x + 0 – x | ||

= x |

So this cannot be a magnetic field.

(c) |
| = ¶ B |

= ¶ (y | ||

= 0 + 0 + 0 | ||

= 0 |

So this may be a magnetic field.

(d) |
| = ¶ B |

= ¶ | ||

= 1/x + 1/x + 0 | ||

= 2/x |

So this cannot be a magnetic field.

Examining Figure 2, we immediately see that

**A** = a*sin*(β ) ** i** + 0

**+ a**

*j**cos*(β

**)**.

*k*By geometry, we see that d** l** is angle β below the positive x axis. As well, the length of the arc is dl = | d

**| = a dβ . Hence**

*l*d** l** = adβ

*cos*(β )

**+ 0**

*i***- adβ**

*j**sin*(β

**)**.

*k*We are told **P** is in the yz plane so

**P** = 0 ** i** + R

*cos*(α)

**+ R**

*j**sin*(α)

**.**

*k*Vector **r** is the vector from **A** to **P**, so

**r** = **P** - **A** = -a*sin*(β ) ** i** + R

*cos*(α)

**+ [R**

*j**sin*(α) - a

*cos*(β

**)**]

**.**

*k*d r | = adβ { - jsin(β ) } k´ {-a cos(α) + [Rjsin(α)-acos(β )]}k |

= adβ { R sin(α)cos(β )] + Rjcos(α)cos(β ) }k |

d** B** = (μ

_{0}/4π)(I/r

^{3}) d

**´**

*l***r**.

We first need to find r = | **r** |,

Then we observe that we need to integrate around the entire circumference of the loop from β = 0 to β = 2π . So the field is

Expanding the integral’s denominator, we have

We neglect terms of order (a/R)^{2} and higher. Thus the x and y components of the magnetic field are

and

where we have used MAPLE to evaluate the integrals.

B_{r} = B_{y}*cos*(α) + B_{z}*sin*(α) .

After using some trigonometric identities, we find

B_{r} = 2μ_{0}m*cos*(α)/4πR^{2} .

Similarly, the magnetic field along the θ axis is

B_{θ} = -B_{y}*sin*(α) + B_{z}cos(α) ,

which reduces to

B_{θ} = μ_{0}m*sin*(α)/4π R^{2} .

The B_{φ} term is zero since it is at right angles to B_{r} and B_{θ} which in turn makes it perpendicular to the B_{y} and B_{z} terms.

Questions?[email protected]

A law of physics which states that the magnetic flux density (magnetic induction) near a long, straight conductor is directlyproportional to the current in the conductor and inversely proportional to the distance from the conductor. The field near astraight conductor can be found by application of Ampère's law. The magnetic flux density near a long, straight conductor isat every point perpendicular to the plane determined by the point and the line of the conductor. Therefore, the lines ofinduction are circles with their centers at the conductor. Furthermore, each line of induction is a closed line. Thisobservation concerning flux about a straight conductor may be generalized to include lines of induction due to a conductor ofany shape by the statement that every line of induction forms a closed path.

Biot-Savart law is a fundamental quantitative relationship between an electric current and the magnetic field it produces, based on the experiments in 1820 of the French scientists Jean-Baptiste Biot and Félix Savart. The Biot-Savart law is applied in a specific case by adding up the contributions to the magnetic field at a given point from the whole series of short current segments that constitute a specific conductor of whatever shape. For instance, with a very long straight wire carrying current, the value of the magnetic field at a point nearby is just directly proportional to the value of the current and inversely proportional to the perpendicular distance from the wire to the given point. Compare Ampère’s law.

### Biot Savart Law Explained

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Therefore final Biot Savart law derivation is,