### The Biot-Savart law

#### Problem:

(a)  A circular loop of wire of radius R carries a current I.  Find the magnetic induction B on the axis of the loop, as a function of the distance z from the center of the loop.
(b)  Use the result to find B at points on the axis of a solenoid of radius R and length L wound with n turns per unit length.
(c)  Use this result to find the self-inductance of a very long solenoid (L >> R).

Solution:

• Concepts:
The Biot-Savart law, the principle of superposition, self inductance
• Reasoning:
We find the field of a single loop using the Biot-Savart law.  The principle of superposition then yields the field of many loops.
• Details of the calculation:

(a)  From the Biot-Savart law we have
dBz(r) = (μ0/(4π))[I dl' x (r-r')/|r-r'|3] ∙ k.
On the z-axis r  = zk.
Symmetry dictates that B = Bzk on the z-axis.
dl = Rdφ (φ/φ), r - r' = z k - R (ρ/ρ),
dl × (r - r') = zRdφ (ρ/ρ) - R2dφ  k.
dBz = (μ0/(4π))[IR2dφ/(R2+z2)3/2].
Bz = ∫0dBz = ½μ0IR2/(R2 + z2)3/2.

At the center of the loop Bz = μ0I/(2R), and for z >> R we have Bz = μ0IR2/(2z3).
For z >> R the field is proportional to 1/z3.
(b) Assume N loops are stacked up, from z' = -z1 to z' = z1, n per unit length.
Bz = ½μ0IR2-z1z1ndz'/(R2 + (z-z')2)3/2 = ½μ0IR2n∫z-z1z+z1dz''/(R2 + z''2)3/2.
Bz = ½μ0IR2n{(z + z1)/[R2(R2 + (z+z1)2)½] - (z - z1)/[R2(R2 + (z-z1)2)½]}.

cosθ1 = (z + z1)/(R2 + (z + z1)2)½,  cosθ2 = (z - z1)/(R2 + (z - z1)2)½.
Bz = (μ0In/2)(cosθ1 - cosθ2)
For the solenoid n = N/L, z1 = L/2.
(c)  At the center of the solenoid
B
= Bzk.  Bz = (μ0In/2)(2z1/(R2 + z12)½) = (μ0IN/2)((R2 + (L/2)2)).
If L >> R then Bz ≈ (μ0In).
The field is approximately uniform inside the solenoid and approximately zero outside the solenoid.
F = L1I, F = Flux, L1 = self inductance.
F = BπR2N = μ0InπR2N = L1I, L1 = μ0n2πR2L = μ0n2V,  V = volume of the solenoid.

#### Problem:

A circuit in the form of a regular polygon of n sides is circumscribed about a circle of radius a.
(a)  If it is carrying a current I, find the magnitude of the magnetic field B at the center of the circle in terms of μ0, n, I, and a.
(b)  Find B at the center of the circle as n is indefinitely increased.

Solution:

• Concepts:
The Biot-Savart law, the principle of superposition
• Reasoning:
We find the field of a single segment using the Biot-Savart law.
The principle of superposition then lets us add the fields of the  different segments.
• Details of the calculation:
(a)  The field B at the center of the circle due to the current in the wire from -l to +l is
B(r) = (μ0/(4π))∫vI dl' × (r-r')/|r-r'|3.   (SI units).

Here r = 0,  r' = xi + aj,  dl' = dxi,  |r-r'| = (x2 + a2)½
dl' × (r-r') = -(a dx) k.
B
= -[μ0I/(4πa)]2l/(l2 + a2)½ k = -[μ0I/(2πa)]sin(α) k.
For a regular plane polygon of n sides we have n wire segments.
Example:  hexagon

For each segment 2α = 2π/n,  α = π/n.
Symmetry dictates that B(0) due to each segment is the same.
B(0) = -[nμ0I/(2πa)]sin(π/n) k.  B(0) = -[nμ0I/(2πa)]sin(π/n).
b)  As n --> ∞, sin(π/n) --> π/n,  B(0) --> nμ0I/(2πa)](π/n) = μ0I/(2a).
This is the magnetic field at the center of a circular current loop of radius a.

Similar problems

#### Problem:

Find the magnetic field at the center of a circular loop of radius R that is formed in a long straight thin wire that carries current I; use the SI system of units.

Solution:

• Concepts:
The magnetic field of a long straight wire and a current loop.
• Reasoning:
We use the principle of superposition.
• Details of the calculation:
B(r) = [μ0I/(2πR)][π + 1]  out of the page

#### Problem:

In practical magnetic structures, uniform magnetic fields are frequently necessary.  The uniformity of the field produced by Helmholtz coils, or two co-axial loops which carry currents in the same direction, is one of their most important characteristics.  Assume that the coils have a radius a, have axes on the x-axis, carry current I each, and are separated by a distance b.

(a)  Find the magnetic field at a point P on the axis of the loops and a distance x from the midpoint O.
(b)  Expand the expression for the field in a power series, retaining terms to order x2.
(c)  What relationship must exist between a and b such that the x2 terms vanish?  What is the significance of this?
(d)  Show that the field created by the coils to this order and under the conditions established in part c is given by Bx = 8I/(53/20c2).

Solution:

• Concepts:
The Biot-Savart law, a series expansion
• Reasoning:
The magnetic field on the axis of a current loop is found from the Biot-Savart law.  Expanding the field near x = 0 in terms of x, we can find the condition for the second order term to vanish.  The first order term is be zero due to the symmetry of the situation, so B(x) = B(0) to second order if the term proportional to x2 vanished.
• Details of the calculation:
(a)  B = B(x)i on the x-axis.
B(x) = (μ0Ia2/2)[a2 + (d-x)2]-3/2 + (μ0Ia2/2)[a2 + (d+x)2]-3/2
Here d = b/2.
(b)  B(x) = B(0) + x(∂B/∂x)|0 + (x2/2)(∂2B/∂x2)|0 + ...
∂B/∂x =(μ0Ia2/2)[-(3/2)2(d+x)[a2 + (d+x)2]-5/2 + (3/2)2(d-x)[a2 + (d-x)2]-5/2,
(∂B/∂x)|0 = 0.
2B/∂x2 = (3μ0Ia2/2)[5(d+x)2[a2 + (d+x)2]-7/2 - [a2 + (d+x)2]-5/2
+ 5(d-x)2[a2 + (d-x)2]-7/2 - [a2 +( d-x)2]-5/2].
(∂2B/∂x2)|0 = (3μ0Ia2/2)[10d2[a2 + d2]-7/2 - 2[a2 + d2]-5/2]
= (3μ0Ia2/2)[(8d2-2a2)[a2 + d2]-7/2].
(c)  For the second order term to vanish we need 8d2-2a2 = 0, d = a/2,  b = a.  The field is then uniform to second order.
(d)  B(x) = B(0) = (μ0Ia2)[a2 + d2]-3/2
= (μ0I/a)[1 + ¼]-3/2 = (μ0I/a)(4/5)3/2 = 8I/(53/20c2).

#### Problem:

An AC voltage source V(t) = V0sinωt is connected to a wire of resistance R which forms a circular loop of radius a.  The wire loop rotates about the z-axis with angular frequency ω.  Find the instantaneous and the average magnetic field B (magnitude and direction) at the center of the loop.  Neglect the self-inductance of the loop.

Solution:

• Concepts:
The magnetic field at the center of a current loop
• Reasoning:
The magnetic field at the center of the loop has time varying x and y components.
• Details of the calculation:
I(t) = V(t)/R = (V0/R)sinωt = I0sinωt.
B(t) = μ0I(t)/(2a) = μ0I0sinωt/(2a) = B0sinωt.  B0 = μ0V0/(2a).
B(t) is the magnetic field at the center of the loop along the n direction.
B(t) = B0sinωt(cos(ωt)i + sin(ωt) j)  is the instantaneous magnetic field at the center of the loop expressed in terms of its components along the axes of the coordinate system.
The average values of its components are
<Bx> = B0<sin(ωt) cos(ωt)> = 0,
<By> = B0<sin2(ωt)> = B0/2.
The average magnetic field is B = (B0/2) j.