Question

A tubular conductor (pipe) as shown below carries a current I in the z direction. The inner and outer radii are given by a and b, respectively, while is length is given by L. a) Describe the procedure to obtain the magnetic field distribution (H) produced by this arrangement along the radial distance any point along p). b) Use the result of a) to evaluate the magnetic field of the conductor if a = 0.05 m, b = 0.06 m and L = 1 m at the following points: p=0.045 m, p=0.055 m, p=0.065 m. The surrounding medium is air and the conductor is centered at the origin. Starting from the result of a), obtain a formula for the total inductance L of the conductor (internal of the tubular conductor Lint plus external to the outside air Lext). You can use either the method of flux linkage or the method of energy to calculate the inductance analytically (10 extra points for using both correctly).

Answer 1

We can use the Biot-Savart law to find the magnetic field due to a current. We first consider arbitrary segments on opposite sides of the loop to qualitatively show by the vector results that the net magnetic field direction is along the central axis from the loop. From there, we can use the Biot-Savart law to derive the expression for magnetic field.

Let P be a distance y from the center of the loop. From the right-hand rule, the magnetic field dB⃗ dB→ at P, produced by the current element Idl⃗ ,Idl→, is directed at an angle θθ above the y-axis as shown. Since dl⃗ dl→ is parallel along the x-axis and rˆr^ is in the yz-plane, the two vectors are perpendicular, so we have

dB=μ04πIdlsinπ/2r2=μ04πIdly2+R2dB=μ04πIdlsinπ/2r2=μ04πIdly2+R2

where we have used r2=y2+R2.r2=y2+R2.

Now consider the magnetic field dB⃗ ′dB→′ due to the current element Idl⃗ ′,Idl→′, which is directly opposite Idl⃗ Idl→ on the loop. The magnitude of dB⃗ ′dB→′ is also given by Equation above, but it is directed at an angle θθ below the y-axis. The components of dB⃗ dB→ and dB⃗ ′dB→′ perpendicular to the y-axis therefore cancel, and in calculating the net magnetic field, only the components along the y-axis need to be considered. The components perpendicular to the axis of the loop sum to zero in pairs. Hence at point P:

B⃗ =jˆ∫loopdBcosθ=jˆμ0I4π∫loopcosθdly2+R2.B→=j^∫loopdBcosθ=j^μ0I4π∫loopcosθdly2+R2.

For all elements dl⃗ dl→ on the wire, y, R, and cosθcosθ are constant and are related by

cosθ=Ry2+R2−−−−−−−√.cosθ=Ry2+R2.

Now from Equation 12.14, the magnetic field at P is

B⃗ =jˆμ0IR4π(y2+R2)3/2∫loopdl=μ0IR22(y2+R2)3/2jˆB→=j^μ0IR4π(y2+R2)3/2∫loopdl=μ0IR22(y2+R2)3/2j^

where we have used ∫loopdl=2πR.∫loopdl=2πR. As discussed in the previous chapter, the closed current loop is a magnetic dipole of moment μ⃗ =IAnˆ.μ→=IAn^. For this example, A=πR2A=πR2 and nˆ=jˆ,n^=j^, so the magnetic field at P can also be written as

B⃗ =μ0μjˆ2π(y2+R2)3/2.B→=μ0μj^2π(y2+R2)3/2.

(12.16)

By setting y=0y=0 in Equation 12.16, we obtain the magnetic field at the center of the loop:

B⃗ =μ0I2Rjˆ.B→=μ0I2Rj^.

This equation becomes B=μ0nI/(2R)B=μ0nI/(2R) for a flat coil of n loops per length. It can also be expressed as

B⃗ =μ0μ⃗ 2πR3.B→=μ0μ→2πR3.

If we consider y≫Ry≫R in Equation 12.16, the expression reduces to an expression known as the magnetic field from a dipole:

B⃗ =μ0μ⃗ 2πy3.B→=μ0μ→2πy3.

The calculation of the magnetic field due to the circular current loop at points off-axis requires rather complex mathematics, so we’ll just look at the results. Notice that one field line follows the axis of the loop. This is the field line we just found. Also, very close to the wire, the field lines are almost circular, like the lines of a long straight wire.