Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 4.1.7

Interactive: Ampère's Law Profiles

Problem 4.1.7

A given field: find the source

Current flows inside the infinitely long cylindrical region $ρ \leq b$ . The magnetic flux density in this region is given as

B = 200 μ_{0} a_{ϕ} [T] .

There is no surface current at $ρ = b$ .

a) Find the current density $J$ . b) Find the total current $I$ . c) Find the total current $I$ by using Ampère's formula. d) Determine the magnetic flux density for $ρ > b$ .

Try it yourself first.

Strategy

Here the field is given and the source is asked for: Ampère's law used backwards, in its differential form $J = \nabla \times H$ and in its integral form $I = \oint_{C} H \cdot d l$ . The two routes must agree.

a) The current density

In vacuum $H = B / μ_{0} = 200 a_{ϕ}$ [A/m] for $ρ \leq b$ . With only a $ϕ$ -component that depends on neither $ϕ$ nor $z$ , the curl in cylindrical coordinates (Section 2.3) reduces to

J = \nabla \times H = - \frac{\partial H _{ϕ}}{\partial z} a_{ρ} + \frac{1}{ρ} \frac{\partial ( ρ H _{ϕ} )}{\partial ρ} a_{z} = \frac{1}{ρ} \frac{\partial ( 200 ρ )}{\partial ρ} a_{z},

Answer

J = \frac{200}{ρ} a_{z} [A/ m^{2}], ρ < b

b) The total current from J

Integrate $J$ over the cross-section, $d S = ρ d ρ d ϕ a_{z}$ :

I = \int_{0}^{2 π} \int_{0}^{b} \frac{200}{ρ} ρ d ρ d ϕ = 200 (2 π) (b) = 400 π b [A] .

c) The total current from Ampère's formula

On the circle $ρ = b$ , $H$ is tangential and constant, so

I = \oint_{C} H \cdot d l = 200 (2 π b) = 400 π b [A],

the same answer: (b) integrates the source, (c) integrates the field round it, and Stokes' theorem guarantees that they agree.

d) The field outside

For $ρ > b$ no further current is enclosed, and there is no surface current at $ρ = b$ , so $I_{enc} = 400 π b$ for every $ρ > b$ :

2 π ρ H_{ϕ} = 400 π b ⟹ H_{ϕ} = \frac{200 b}{ρ} .

Answer

B = \frac{200 μ _{0} b}{ρ} a_{ϕ} [T], ρ > b

Problem 4.1.7. The given field is uniform inside the current region and continues as 200b/\rho outside, continuous at \rho=b because there is no surface current. The source |\vec J|=200/\rho grows towards the axis (dashed, shown for \rho>b/2), yet the enclosed current 400\pi\rho still goes to zero there. — **Figure 4.1.24.** Problem 4.1.7. The given field is uniform inside the current region and continues as $200 b / ρ$ outside, continuous at $ρ = b$ because there is no surface current. The source $∣ J ∣ = 200/ ρ$ grows towards the axis (dashed, shown for $ρ > b /2$ ), yet the enclosed current $400 π ρ$ still goes to zero there.

Check that $I = 400 π b$ both from $J$ and from the circulation, and that $H$ is continuous at $b$ .

InteractiveLoading simulation…

Check

$H_{ϕ}$ is continuous at $ρ = b$ , and that is the statement "no surface current". Although $∣ J ∣ \propto 1/ ρ$ grows towards the axis, the enclosed current $400 π ρ$ goes to zero there: nothing singular hides on the axis (Figure 4.1.24). Units: $200$ A/m divided by metres gives A/m $^{2}$ . Numbers: inside, $B = 200 μ_{0} = 200 \times 4 π \times 1 0^{- 7} = 251$ μT, and a wire with $b = 1$ cm carries $I = 400 π \times 0.01 = 12.6$ A. Whenever a field is prescribed, check first that it is a possible magnetostatic field: $\nabla \cdot B = 0$ holds here automatically for a $B_{ϕ}$ that depends on $ρ$ only.

Problem 4.1.7: A given field: find the source