Electromagnetic Waves, Antennas, and RF: Ampère’s Law and Symmetry - Exercises and Complete Worked Solutions

This companion article provides self-study exercises for EM12, Ampère’s Law and Symmetry. All exercises are stated first. Complete worked solutions follow in Part II.

The central magnetostatic circulation law is

(1)

For an infinitely long straight wire,

(2)

For a cylindrical wire of radius a carrying uniformly distributed current I,

(3)

and

(4)

For an ideal long solenoid,

(5)

and for an ideal toroid within its winding region,

(6)

The equivalent local magnetostatic law is

(7)

These are the same definitions and conventions developed in EM12 [1, 2, 3, 5].

How to use this problem set

Attempt every exercise in Part I before consulting Part II. For each Ampère-law problem, identify the symmetry first. Then decide whether the proposed Amperian path makes the field tangent to the path and whether the magnitude of B is constant on the contributing segment. Only after those geometric steps should the circulation integral be simplified.

Part I: Exercises

Exercise 1: magnetic circulation around a known circular field

A magnetic field has constant magnitude

(8)

and is everywhere tangent to a circular path of radius

(9)

The field points in the same direction as the path traversal. Find

(10)

Exercise 2: signed enclosed current

A closed Amperian path is traversed counterclockwise as viewed from the +z side. Three currents pierce the bounded surface:

(11)

(12)

and

(13)

Find the signed enclosed current and the resulting magnetic circulation.

Exercise 3: field around a long straight wire

An infinitely long straight wire carries

(14)

Find the magnetic-field magnitude at perpendicular distance

(15)

State why a circular Amperian loop centered on the wire is a symmetry-matched path.

Exercise 4: infer current from a measured field

A magnetic-field sensor measures

(16)

at perpendicular distance

(17)

from a very long straight wire. Find the wire current.

Exercise 5: enclosed current inside a uniformly current-filled wire

A cylindrical wire has radius

(18)

and carries total current

(19)

uniformly over its cross section. An Amperian circle has radius

(20)

Find:

1. the enclosed current I_enc;
2. the magnetic-field magnitude at that radius.

Exercise 6: where is the field largest in a uniform-current wire?

A uniformly current-filled wire has radius

(21)

and carries current

(22)

Use the inside and outside formulas to show that the magnetic field is continuous at s = a. Find the field magnitude there and explain why this is the maximum field magnitude for this idealized wire.

Exercise 7: field outside a finite-radius wire

A cylindrical wire carries total current

(23)

Find the magnetic field at radius

(24)

where the observation point lies outside the wire. Explain why the outside field does not depend on the wire radius when the total current is fixed.

Exercise 8: ideal long-solenoid field

An ideal long solenoid has

(25)

and carries current

(26)

Find the interior magnetic-field magnitude.

Exercise 9: design the turn density of a solenoid

An ideal long solenoid must produce

(27)

with current

(28)

Find the required turn density n. If the solenoid length is

(29)

estimate the required number of turns N = nL.

Exercise 10: ideal-toroid field

An ideal toroid has

(30)

turns and carries current

(31)

Find the magnetic field within the winding region at

(32)

Exercise 11: choose the useful Amperian path

For each source below, state whether the proposed path is useful for solving directly for B, and explain why.

1. A circle centered on an infinitely long straight wire.
2. A circle displaced sideways from an infinitely long straight wire.
3. A rectangular path with one long side inside an ideal long solenoid and one long side outside.
4. An arbitrary irregular loop around two separated finite wires.

Your answer should distinguish between “Ampère’s law is true” and “the path lets us pull B out of the integral.”

Exercise 12: derive the inside-wire field from current density

A cylindrical Conductor of radius a carries uniform current density

(33)

Using an Amperian circle of radius s < a, derive the magnetic-field magnitude in terms of J₀ and s. Show that

(34)

Exercise 13: differential Ampère law from a specified field

Suppose

(35)

where

(36)

Find:

1. ∇× B;
2. the corresponding current density J using

   (37)

Exercise 14: magnetostatic limitation and Maxwell’s correction

Explain why

(38)

is not the complete law for genuinely time-varying electromagnetic fields. State the additional term that Maxwell introduced and explain physically what quantity it depends on.

Part II: Complete Worked Solutions

Solution 1: magnetic circulation around a known circular field

Because B is tangent to the path, points in the same direction as the traversal, and has constant magnitude,

(39)

Substitute the values:

∮ CB ⋅ dℓ	= (1.5 × 10−4)2π(6.0 × 10−2)	(40)
	= 5.65 × 10−5 T m.	(41)

Therefore,

(42)

Solution 2: signed enclosed current

Counterclockwise traversal as seen from +z corresponds, by the right-hand rule, to a positive surface normal in the +z direction.

Therefore currents in +z count positive and currents in −z count negative:

Ienc	= I1 − I2 + I3	(43)
	= 5.0 − 2.0 + 1.5	(44)
	= 4.5 A.	(45)

Thus,

(46)

Ampère’s law then gives

∮ CB ⋅ dℓ	= μ0Ienc	(47)
	= (4π × 10−7)(4.5)	(48)
	= 5.65 × 10−6 T m.	(49)

Therefore,

(50)

Solution 3: field around a long straight wire

For an infinitely long straight wire,

(51)

Substitute:

B	=	(52)
	= 1.20 × 10−4 T.	(53)

Hence,

(54)

The centered circle is useful because cylindrical symmetry guarantees that B is tangent to the circle and that its magnitude is the same everywhere on that circle.

Solution 4: infer current from a measured field

Rearrange

(55)

to obtain

(56)

Substitute:

I	=	(57)
	= 12 A.	(58)

Therefore,

(59)

Solution 5: enclosed current inside a uniformly current-filled wire

Uniform current density means the enclosed current scales with enclosed cross-sectional area:

(60)

Substituting,

Ienc	= 122	(61)
	= 122	(62)
	= 1.33 A.	(63)

Thus,

(64)

Now apply Ampère’s law:

(65)

Therefore,

B	=	(66)
	=	(67)
	= 1.33 × 10−4 T.	(68)

Hence,

(69)

Solution 6: where is the field largest in a uniform-current wire?

Inside the wire,

(70)

At s = a,

(71)

Outside the wire,

(72)

so at s = a,

(73)

Thus the two expressions agree at the surface; the field is continuous there.

Numerically,

B(a)	=	(74)
	= 4.0 × 10−4 T.	(75)

Therefore,

(76)

Inside, B ∝ s, so the field rises with radius. Outside, B ∝ 1∕s, so it falls with radius. Therefore the maximum occurs at the surface s = a.

Solution 7: field outside a finite-radius wire

Outside the wire, the Amperian loop encloses the full current. Therefore,

(77)

Substitute:

B	=	(78)
	= 1.33 × 10−4 T.	(79)

So,

(80)

Once the observation point lies outside the conductor, Ampère’s law depends only on the total enclosed current and the observation radius. The detailed distribution of that current inside the wire does not affect this symmetry result.

Solution 8: ideal long-solenoid field

For an ideal long solenoid,

(81)

Substitute:

B	= (4π × 10−7)(1500)(0.60)	(82)
	= 1.13 × 10−3 T.	(83)

Therefore,

(84)

Solution 9: design the turn density of a solenoid

From

(85)

solve for n:

(86)

Substituting,

n	=	(87)
	≈ 1.66 × 103 turns/m.	(88)

Thus,

(89)

The turn count for L = 0.30 m is

N	= nL	(90)
	≈ (1.658 × 103)(0.30)	(91)
	≈ 497.	(92)

Therefore about

(93)

are required.

Solution 10: ideal-toroid field

For an ideal toroid,

(94)

Substitute:

B	=	(95)
	= 6.67 × 10−4 T.	(96)

Thus,

(97)

Solution 11: choose the useful Amperian path

1. Useful. A circle centered on an infinitely long straight wire matches cylindrical symmetry. The field is tangent to the circle and has constant magnitude on it.
2. Not useful for directly solving for B. Ampère’s law remains true, but the distance to the wire varies around the off-center circle. Thus the field magnitude is not constant and the field is not everywhere tangent to that circle.
3. Useful under the ideal long-solenoid approximation. The interior long side is parallel to the nearly uniform interior field, the exterior long side lies where the idealized field is negligible, and the short sides are perpendicular to B.
4. Generally not useful for directly solving the local field. Ampère’s law still gives the total circulation, but neither field direction nor magnitude is generally simple enough along an arbitrary irregular loop to extract a single B.

The distinction is important: the law remains valid under magnetostatic conditions even when a particular path does not simplify the unknown field.

Solution 12: derive the inside-wire field from current density

For uniform current density J₀, the enclosed current within radius s < a is

(98)

Cylindrical symmetry gives a magnetic field tangent to the circular Amperian loop and constant in magnitude along it. Thus,

(99)

Cancel one factor of πs:

(100)

Therefore,

(101)

This form makes the linear dependence on radius especially clear.

Solution 13: differential Ampère law from a specified field

The field components are

(102)

Only the z component of the curl is nonzero:

(∇× B)z	= −	(103)
	= α − (−α)	(104)
	= 2α.	(105)

Therefore,

(106)

With α = 3.0 × 10⁻⁴ T/m,

(107)

Using

(108)

we obtain

J	= z	(109)
	= z	(110)
	≈ 4.77 × 102z A/m2.	(111)

Thus,

(112)

Solution 14: magnetostatic limitation and Maxwell’s correction

The simple form

(113)

is the magnetostatic form of Ampère’s law. It assumes steady currents and therefore does not contain all sources of magnetic-field circulation in a time-varying electromagnetic system.

Maxwell’s correction adds a term proportional to the time rate of change of electric flux:

(114)

The extra term depends on

(115)

the rate at which electric flux through a surface bounded by C changes with time. This term allows a changing electric field to act as a source of magnetic-field circulation even where no conduction current crosses the surface.

That correction is essential for the later development of electromagnetic waves and antenna radiation.

Common mistakes

• Mistake: assuming every closed path makes Ampère’s law easy to solve. The law gives circulation; symmetry determines whether B can be removed from the integral.
• Mistake: using total current when the loop encloses only part of a distributed current.
• Mistake: ignoring the orientation sign of current through the chosen surface.
• Mistake: assuming an off-center circular path around a straight wire has constant B.
• Mistake: treating ideal solenoid and toroid expressions as exact for every finite winding geometry.
• Mistake: applying the magnetostatic law without modification to time-varying antenna currents.

What EM12E reinforces

The key working sequence is

(116)

For magnetostatics,

(117)

and

(118)

are two forms of the same physical law, linked by Stokes’ theorem.

The next stage of the series can now move into time-varying fields, where Maxwell’s correction becomes essential.

References