Electromagnetic Waves, Antennas, and RF: Magnetic Fields Produced by Currents - Exercises and Complete Worked Solutions

This companion article provides self-study exercises for EM11, Magnetic fields Produced by Currents. All exercises are stated first. Complete worked solutions follow in Part II.

The central magnetostatic source law is the Biot–Savart law:

(1)

where

(2)

For an infinitely long straight wire,

(3)

and for a circular loop of radius a,

(4)

(5)

For a volume current density,

(6)

These are the same definitions and conventions developed in EM11 [1, 2, 3, 5].

How to use this problem set

Attempt all exercises in Part I before consulting Part II. Keep source and observation quantities separate. In current-source problems, a common error is to mix up the source coordinate r′ and the observation point r. In direction problems, determine the cross-product direction first and substitute numerical values second.

Part I: Exercises

Exercise 1: direction from a single current element

A small source element has direction

(7)

and the observation point lies directly in the +y direction from the element, so

(8)

Determine the direction of dB.

Exercise 2: magnitude from one current element

A current element has

(9)

The observation point is at distance

(10)

and the angle between dℓ′ and R is 30^∘. Find the magnitude of dB.

Exercise 3: field of a long straight wire

An infinitely long straight wire carries current

(11)

Find the magnetic-field magnitude at perpendicular distance

(12)

Also state the field-line geometry.

Exercise 4: inverse-distance scaling for a straight wire

At a distance s from a long straight wire, the magnetic field has magnitude B₁. What is the field magnitude at distance 2s? What is the ratio B(2s)∕B(s)?

Exercise 5: finite straight-wire field

A finite straight wire carries current

(13)

At the observation point, the perpendicular distance to the wire is

(14)

and the end angles are

(15)

Using

(16)

find the magnetic-field magnitude.

Exercise 6: infer current from a measured field

The magnetic field near a long straight wire is measured to be

(17)

at perpendicular distance

(18)

Find the current in the wire.

Exercise 7: field at the center of a circular loop

A single circular loop of radius

(19)

carries current

(20)

Find the magnetic-field magnitude at the center of the loop.

Exercise 8: field at the center of an N-turn loop

A 25-turn circular coil has radius

(21)

and carries current

(22)

Find the field magnitude at the center.

Exercise 9: field on the axis of a circular loop

A single circular loop has radius

(23)

and current

(24)

Find the magnetic field at the point on the axis located at

(25)

from the center.

Exercise 10: superposition from two long parallel wires

Two infinitely long parallel wires are separated by distance

(26)

Each carries current of magnitude

(27)

Find the magnetic-field magnitude at the midpoint between the wires for the following two cases:

1. both currents are in the same direction;
2. the currents are in opposite directions.

Exercise 11: using current density to find the field outside a wire

A cylindrical wire of radius

(28)

carries a uniform current density

(29)

Find:

1. the total current I in the wire;
2. the magnetic-field magnitude at a point outside the wire at distance

   (30)

   from the axis.

Exercise 12: magnetostatic limitation of Biot–Savart

Explain why the magnetostatic Biot–Savart law developed in EM11 is not, by itself, the complete field law for a time-varying antenna current. Your answer should mention the steady-current assumption and the role of time-dependent electromagnetic propagation.

Part II: Complete Worked Solutions

Solution 1: direction from a single current element

From the Biot–Savart law,

(31)

Here,

(32)

Therefore,

(33)

So the field contribution points in the +z direction:

(34)

Solution 2: magnitude from one current element

Use

(35)

Substitute the given values:

dB	= 10−7	(36)
	= 10−7	(37)
	= 10−7	(38)
	= 3.125 × 10−7 T.	(39)

Thus,

(40)

Solution 3: field of a long straight wire

For an infinitely long straight wire,

(41)

Substitute the values:

B	=	(42)
	=	(43)
	= 4.0 × 10−5 T.	(44)

Therefore,

(45)

The field lines are circles centered on the wire, with direction determined by the right-hand rule.

Solution 4: inverse-distance scaling for a straight wire

Because

(46)

we see directly that B ∝ 1∕s.

At distance 2s,

(47)

Hence,

(48)

and the ratio is

(49)

Solution 5: finite straight-wire field

Use the finite-wire expression:

(50)

Substituting,

B	= 10−7(sin 40∘ + sin 55∘)	(51)
	= (0.6428 + 0.8192)	(52)
	= 3.333 × 10−5(1.4620)	(53)
	= 4.873 × 10−5 T.	(54)

Thus,

(55)

Solution 6: infer current from a measured field

From the long-wire formula,

(56)

Solve for I:

(57)

Now substitute:

I	=	(58)
	=	(59)
	= 4.0 A.	(60)

Therefore,

(61)

Solution 7: field at the center of a circular loop

For a single loop,

(62)

Substitute the values:

B	=	(63)
	=	(64)
	= 3.77 × 10−5 T.	(65)

Thus,

(66)

Solution 8: field at the center of an N-turn loop

An N-turn coil multiplies the single-turn center field by N:

(67)

Substitute the values:

B	=	(68)
	=	(69)
	= 7.85 × 10−5 T.	(70)

Therefore,

(71)

Solution 9: field on the axis of a circular loop

Use the on-axis formula:

(72)

With

(73)

we have

(74)

Then

Bz	=	(75)
	=	(76)
	=	(77)
	= 4.44 × 10−6 T.	(78)

So,

(79)

The direction is along the loop axis, determined by the right-hand rule.

Solution 10: superposition from two long parallel wires

The midpoint lies at distance

(80)

from each wire.

Each wire individually contributes

(81)

For case (a), same current direction, the two field directions at the midpoint are opposite. They cancel:

(82)

For case (b), opposite current directions, the two field directions at the midpoint are the same. They add:

(83)

Hence,

(84)

Solution 11: using current density to find the field outside a wire

The total current is current density times cross-sectional area:

(85)

Substitute the values:

I	= (5.0 × 105)π(2.0 × 10−3)2	(86)
	= (5.0 × 105)π(4.0 × 10−6)	(87)
	= 2π A	(88)
	≈ 6.28 A.	(89)

Therefore,

(90)

Now use the outside-wire straight-current formula at s = 1.0 × 10⁻² m:

B	=	(91)
	=	(92)
	= 1.2566 × 10−4 T.	(93)

Thus,

(94)

Solution 12: magnetostatic limitation of Biot–Savart

The Biot–Savart law developed in EM11 assumes a steady current distribution. That means the source does not change with time, so the magnetic field is treated as a magnetostatic field.

For a time-varying antenna current, however, the fields do not adjust instantaneously everywhere in space. Electromagnetic influences propagate at finite speed, and changing electric and magnetic fields become coupled through the full Maxwell equations. In that regime, one must use the time-dependent field laws with retarded dependence on the source rather than the simple steady-current Biot–Savart expression.

So the essential point is:

(95)

Closing summary

These exercises reinforce four core ideas from EM11:

1. current elements produce magnetic-field contributions through the cross product dℓ′× R;
2. straight-wire fields scale as 1∕s;
3. loop fields are obtained by systematic superposition and symmetry;
4. the magnetostatic Biot–Savart law is foundational but limited to steady-current situations.

These skills prepare directly for the next topic: circulation laws and the stronger symmetry-based machinery of Ampère’s law.

References