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[parent] Electromagnetic Waves: Electric Current and Current Density - Exercises and Complete Worked Solutions (Example)

Electromagnetic Waves, Antennas, and RF: Electric Current and Current Density - Exercises and Complete Worked Solutions

This companion article provides self-study exercises for EM08, Electric Current and Current Density. All exercises are stated first. Complete worked solutions follow in Part II.

The central relations are

|-------|
|    dq |
I =  ---|
-----dt--
(1)

and

|----∫---------|
|              |
|I =    J ⋅ dA.|
------S--------
(2)

For uniform current density normal to a flat cross section,

|--------|
-I =-JA.--
(3)

For moving continuous charge,

|--------|
-J-=-ρv,-|
(4)

and for mobile carriers of number density n, charge q, and drift velocity vd,

|----------|
-J-=-nqvd.-|
(5)

Charge conservation is expressed globally by

|--∫------------∮--------|
|d-   ρ dV =  −    J ⋅ dA|
-dt-V------------S--------
(6)

and locally by the continuity equation

|----------------|
|∂ρ              |
|∂t-+  ∇ ⋅ J = 0.|
-----------------
(7)

These are the same definitions and sign conventions developed in EM08 [1235].

How to use this problem set

Attempt every problem in Part I before consulting Part II. For current-density problems, identify the surface orientation before doing algebra. For continuity problems, decide first whether you are using a global control-volume statement or the local differential equation.

Part I: Exercises

Exercise 1: charge transported by steady current

A current of

I = 3.5A
(8)

flows for

Δt = 12 s.
(9)

How much charge crosses the selected surface?

Exercise 2: time-dependent charge gives time-dependent current

The net charge that has crossed a surface by time t is

q (t) = 2.0t2 + 0.50t3,
(10)

where q is in coulombs and t is in seconds.

Find:

  1. the instantaneous current I(t);
  2. the current at t = 2.0 s.

Exercise 3: uniform current density in a wire

A cylindrical Conductor carries

I = 4.0A
(11)

through a circular cross section of radius

R =  0.75 mm.
(12)

Assuming uniform current density perpendicular to the cross section, find J.

Exercise 4: current through a tilted surface

A uniform current density has magnitude

J = 8.0A/m2.
(13)

It crosses a flat surface of area

A  = 0.30m2.
(14)

The angle between J and the chosen surface normal is

      ∘
𝜃 = 60 .
(15)

Find the signed current through the surface.

PIC

Figure. The sign and magnitude of current through a surface follow from the projection J n.

Exercise 5: vector current density and surface orientation

Let

J = (3ˆx − 4yˆ+  2ˆz)A/m2.
(16)

A flat surface has area

A = 0.50 m2
(17)

and unit normal

      1
ˆn =  √--(2ˆx + ˆz).
       5
(18)

Assuming J is uniform, find the signed current through the surface.

Exercise 6: current density from moving volume charge

A continuous charge density

ρ = − 3.0 × 10 −6C/m3
(19)

moves with velocity

v = 5.0ˆx m/s.
(20)

Find J and explain the direction of the conventional current density relative to the charge motion.

Exercise 7: carrier form and electron drift direction

A metal contains mobile electrons with number density

n = 8.0 × 1028m − 3.
(21)

Their average drift velocity is

               −4
vd = − 2.0 × 10  ˆx m/s.
(22)

Using

                          −19
q = − e,    e = 1.602 × 10   C,
(23)

find J.

PIC

Figure. For negative mobile carriers, conventional current density points opposite the electron drift velocity.

Exercise 8: drift speed from current

A wire carries

I = 2.0A
(24)

with

             28   −3                  −6  2
n =  8.5 × 10  m   ,    A  = 1.2 × 10  m  .
(25)

Assuming one mobile electron charge of magnitude e = 1.602 × 1019 C per carrier, find the magnitude of the electron drift velocity.

Exercise 9: nonuniform current density through a circular cross section

A circular conductor of radius R carries the axial current-density distribution

J(s) = J  sˆz,
        0 R
(26)

where s is distance from the axis.

Find the total current through the cross section.

PIC

Figure. For nonuniform current density, the total current must be obtained by integrating J dA over the cross section.

Exercise 10: local continuity equation

A current-density field is

J = αx ˆx + 2αy ˆy − αz ˆz,
(27)

where α is constant.

Find:

  1. ∇⋅ J;
  2. ∂ρ∕∂t from the continuity equation;
  3. whether charge density locally increases or decreases when α > 0.

Exercise 11: a steady but nonuniform current density

Consider

J =  J0(− y ˆx + xˆy),
(28)

where J0 is constant.

Compute ∇⋅ J. Is this field compatible with steady charge density according to the continuity equation? Does zero divergence mean the current density is spatially uniform?

Exercise 12: global charge conservation in a control volume

A fixed volume initially contains

Q (0) = 10 mC.
(29)

A constant net outward current of

Iout = 2.0 mA
(30)

crosses its boundary for 3.0 s, with no other current crossing the surface.

Find the charge remaining inside the volume after 3.0 s.

PIC

Figure. For a fixed control volume, positive net outward current reduces the charge stored inside.

Exercise 13: identify and correct misconceptions

For each statement, decide whether it is correct. If it is incorrect, rewrite it accurately.

  1. “A current of 5 A means 5 C of charge are present in the wire.”
  2. “Current density J and current I have the same units.”
  3. “If electrons drift toward x, conventional current in a metal points toward +x.”
  4. “If ∇⋅ J = 0, then J must be constant everywhere.”
  5. “Positive net outward current from a fixed volume causes its enclosed charge to decrease.”

Exercise 14: synthesis from current distribution to charge conservation

A cylindrical conductor of radius R carries the time-dependent current-density distribution

                   (      2 )
J(s,t) = J  cos(ωt)  1 − s--  ˆz.
          m              R2
(31)

Find:

  1. the total current I(t) through a cross section normal to z;
  2. the peak current amplitude;
  3. the average current over one full cycle;
  4. the physical reason a time-varying current distribution such as this is relevant to the later antenna sections of the series.

Part II: Complete Worked Solutions

Solution 1: charge transported by steady current

For constant current,

Δq  = I Δt.
(32)

Therefore,

Δq = (3.5 A)(12 s) (33)
= 42 C. (34)

Thus,

|-----------|
Δq--=-42-C.--
(35)

The current is a rate; the accumulated charge grows with elapsed time.

Solution 2: time-dependent charge gives time-dependent current

Current is

I(t) =  dq.
       dt
(36)

Differentiate:

I(t) = d
--
dt(             )
 2.0t2 + 0.50t3 (37)
= 4.0t + 1.5t2. (38)

Hence,

|----------------2----|
I(t)-=-4.0t +-1.5t-A.--
(39)

At t = 2.0 s,

I(2) = 4.0(2) + 1.5(2)2 (40)
= 8 + 6 (41)
= 14 A. (42)

Therefore,

|----------------|
-I(2.0s)-=-14-A.-|
(43)

Solution 3: uniform current density in a wire

The cross-sectional area is

A = πR2 (44)
= π(0.75 × 103)2 (45)
1.767 × 106 m2. (46)

For uniform perpendicular current density,

J =  I-.
     A
(47)

Thus,

J = -----4.0-----
1.767 × 10 −6 (48)
2.26 × 106 A/m2. (49)

Therefore,

|----------------------|
|J ≈  2.26 × 106 A/m2.  |
-----------------------
(50)

Solution 4: current through a tilted surface

For uniform J over a flat surface,

I = J A cos𝜃.
(51)

Substituting,

I = (8.0)(0.30) cos 60 (52)
= 2.4(0.5) (53)
= 1.2 A. (54)

Hence,

|----------|
-I-=-1.2A.--
(55)

The result is positive because the angle with the chosen normal is less than 90.

Solution 5: vector current density and surface orientation

For uniform current density,

I = AJ  ⋅ ˆn.
(56)

The dot product is

J n = (3,4, 2) √1--
  5(2, 0, 1) (57)
= 6 + 2
-√---
   5 (58)
= -8--
√5-- A/m2. (59)

Therefore,

I = (0.50) 8
√---
  5 (60)
= √4--
  5 A (61)
1.79 A. (62)

Thus,

|-----------|
I-≈-1.79-A.--
(63)

Solution 6: current density from moving volume charge

Use

J = ρv.
(64)

Then

J = (3.0 × 106)(5.0x) (65)
= 1.5 × 105x A/m2. (66)

Therefore,

|------------------------|
|J = − 1.5 × 10 −5ˆxA/m2.  |
--------------------------
(67)

The negative charge moves toward +x, so conventional current density points toward x.

Solution 7: carrier form and electron drift direction

For electrons,

J = n (− e)vd.
(68)

Substitute the data:

J = (8.0 × 1028)(1.602 × 1019)(2.0 × 104x) (69)
2.56 × 106x A/m2. (70)

Hence,

|----------------------|
|J ≈ 2.56 × 106ˆx A/m2. |
------------------------
(71)

The electron drift is toward x, but conventional current density is toward +x.

Solution 8: drift speed from current

The drift-speed magnitude is

      I
vd = ----.
     neA
(72)

Substituting,

vd =                  2.0
--------28------------−19----------−6-
(8.5 × 10  )(1.602 × 10   )(1.2 × 10   ) (73)
1.22 × 104 m/s. (74)

Thus,

|----------------------|
|              −4      |
-vd-≈-1.22-×-10---m/s.-
(75)

This very small speed reinforces the distinction between carrier drift and rapid electromagnetic signal propagation.

Solution 9: nonuniform current density through a circular cross section

The surface is normal to z, so

J ⋅ dA = J -s dA.
          0R
(76)

In polar coordinates on the disk,

dA  = sds dϕ.
(77)

Therefore,

I = 02π 0RJ 0s
--
Rsdsdϕ (78)
= 2πJ0-
 R 0Rs2 ds (79)
= 2πJ
---0-
 R[s3]
 --
  30R (80)
= 2πJ0R2
--------
   3. (81)

Hence,

|------------|
|    2πJ0R2  |
I =  -------.|
--------3-----
(82)

Because J varies across the cross section, using I = JA with the edge value J0 would be incorrect.

Solution 10: local continuity equation

The current density is

J = αx ˆx + 2αy ˆy − αz ˆz.
(83)

Its divergence is

∇⋅ J = ∂(αx )
------
  ∂x + ∂(2αy )
-------
  ∂y + ∂(− αz )
--------
   ∂z (84)
= α + 2α α (85)
= 2α. (86)

Thus,

|------------|
-∇-⋅-J-=-2α.-|
(87)

The continuity equation gives

∂ ρ
--- = − ∇ ⋅ J,
∂t
(88)

so

|-----------|
∂-ρ = − 2α. |
-∂t----------
(89)

If α > 0, the divergence is positive, meaning net current leaves a small region. The local charge density therefore decreases.

Solution 11: a steady but nonuniform current density

For

J =  J0(− y ˆx + xˆy),
(90)

we have

Jx = − J0y,     Jy = J0x,     Jz = 0.
(91)

Therefore,

∇⋅ J = ∂(−-J0y)-
   ∂x + ∂(J0x-)
  ∂y + 0 (92)
= 0. (93)

Hence,

|----------|
|∇ ⋅ J = 0.|
-----------
(94)

The continuity equation then permits

∂ρ-
∂t =  0.
(95)

So the field is compatible with steady charge density. However, J is clearly not spatially uniform: both its magnitude and direction vary with position. Zero divergence does not mean constant vector field.

Solution 12: global charge conservation in a control volume

Global conservation gives

dQ
--- = − Iout.
dt
(96)

For constant outward current,

Q (t) = Q (0) − Ioutt.
(97)

Using milliamperes and seconds gives millicoulombs directly:

Q(3.0 s) = 10 mC (2.0 mA)(3.0 s) (98)
= 10 mC 6.0 mC (99)
= 4.0 mC. (100)

Therefore,

|------------------|
-Q(3.0s)-=-4.0-mC.--
(101)

Solution 13: identify and correct misconceptions

  1. Incorrect. A current of 5 A means charge crosses the chosen surface at a rate of 5 C/s; it does not specify how much charge is present in the wire.
  2. Incorrect. Current has units of amperes, while current density has units of amperes per square meter.
  3. Correct. Electron drift toward x corresponds to conventional current toward +x.
  4. Incorrect. Zero divergence means there is no local net source or sink of current density. The vector field may still vary strongly with position.
  5. Correct. The global continuity equation contains a minus sign: positive net outward current decreases enclosed charge.

Solution 14: synthesis from current distribution to charge conservation

The current density is

                   (        )
                         s2-
J(s,t) = Jm cos(ωt)  1 − R2   ˆz.
(102)

For a cross section normal to z,

dA  = sds dϕ.
(103)

Therefore,

I(t) = 02π 0RJ m cos(ωt)(     s2 )
  1 − -2-
      Rsdsdϕ (104)
= 2πJm cos(ωt)[         ]
 s2    s4
 2- − 4R2-0R (105)
= 2πJm cos(ωt)(         )
  R2-  R2-
  2  −  4 (106)
=      2
πJmR---
  2 cos(ωt). (107)

Thus,

|------------2---------|
|I(t) =  πJmR---cos(ωt).|
----------2-------------
(108)

The peak current amplitude is

|----------------|
|       πJmR2    |
|Imax = -------. |
-----------2-----
(109)

The average of cos(ωt) over one full cycle is zero, so

|--------|
-⟨I⟩ =-0.|
(110)

A zero cycle-average current does not mean nothing happens. Charges oscillate back and forth, producing time-varying current density. Later in the series, time-varying antenna currents will act as sources of time-varying electromagnetic fields and radiation.

Common mistakes

  • Treating current as stored charge. Current is a rate of charge transport.
  • Treating I and J as interchangeable. Current is a signed scalar through a selected surface; current density is a vector field.
  • Using I = JA when J varies over the surface. The general relation is I = SJdA.
  • Ignoring surface orientation. Reversing the chosen normal reverses the sign of current through the same surface.
  • Forgetting the sign of carrier charge in J = nqvd. Electron current density points opposite electron drift.
  • Assuming ∇⋅ J = 0 means J is constant. It means only zero local net outflow.
  • Dropping the minus sign in charge conservation. Positive outward current causes enclosed charge to decrease.

What EM08E reinforces

The fundamental distinction is

|------|
|I ⁄= J.|
--------
(111)

Current through a surface is obtained from the local current-density field by

|--------------|
|    ∫         |
|I =    J ⋅ dA.|
------S--------
(112)

Microscopically,

|----------|
|J = nqvd, |
-----------
(113)

and charge conservation requires

|----------------|
|∂ρ-+  ∇ ⋅ J = 0.|
-∂t--------------|
(114)

The next main lesson, EM09, introduces magnetic fields and the magnetic force on moving charge.

References

[1]   David J. Griffiths, Introduction to Electrodynamics, 4th ed., Cambridge University Press, 2017.

[2]   Samuel J. Ling, Jeff Sanny, and William Moebs, University Physics, Volume 2, OpenStax, 2016, sections on electric current, current density, and charge conservation.

[3]   Edward M. Purcell and David J. Morin, Electricity and Magnetism, 3rd ed., Cambridge University Press, 2013.

[4]   Richard P. Feynman, Robert B. Leighton, and Matthew Sands, The Feynman Lectures on Physics, Volume II, Addison-Wesley, 1964, chapters on current, charge conservation, and electromagnetic fields.

[5]   Massachusetts Institute of Technology, 8.02 Physics II: Electricity and Magnetism, MIT OpenCourseWare, materials on current, current density, magnetic fields, and Maxwell’s equations.

[6]   John D. Jackson, Classical Electrodynamics, 3rd ed., Wiley, 1999, sections on charge conservation and current density.


"Electromagnetic Waves: Electric Current and Current Density - Exercises and Complete Worked Solutions" is owned by bloftin.
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Classification:
Physics Classification41.20.-q (Applied classical electromagnetism)
 03.50.De (Classical electromagnetism, Maxwell equations )
 72.10.Bg (General formulation of transport theory)
 41.20.Jb (Electromagnetic wave propagation; radiowave propagation )
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