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In this Physics tutorial, you will learn:

- What does Gauss Law for electric field say?
- The same for magnetic field?
- How can we induce a magnetic field inside the plates of a charged capacitor?
- What does Ampere-Maxwell Law say?
- What are the special cases of Ampere-Maxwell Law?
- How does the magnetic field inside the capacitor change with the distance from the centre of plates?
- What is the displacement current? How is it related to the real current used to charge the capacitor?
- How to Find the Induced Magnetic Field?
- What are the Maxwell Equations? What do we find using them?
- Why Maxwell Equations are so important?

Φ_m = ∮▒〖B*⃗* ∙ dA*⃗* 〗 = 0

This outcome is different from the Gauss Law in electric fields. Recall that in electric fields, we have Φ_e = ∮▒〖E*⃗* ∙ dA*⃗* 〗 = Q/ε_{0}

Unlike in electric fields in which the net electric flux through a closed Gaussian surface is proportional to the charge, in magnetic field the net magnetic flux through a Gaussian surface is zero. This is because the net magnetic charge in all magnets is zero because the charge is balanced; in magnets, we simply have a certain regular alignment of dipoles but the net charge is zero. The figure below gives a clearer idea on this point. As you see from the figure, there are two lines entering the loop and two lines leaving the loop. Therefore, the net flux flowing through the Amperian loop is zero. Example: In which of the points shown in the figure a) the magnetic field is the greatest? b) the magnetic flux is the greatest? Solution The magnetic field is strongest near the poles as the magnetic field lines have a higher density in those regions. In addition, the magnetic field decreases with the increase in distance from the magnet. Therefore, the order of magnetic fields in the four given points from the smallest to the largest is B_{d}**c****b****a**

Hence, the magnetic field is the strongest at the point A. As for the magnetic flux, it is zero everywhere as the number of magnetic field lines entering from up at any closed Amperian loop around the given points is equal to the number of lines leaving the loop, regardless the numbers vary according the corresponding magnetic field strengths. For example, we may have 10 lines entering a closed loop around the point A and 10 lines leaving the loop, and for example 7 lines entering the same closed loop around the point B and 7 lines leaving the loop. In both cases the total flux is zero, regardless the magnetic field is different. Induced Magnetic Fields In the tutorial 16.7, we explained that a changing magnetic flux induces an electric field in a loop, which on the other hand generates an emf (and current) in the loop. This represents the Faraday's Law and its integral form is ∮▒E*⃗* dL*⃗* = -(dΦ_{b})/dt

The reverse is also true. A changing electric flux can induced a magnetic field. The equation representing this property is known as Maxwell Law of Induction and the corresponding equation is ∮▒B*⃗* dL*⃗* = μ_{0} ∙ ε_{0} ∙ (dΦ_E)/dt

Here B∮▒B*⃗* dL*⃗* = μ_{0} ∙ ε_{0} ∙ (dΦ_E)/dt+μ_{0} ∙ i_encl

As special cases, we can consider the two following scenarios: When there is a current but no change in electric flux (such as a wire carrying a constant current), the first term on the right side of the Ampere-Maxwell Law is zero. As a result, the Ampere's Maxwell Law reduces to the Ampere's Law of Induction we have seen in tutorial 16.6. ∮▒B*⃗* dL*⃗* = μ_{0} ∙ i_encl

When there is a change in electric flux but no current (such as inside or outside the gap of a charging capacitor), the second term on the right side of the Ampere-Maxwell Law becomes zero, and so the Ampere's Maxwell Law reduces to the Maxwell's Law of induction ∮▒B*⃗* dL*⃗* = μ_{0} ∙ ε_{0} ∙ (dΦ_E)/dt

Example: A circular-plate capacitor of plate radius R is being charged as shown in the figure below. What is the magnetic field in terms of electric field for any r < R? Calculate the magnetic field magnitude for r = 1/3 R = 6 cm and dE/dt = 4 × 1012 V/(m·s) Calculate the maximum magnitude of magnetic field inside the plates of capacitor Derive an expression for the magnetic field outside the capacitor (r > R) and calculate the magnitude of magnetic field at r = 25 cm from the centre of plates Solution From the equation representing the Ampere-Maxwell Law, we see that a magnetic field can be set up either by a current or by induction due to a changing electric flux (or from both factors combined together). In the specific case, we don't have any current flowing between the capacitor plates but anyway, the electric flux is changing due to the increase of opposite charges in both plates. Hence, the Ampere-Maxwell Law equation ∮▒B*⃗* dL*⃗* = μ_{0} ∙ ε_{0} ∙ (dΦ_E)/dt+μ_{0} ∙ i_encl

reduces to the Maxwell's Law of induction ∮▒B*⃗* dL*⃗* = μ_{0} ∙ ε_{0} ∙ (dΦ_E)/dt

where B is the magnetic field, L is the length (perimeter) of the Amperian loop, ε0 and μ0 are the electric and magnetic field constants respectively, and dΦE/dt is the change in the electric flux in the unit of time. To derive an expression for the magnetic field in terms of electric field, we must consider separately the left and the right part of Maxwell's Law of induction. For the left side, we consider an Amperian loop of radius r < R as shown in the figure, as we want to calculate the magnetic field inside the capacitor, i.e. for r ≤ R. As seen in theory, the magnetic field along the loop is tangent to it, in the same direction to the path element dL. They can either be parallel antiparallel but for simplicity we will assume them as parallel (cos θ = 00), because this does not affect the result. Thus, we can write: ∮▒B*⃗* dL*⃗* = ∮▒B dL ∙ cos〖0^0 〗 = ∮▒B dL

Since the magnetic field is constant (the loop is symmetric), we obtain ∮▒B dL = B ∙ ∮▒dL

The expression inside the integral is simply the circumference of the circle formed by the Amperian loop considered. Thus, we have for the left side of the Maxwell Law of Induction applied in the specific case: B ∙ ∮▒dL = B ∙ (2π ∙ r)

As for the right side of the equation representing the Maxwell law of induction, we assume the electric field E is uniform between the capacitor plates and perpendicular to them. This means the electric flux is the product of two constants: E and A where A is the area of the Amperian loop considered above. Thus, we can write: Φ_E = E ∙ A

and the right side of the Maxwell Law of Induction becomes μ_{0} ∙ ε_{0} ∙ (dΦ_E)/dt = μ_{0} ∙ ε_{0} ∙ (d(E·A))/dt

Substituting the expressions obtained for each side of the Maxwell Law of Induction, we obtain B ∙ (2π ∙ r) = μ_{0} ∙ ε_{0} ∙ (d(E·A))/dt

Since the area of the Amperian loop considered here is constant and it is calculated by A = π ∙ r^{2}

we can write B ∙ (2π ∙ r) = μ_{0} ∙ ε_{0} ∙ (π ∙ r^{2} ) ∙ dE/dt

In this way, we can now derive an expression for magnetic field in terms of electric field (more precisely, in terms of the rate of change of the electric field): B = (μ_{0} ∙ ε_{0} ∙ r)/2 ∙ dE/dt

The above expression indicates that the magnetic field inside the capacitor increases linearly with the change in distance from the centre of the plate (where the magnetic field is zero because r = 0) to the maximum value obtained at the outer frame of the plates (where r = R). In this part of exercise we have the following clues: r = 6 cm = 6 × 10-2 m R = 3r = 18 × 10-2 m = 1.8 × 10-1 m dE/dt = 4 × 1012 V/(m·s)

Also, we know that ε0 = 8.85 × 10-12 C2/(N·m2) μ0 = 4π × 10-7 N/A2

Thus, from the expression derived in (a), we obtain for the magnetic field B at the given distance r from the centre of capacitor plates: B = (μ_{0} ∙ ε_{0} ∙ r)/2 ∙ dE/dt = ((4 ∙ 3.14 × 10^{-7} N/A^{2} ) ∙ (8.85 × 10^{-12} C^{2}/(N ∙ m^{2} )) ∙ (6 × 10^{-2} m))/2 ∙ (4 × 10^{1}2 V/(m ∙ s)) = 1.334 × 10^{-6} T

The maximum magnetic field inside the plates is obtained for r = R. Thus, substituting the values in the equation used in (b), we obtain B = (μ_{0} ∙ ε_{0} ∙ R)/2 ∙ dE/dt

It is not necessary to do again the operations as it is obvious that since the radius increases by a factor of 3, the magnetic field increases by the same factor as well. Thus, we obtain for the maximum magnetic field between the capacitor plates is B_max = 3 ∙ B = 3 ∙ 1.334 × 10^{-6} T = 4.0 × 10^{-6} T

From previous tutorials, we known that electric field exists only between the plates of capacitor. This means that despite having a distance r > R from the centre of plates, the area to consider in the Maxwell equation is only A = π · R2. Thus, since the left part of the equation B ∙ (2π ∙ r) = μ_{0} ∙ ε_{0} ∙ (π ∙ r^{2} ) ∙ dE/dt

does not change, we obtain after substituting r = R in the right side: B ∙ (2π ∙ r) = μ_{0} ∙ ε_{0} ∙ (π ∙ R^{2} ) ∙ dE/dt B = (μ_{0} ∙ ε_{0} ∙ (π ∙ R^{2} ) ∙ dE/dt)/((2π ∙ r) ) = (μ_{0} ∙ ε_{0} ∙ R^{2} ∙ dE/dt)/2r

Substituting the known values (here, r = 25 cm = 2.5 × 10-1 m), we obtain for the magnetic field B: B = ((4 ∙ 3.14 × 10^{-7} N/A^{2} ) ∙ (8.85 × 10^{-12} C^{2}/(N ∙ m^{2} )) ∙ (1.8 × 10^(-1 ) m)^{2} ∙ (4 × 10^{1}2 V/(m ∙ s)))/((2 ∙ 2.5 × 10^(-1 ) m) ) = 2.88 × 10^{-6} T

As you see from the results, the magnetic field outside the capacitor decreases when the distance from the centre of plates increases. However, even between the plates the magnetic field is very small when compared to the magnitude of electric field. This is the reason why capacitors are considered as electric devices and not as magnetic ones. Displacement Current When analysing the equation derived from the Ampere-Maxwell Law ∮▒B*⃗* dL*⃗* = μ_{0} ∙ ε_{0} ∙ (dΦ_E)/dt+μ_{0} ∙ i_encl

it is obvious that the expression ε_{0} ∙ (dΦ_E)/dt

must have the dimension of current. This current is known as the "displacement current, id", despite no current is really being displaced. Thus, the above equation becomes ∮▒B*⃗* dL*⃗* = μ_{0} ∙ i_(d,encl)+μ_{0} ∙ i_encl

Let's consider again the charging capacitor of the previous section, shown in the figure below. Let's find a way to relate the real current I that is used to charge the capacitor plates to the fictitious displacement current id which is associated to the change in the electric field between the plates. From the Gauss Law for electric field Φ_e = ∮▒〖E*⃗* ∙ dA*⃗* 〗 = Q/ε_{0}

we obtain for the charge Q on the plates at any time: Q(t) = ε_{0} ∙ E(t) ∙ A

where E is the magnitude of electric field between the plates at that time. Differentiating the above equation with respect to time, we obtain for the real current I which charges the capacitor: dQ(t)/dt = I = ε_{0} ∙ A ∙ (dE(t))/dt

As for the displacement current id, we have i_{d} (t) = ε_{0} ∙ (dΦ_E)/dt = ε_{0} ∙ d(E ∙ A)/dt = ε_{0} ∙ A ∙ (dE(t))/dt

As you see, we obtained the same expression for the displacement current to the expression obtained for the real current. Thus, we can consider the fictitious displacement current id simply as a continuation of the real current I from one plate of capacitor to the other plate flowing through the gap between plates. Although no charge actually moves across the gap between the plates, the idea of the fictitious current id can help us to quickly find the direction and magnitude of an induced magnetic field as we will see in the next paragraph. How to Find the Induced Magnetic Field? In tutorial 16.2 "Magnetic Field Produced by Electric Currents" we explained that the direction of magnetic field produced by a current-carrying wire is found by using the right-hand rule, i.e. we grasp the wire with the right hand outstretching the thumb in the direction of current. In this case, the other four curled fingers show the direction of magnetic field. We can apply the same rule for the magnetic field produced by the displacement current as well. In this case, we consider the cylinder formed by the capacitor plates and the space between them as a cylindrical conducting wire of radius R. Hence, applying the known equations derived in the tutorial 16.2, we obtain for the magnetic field produced by the displacement current inside the capacitor plates, at a distance r from the central axis (r < R), B = (μ_{0} ∙ i_{d})/(2π ∙ R^{2} ) ∙ r

and for the magnetic field outside the plates (at r > R from the central axis), B = (μ_{0} ∙ i_{d})/(2π ∙ r)

Example: A circular parallel-plate capacitor with plate radius R = 4 cm is being charged by a current I = 5A. What is the magnitude of ∮▒〖B ∙ dL〗 at a distance of R/7 from the central point of capacitor plates? What is the magnitude of magnetic field at this distance? Solution First, let's calculate the magnitude of the displacement current in the given section between the capacitor plates. Since the magnitude of the displacement current id represents a part of the real current I which charges the capacitor, the ratio between the area encircled by the loop of radius r to the total area between the plates is equal to the ratio of the two abovementioned currents. Thus, we have i_{d}/I = (π ∙ r^{2})/(π ∙ R^{2} )

Hence, the displacement current at the given position is i_{d} = I ∙ r^{2}/R^{2} = I ∙ ((R/7)/R)^{2} = I/49 = ((5 A))/49 = 0.102 A

Therefore, since at the given distance ∮▒BdL = μ_{0} ∙ i_{d}

we obtain for the value of integral ∮▒BdL = (4π × 10^{-7} N/A^{2} ) ∙ (0.102 A) = 1.28 × 10^{-7} N/A

Since r = R/5 represents a location inside the plates, we have for the magnetic field B: B = (μ_{0} ∙ i_{d})/(2π ∙ R^{2} ) ∙ r = ((4π × 10^{-7} N/A^{2} ) ∙ (0.102 A))/(2π ∙ (4 × 10^{-2} m)^{2} ) ∙ ((4 × 10^{-2} m)/5) = 1.02 × 10^{-3} T

Maxwell Equations The four equations shown in the table below are known as Maxwell Equations. As explained at the beginning of this tutorial, these equation represent the key relationships between quantities in Electromagnetism. Thus, the four Maxwell Equations expressed in the integral form, are: Summary The Gauss Law for the magnetic field implies that magnetic monopoles do not exist. Mathematically, this law means that the net magnetic flux Φm through any closed Gaussian surface is zero. The Gauss Law formula for magnetic field is Φ_m = ∮▒〖B*⃗* ∙ dA*⃗* 〗 = 0

The Gauss law for electric fields on the other hand, is Φ_e = ∮▒〖E*⃗* ∙ dA*⃗* 〗 = Q/ε_{0}

A changing magnetic flux induces an electric field in a loop, which on the other hand generates an emf (and current) in the loop. This represents the Faraday's Law of Induction the integral form of which, is ∮▒E*⃗* dL*⃗* = -(dΦ_{b})/dt

The reverse is also true. A changing electric flux can induced a magnetic field. The equation representing this property is known as Maxwell Law of Induction and the corresponding equation is ∮▒B*⃗* dL*⃗* = μ_{0} ∙ ε_{0} ∙ (dΦ_E)/dt

Here B∮▒B*⃗* dL*⃗* = μ_{0} ∙ ε_{0} ∙ (dΦ_E)/dt+μ_{0} ∙ i_encl

As special cases of this law, we can consider the two following scenarios: When there is a current but no change in electric flux (such as a wire carrying a constant current), the first term on the right side of the Ampere-Maxwell Law is zero. As a result, the Ampere's Maxwell Law reduces to the Ampere's Law of Induction we have seen in tutorial 16.6. ∮▒B*⃗* dL*⃗* = μ_{0} ∙ i_encl

When there is a change in electric flux but no current (such as inside or outside the gap of a charging capacitor), the second term on the right side of the Ampere-Maxwell Law becomes zero, and so the Ampere's Maxwell Law reduces to the Maxwell's Law of induction ∮▒B*⃗* dL*⃗* = μ_{0} ∙ ε_{0} ∙ (dΦ_E)/dt

The magnetic field outside the plates of a charging capacitor decreases when the distance from the centre of plates increases. However, even between the plates the magnetic field is very small when compared to the magnitude of electric field. This is the reason why capacitors are considered as electric devices and not as magnetic ones. When analysing the equation derived from the Ampere-Maxwell Law ∮▒B*⃗* dL*⃗* = μ_{0} ∙ ε_{0} ∙ (dΦ_E)/dt+μ_{0} ∙ i_encl

it is obvious that the expression ε_{0} ∙ (dΦ_E)/dt

must have the dimension of current. This current is known as the "displacement current, id", despite no current is really being displaced. Thus, the above equation becomes ∮▒B*⃗* dL*⃗* = μ_{0} ∙ i_(d,encl)+μ_{0} ∙ i_encl

The real current I which charges the capacitor is dQ(t)/dt = I = ε_{0} ∙ A ∙ (dE(t))/dt

As for the displacement current id, we have i_{d} (t) = ε_{0} ∙ (dΦ_E)/dt = ε_{0} ∙ d(E ∙ A)/dt = ε_{0} ∙ A ∙ (dE(t))/dt

Thus, since the expressions for the true and fictitious currents are identical, we can consider the fictitious displacement current id simply as a continuation of the real current I from one plate of capacitor to the other plate flowing through the gap between plates. Although no charge actually moves across the gap between the plates, the idea of the fictitious current id can help us to quickly find the direction and magnitude of an induced magnetic field. The magnetic field produced by the displacement current inside the capacitor plates, at a distance r from the central axis (r < R), is B = (μ_{0} ∙ i_{d})/(2π ∙ R^{2} ) ∙ r

and the magnetic field outside the plates (at r > R from the central axis), is B = (μ_{0} ∙ i_{d})/(2π ∙ r)

The four Maxwell Equations are: Gauss' Law for electricity: ∮▒〖E*⃗* ∙ dA*⃗* 〗 = Q_encl/ε_{0}

It shows the relationship between net electric flux and net enclosed electric charge. Gauss' Law for magnetism: ∮▒〖B*⃗* ∙ dA*⃗* 〗 = 0

It shows the relationship between net magnetic flux and net enclosed magnetic charge. Faraday's Law of induction ∮▒E*⃗* dL*⃗* = -(dΦ_{b})/dt

It shows the relationship between induced electric field and changing magnetic flux. Ampere-Maxwell Law ∮▒B*⃗* dL*⃗* = μ_{0} ∙ ε_{0} ∙ (dΦ_E)/dt+μ_{0} ∙ i_encl

It shows the relationship between induced magnetic field and changing electric flux and to current. **1)** A circular-plate capacitor of plate radius R is being charged as shown in the figure below.

What is the magnitude of magnetic field at 4 cm from the centre of plates if the diameter of each plate is 10 cm and dE/dt = 5 × 1012 V/(m·s).

Take ε0 = 8.85 × 10-12 C2/(N·m2) and μ0 = 4π × 10-7 N/A2.

- 2.22 μT
- 1.11 μT
- 5.55 μT
- 1.33 μT

**Correct Answer: B**

**2)** A parallel-plate capacitor of circular plates is being charged at constant current. The capacitor plates are circular and the radius of each plate is 2 cm. What is the rate of electric field change between the plates if the maximum magnetic field produced during this process is Bmax = 8 μT?

Take ε0 = 8.85 × 10-12 C2/(N·m2) and μ0 = 4π × 10-7 N/A2.

- 7.2 × 1013 V/(m·s)
- 7.2 × 1012 V/(m·s)
- 8 × 1015 V/(m·s)
- 8 × 1012 V/(m·s)

**Correct Answer: A**

**3)** A parallel plate capacitor of circular plates the area of which is 30 mm2 each, is being charged by a constant current I = 2 A. What is the magnetic field induced at 5 mm from the centre of plates?

Take μ0 = 4π × 10-7 N/A2.

- 8 × 10-4 T
- 7.2 × 10-4 T
- 2 × 10-4 T
- 4 × 10-4 T

**Correct Answer: D**

We hope you found this Physics tutorial "Maxwell Equations" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In our next tutorial, we move into our next section "Electronics" and start by examining Electronic Essentials: Analogue and Digital Signals, Binary Operations and Logic Gates.

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