Pearson Edexcel Physics

Unit 4: Further Mechanics, Fields and Particles

4.4 Electric and Magnetic Fields

Pearson Edexcel Physics Unit 4: Further Mechanics, Fields and Particles 4.4 Electric and Magnetic Fields Candidates will be assessed on their ability to::
92.	Understand that an electric field (force field) is defined as a region where a charged particle experiences a force
93.	Understand that electric field strength is defined as [math]E = \frac{F}{Q}[/math] and be able to use this equation
94.	Be able to use the equation [math]F = \frac{Q_1 Q_2}{4\pi \varepsilon_0 r^2}[/math] for the force between two charges
95.	Be able to use the equation [math]E = \frac{Q}{4\pi \varepsilon_0 r^2}[/math] for the electric field due to a point charge
96.	Know and understand the relation between electric field and electric potential
97.	Be able to use the equation [math]E = \frac{V}{d}[/math] for an electric field between parallel plates
98.	Be able to use [math]V = \frac{Q}{4\pi \varepsilon_0 r^2}[/math] for a radial field
99.	Be able to draw and interpret diagrams using field lines and equipotential to describe radial and uniform electric fields
100.	Understand that capacitance is defined as [math]C = \frac{Q}{V}/math] and be able to use this equation
101.	Be able to use the equation [math]W = \frac{1}{2} QV[/math] for the energy stored by a capacitor, be able to derive the equation from the area under a graph of potential difference against charge stored and be able to derive and use the equations [math]W = \frac{1}{2} C V^2 \quad \text{and} \quad W = \frac{1}{2} \frac{Q^2}{C}[/math]
102.	Be able to draw and interpret charge and discharge curves for resistor capacitor circuits and understand the significance of the time constant RC
103.	CORE PRACTICAL 11: Use an oscilloscope or data logger to display and analyse the potential difference (p.d.) across a capacitor as it charges and discharges through a resistor
104.	Be able to use the equation [math]Q = Q_0 e^{-\frac{t}{RC}}[/math] and derive and use related equations for exponential discharge in a resistor-capacitor circuit, [math]I = I_0 e^{-\frac{t}{RC}}[/math] , and [math]V = V_0 e^{-\frac{t}{RC}}[/math] and the corresponding log equations [math]\ln Q = \ln Q_0 – \frac{t}{RC}, \quad \ln I = \ln I_0 – \frac{t}{RC} \quad \text{and} \quad \ln V = \ln V_0 – \frac{t}{RC}[/math]
105.	Understand and use the terms magnetic flux density B, flux ∅ and flux linkage N∅
106.	Be able to use the equation F = Bqv sinθ and apply Fleming’s left-hand rule to charged particles moving in a magnetic field
107.	Be able to use the equation F = BIl sinθ and apply Fleming’s left-hand rule to current carrying conductors in a magnetic field
108.	Understand the factors affecting the e.m.f. induced in a coil when there is relative motion between the coil and a permanent magnet
109.	Understand the factors affecting the e.m.f. induced in a coil when there is a change of current in another coil linked with this coil
110.	Understand how to use Faraday’s law to determine the magnitude of an induced e.m.f. and be able to use the equation that combines Faraday’s and Lenz’s laws [math]\varepsilon = -\frac{d(N\Phi)}{dt}[/math]

92. Understand that an electric field (force field) is defined as a region where a charged particle experiences a force
Fast-moving charged particles are used in many devices. For instance, the X-rays in a hospital X-ray equipment are created by high-speed electrons smashing against a metal target.
An electric field is defined as a space that accelerates charged particles.
Figure 1 An electric field accelerates electrons in an X-ray machine.
A charged particle in an electric field will experience a force.
Electric field lines to illustrate the forces produced by the field.
These indicate which way the force generated by the field will push a positively charged particle.
A stronger field is shown by lines that are closer together. This holds true for magnetic fields as well as for other field patterns. Stronger forces result from a stronger field.
The force, F, that acts on a charged particle is the electric field strength (E) multiplied by the amount of charge in coulombs (Q), as given by the equation:
[math]F = EQ[/math]
From this force equation, we can also see how quickly a charge would accelerate. Newton’s second law is that F= ma, so we can equate the two equations:
[math]F = EQ = ma[/math]
So,
[math]a = \frac{EQ}{m}[/math]
93. Understand that electric field strength is defined as [math]E = \frac{F}{Q}[/math] and be able to use this equation
⇒ Electric Field Strength:
The electric field strength at a point in space is defined as:
[math]E = \frac{F}{Q}[/math]
Where:
– E = electric field strength (in newtons per coulomb, N/C)
– F = electric force acting on a test charge (in newtons, N)
– Q = magnitude of the test charge (in coulombs, C)
The electric field strength is the force per unit charge experienced by a small positive test charge placed at a point in the field.
– It’s a vector quantity, meaning it has both magnitude and direction:
– The direction of the electric field is the same as the force experienced by a positive charge.
– It’s opposite for a negative charge.

Use this equation when you know the force acting on a charge and the magnitude of the charge.
The field should not be significantly altered by the test charge (i.e. it should be a small test charge).
Works in uniform fields (e.g., between parallel plates) and non-uniform fields (e.g., around point charges).

Example:
A charge of [math]Q = 2 × 10^{-6} C[/math] experiences a force of [math]F = 0.01 N[/math] What is the electric field strength at that point?
[math]\begin{gather}
E = \frac{F}{Q} \\
E = \frac{0.01}{2 \times 10^{-6}} \\
E = 5000 \ \text{N/C}
\end{gather}[/math]
So, the electric field strength is 5000 N/C, in the direction of the force.
94. Be able to use the equation [math]F = \frac{Q_1 Q_2}{4\pi \varepsilon_0 r^2}[/math] for the force between two charges
95. Be able to use the equation [math]F = \frac{Q}{4\pi \varepsilon_0 r^2}[/math] for the electric field due to a point charge
⇒ Combining Electric Fields:
When many charged objects produce electric fields, the total field at any given location is the vector sum of all each field’s contributions.
The total field is the result of the combined force effects of all component fields.
To calculate the resulting force at each location to determine how a charged particle will be influenced there. The total electric field is the sum of these separate force effects.
Areas surrounding spikes or other points on charged objects have a disproportionately high concentration of charge. This indicates that the field will be strong around them and that their field lines are near together.
Because the concentrated charge will more strongly attract opposing charges, lightning is more likely to strike the lightning conductor than the building it is protecting. This is why lightning conductors are spiked.
Actually, the field surrounding a lightning conductor that has been spiked is frequently so intense that it can induce charge to seep through the conductor.
This occurs prior to a charge building up sufficiently to result in a lightning strike. As a result, there is less chance of lightning striking, further shielding the structure from these threats.
Figure 2 Complex fields also always have their equipotential perpendicular to the field lines at all points.
⇒ Charged particle interactions:
One way to conceptualize the attraction between a proton and an electron is as the proton’s positive charge producing an electric field and the electron feeling the force of the proton’s field.
(Note that the opposite might also be considered, where the electron’s electric field exerts a force on the proton.)
The force between two charged particles, [math]Q_1[/math] and [math]Q_2[/math], which are separated by a distance r, is described by Coulomb’s law and is given by the expression:
[math]F = \frac{Q_1 Q_2}{4\pi \varepsilon_0 r^2} = k \frac{Q_1 Q_2}{r^2}[/math]
Figure 3 A proton’s field will cause a force on a negatively charged electron.
By the electric field:
[math]F = \frac{Q}{4\pi \varepsilon_0 r^2}[/math]
Electric field E created by a point charge Q at a distance r from the charge.
Where:
– E: electric field (in volt per meter, V/m or newtons per coulomb N/m)
– Q: Charge of the point source (in coulombs, C)
– r: Distance from the charge to the point where the field is being calculated (in meters)
– [math]\varepsilon_0[/math] Permittivity of free space, a constant:
[math]\varepsilon_0 = 8.85 \times 10^{-12} \ \text{F/m} \quad \text{(farads per meter)}[/math]
– [math]4 \pi \varepsilon_0 [/math]: A constant that appears often in electrostatics, approximately equal to
[math]4 \pi \varepsilon_0 \approx 1.11 \times 10^{-10} \ \text{C}^2/\text{Nm}^2[/math]
So, for one charge [math]Q_1[/math] creating a radial field in which another charge [math]Q_1[/math] sits at a distance r from [math]Q_1[/math]:
[math]\begin{gather}
F = EQ \\
F = \frac{Q_1 Q_2}{4 \pi \varepsilon_0 r^2}
\end{gather}[/math]
Note that for charged spheres, the calculation of the force between them is also given by Coulomb’s law, but the distance is measured between the centres of the spheres.
96. Know and understand the relation between electric field and electric potential
⇒ Electric Field (E)
A measure of force per unit charge
Vector quantity (has direction)
Units: (N/C or V/m)
Represents how strongly a test charge would be pushed or pulled.
⇒ Electric Potential (V):
A measure of potential energy per unit charge
Scaler quantity (has magnitude only)
Unit: volts (V) or J/C
Represents how much work is done to move a charge
Relationship between Electric Field and Electric Potential:
[math]E = -\frac{dV}{dr}[/math]
– The electric field is the negative gradient (rate of charge) of the electric potential.
– The electric field points in the direction where the electric potential decreases the fastest,
For constant electric fields (like between parallel plates)
[math]E = -\frac{V}{d}[/math]
– E is the electric field strength
– V is the potential difference
– D is the separation between the plates
97. Be able to use the equation [math]E = -\frac{V}{d}[/math] for an electric field between parallel plates
98. Be able to use [math]V = \frac{Q}{4 \pi \varepsilon_0 r}[/math] for a radial field
99. Be able to draw and interpret diagrams using field lines and equipotential to describe radial and uniform electric fields
⇒ Uniform Electric Field Strength between two parallel plates:
Figure 4 The blue arrows show the direction of the uniform electric field produced between parallel plates that have a potential difference between them. The dashed lines show where the potential is constant along equipotential lines.
Any objects at various electrical potentials are subject to an electric field. It is therefore possible to create an electric field between parallel metal plates by connecting them to a power source.
This is seen in Figure 2. If the field lines are parallel throughout the whole field, the field is uniform.
The strength of a uniform electric field is a measure of how rapidly it changes the potential over distance.
The equation which describes this divides the potential difference, V, by the distance over which the potential difference exists, d:
[math]E = \frac{V}{d}[/math]
Example:
In the X-ray machine, there is a potential difference of 45 000 V between a cathode and an anode. These electrode plates are 10 cm apart. What is the electrical field strength between the plates?
Solution:
[math]\begin{gather}
E = \frac{V}{d} \\
E = \frac{4500}{0.1} \\
E = 4.5 \times 10^{5} \ \text{V/m}
\end{gather}[/math]
⇒ Radial Fields:
As seen in figure 5, the electric field will act outward in all directions away from the core of a positively charged sphere or point charge like a proton.
As you go away from the sphere in Figure 5, you’ll notice that the arrows become farther apart. This suggests that the farther you go from the centre, the weaker the field becomes.
As you get farther away, the space between equipotential likewise grows. Because the field is what causes the potential to change, a weaker field causes the potential to change more slowly.
A charged particle inside a charged sphere would experience no resultant force and so there is no electric field.
The overall effect of all the charges on the sphere cancel out within the sphere itself. Distance measurements for use in calculations should always be considered from the centre of a charged sphere.
Figure 5 The radial electric field around a proton (or any positively charged sphere).
⇒ Radial Electric Field strength:
The field lines for a radial field become further apart. This means that the equation for the field strength of a radial electric field must incorporate this weakening of the field with distance from the charge.
The expression for radial field strength at a distance r from a charge Q is:
[math]E = \frac{Q}{4 \pi \varepsilon_0 r}[/math]
This is only truly correct for a field that is produced in a vacuum. This is because the value – the permittivity of free space – is a constant which relates to the ability of the fabric of the Universe to support electric fields. Its value is [math]ε_0 [/math] = 8.85 x 10^-12 F/m.
Other substances, for example, water, may be better or worse at supporting electric fields, so an extra factor (the relative permittivity, e,) would appear in the equation to account for this.
Air is considered to be near enough a vacuum that we use the equation given above for electric fields in air.
Example:
What is the electric field strength at a distance of 1 angstrom (1 x 10^-10 m) from a proton?
[math]\begin{gather}
E = \frac{Q}{4 \pi \varepsilon_0 r^2} \\
E = \frac{1.6 \times 10^{-19}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times (1 \times 10^{-10})^2} \\
E = 1.44 \times 10^{11} \ \text{V/m}
\end{gather}[/math]
⇒ Potential in a radial Electric field:
Electric field strength tells us how quickly the electric potential is changing.
A stronger field will have the equipotential closer together. This equation states that the electric field strength, E, is equal to the rate of change of potential, V, with distance, r.
[math]E = -\frac{dV}{dr}[/math]
This leads to the expression for radial field potential at a distance r from a charge Q:
[math]V = \frac{Q}{4 \pi \varepsilon_0 r}[/math]
Example:
What is the electric field strength at a distance of 1 angstrom (1 x 10^-10 m) from a proton?
[math]\begin{gather}
V = \frac{Q}{4 \pi \varepsilon_0 r} \\
V = \frac{1.6 \times 10^{-19}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times (1 \times 10^{-10})} \\
V = 14.4 \ \text{V}
\end{gather}[/math]
100) Understand that capacitance is defined as [math]C = \frac{Q}{V}[/math] and be able to use this equation
⇒ Storing Charge:
An electric field is set up within the conducting material and this causes electrons to experience a force and thus move through the wires and components of the circuit.
Where there is a gap in a circuit, the effect of the electric field can be experienced by charges across this empty space, but in general, conduction electrons are unable to escape their conductor and move across the empty space. This is why a complete conducting path is needed for a simple circuit to function.
Charge can be made to flow in an incomplete circuit. This can be demonstrated by connecting two large metal plates in a circuit with an air gap between them, as in figure 6.
Figure 6 An electric field will act across a space. You could test this by hanging a charged sphere near the plates and observing the field’s force acting on the sphere.
The circuit shown in figure 7 is similar to the situation shown by the photo in figure 6. When the power supply is connected, the electric field created in the conducting wires causes electrons to flow from the negative terminal towards the positive terminal.
Since the electrons cannot cross the gap between the plates, they collect on the plate connected to the negative terminal, which becomes negatively charged.
Electrons in the plate connected to the positive terminal flow towards the positive of the cell, which results in positive charge being left on that plate.
The attraction between the opposite charges across the gap creates an electric field between the plates, which increases until the potential difference across the gap is equal to the potential difference of the power supply.
Figure 7 A simple capacitor circuit.
A pair of plates such as this with an insulator between them is called a capacitor.
Charge will collect on a capacitor until the potential difference across the plates equals that provided by the power supply to which it is connected. It is then fully charged, and the capacitor is acting as a store of charge.
The amount of charge a capacitor can store, per volt applied across it, is called its capacitance, C, and is measured in farads (F). The capacitance depends on the size of the plates, their separation and the nature of the insulator between them.
Capacitance can be calculated by the equation:
[math]\begin{gather}
\text{capacitance (F)} = \frac{\text{Charge stored (C)}}{\text{Potential difference across capacitor (V)}} \\
C = \frac{Q}{V}
\end{gather}[/math]
Example:
What is the capacitance of a capacitor which can store 18 mC of charge when the p.d. across it is 6 V?
[math]\begin{gather}
C = \frac{Q}{V} \\
C = \frac{18 \times 10^{-3}}{6} \\
C = 3 \times 10^{-3} \ \text{F} \\
C = 3 \ \text{mF}
\end{gather}[/math]
How much charge will be stored on this capacitor if the voltage is increased to 20 V?
[math]\begin{gather}
Q = CV \\
Q = (3 \times 10^{-3})(20) \\
Q = 60 \times 10^{-3} \ \text{C} \\
Q = 0.06 \ \text{C}
\end{gather}[/math]
101. Be able to use the equation [math]W = \frac{1}{2} QV[/math] for the energy stored by a capacitor, be able to derive the equation from the area under a graph of potential difference against charge stored and be able to derive and use the equations
[math]W = \frac{1}{2} C V^2 \quad \text{and} \quad W = \frac{1}{2} \frac{Q^2}{C}[/math]
⇒ Energy stored on a charged capacitor:
A charged capacitor is a store of electrical potential energy. When the capacitor is discharged, this energy can be transferred into other forms.
Our definition of voltage gives the energy involved as [math]E = QV[/math]. However, the energy stored in a charged capacitor is given by
[math]E = \frac{1}{2} Q V[/math]
This is a trick question, because our original equation assumes that the charge and voltage are constant. However, in order to charge a capacitor, it begins with zero charge stored on it and slowly fills up as the p.d. increases, until the charge at voltage V is given by Q. This can be seen on the graph in figure 8.
Each time we add a little extra charge (∆Q) this has to be done by increasing the voltage and pushing the charge on, which takes some energy (we are doing work). By finding the area of each nearly rectangular strip, we find V∆Q which is the amount of extra energy needed for that extra charge.
Therefore, the sum of all the strips, or the area under the line, will give us the total energy stored. This is the area of a triangle, so its area is [math] = \frac{1}{2} \times \text{base} \times \text{height}[/math], which from the graph is [math]\frac{1}{2} QV[/math].
[math]E = \frac{1}{2} QV[/math]
Because Q = CV, you can also find two other versions of this equation for the stored energy.
Figure 8 Graph of potential difference against charge for a capacitor
[math]\begin{gather}
E = \frac{1}{2} QV \\
E = \frac{1}{2} (CV)V \\
E = \frac{1}{2} CV^2 \\
E = \frac{1}{2} QV \\
E = \frac{1}{2} Q \cdot \frac{Q}{C} \\
E = \frac{1}{2} \frac{Q^2}{C}
\end{gather}[/math]
Figure 9 Investigating how the current through a capacitor change over time.
102. Be able to draw and interpret charge and discharge curves for resistor capacitor circuits and understand the significance of the time constant RC
⇒ Capacitor discharge curve:
The capacitor will be at 6 V if it is completely charged in Figure 9, and we can infer from [math]Q = CV[/math] that it will be holding 0.6 mC of charge.
The capacitor’s electrons can migrate to its positive side by discharging through the bulb if the two-way switch in Figure 8 is shifted to position B.
The lamp will impede their journey because of its 100[math]\Omega[/math] resistance, but they will still discharge, and as long as there is current flowing through it, the lamp will remain lit.
The current begins at its maximum when the capacitor discharges, causing the rush of electrons to be as great as possible. Using Ohm’s law, we can determine that this current is 0.06 A.
The p.d. across the capacitor decreases and the electric field, and therefore the push on the remaining electrons, is smaller after some electrons have discharged. The light will be weaker and the current is lower.
After a while, the electron flow is so little that the current [math]\frac{V}{R}[/math] is reduced to a trickle, and the lamp will appear so faint that it may even be off.
When the capacitor is completely depleted, there won’t be any more electrons flowing between its two sides; the current will be zero.
The discharging current, p.d. across the capacitor, and charge left on the capacitor will all exhibit the patterns seen on the three graphs in Figure 10 if we piece this tale together over time.
Figure 10 Discharge curves for a capacitor through a light bulb.
⇒ The time constant:
There are two possibilities:
– Store more charge on the capacitor
– Decrease the rate at which the capacitor discharges
For the same maximum p.d., increasing the capacitance, C, will increase the charge stored, as Q = CV. Alternatively, the charge would flow more slowly if the bulb’s resistance, R, was greater.
An overall impression of the rate of discharge of a capacitor can be gained by working out the time constant, [math]\tau[/math].
This is calculated from [math]\tau = RC[/math], and with resistance in ohms and capacitance in farads, the answer is in seconds.
In fact, the time constant tells you how many seconds it takes for the current to fall to 37% of its starting value.
⇒ Capacitor Charging Curves:
By considering the charging process in the same way as we did the discharge of the capacitor in fig 8, we can quickly work out that the charging process produces graphs such as those in fig 11
Figure 11 Charging curves for a capacitor connected to a 6V supply.
When charging a capacitor through a resistor, the time constant RC has exactly the same effect. A greater resistance or a larger capacitance, or both, means the circuit will take longer to charge up the capacitor.
CORE PRACTICAL 11: Use an oscilloscope or data logger to display and analyse the potential difference (p.d.) across a capacitor as it charges and discharges through a resistor
⇒ Investigating current flow through a capacitor:
Components:
– Parallel attached capacitor to a 6V battery and 100[math][/math]
– A current Sensor (which is computer datalogging current)
– Two terminals which is point A is connected with current sensor and point A is attached with 100[math]\Omega[/math] resistor
Circuit:
Figure 12 examining the variations in a capacitor’s current over time.
The capacitor will be at 6 V if it is completely charged in Figure 8, and we can infer from [math]Q = CV[/math] that it will be holding 0.6 mC of charge.
The capacitor’s electrons can migrate to its positive side by discharging through the bulb if the two-way switch in Figure 8 is shifted to position B.
The lamp will impede their journey because of its 100[math]\Omega[/math] resistance, but they will still discharge, and as long as there is current flowing through it, the lamp will remain lit.
The current begins at its maximum when the capacitor discharges, causing the rush of electrons to be as great as possible. Using Ohm’s law, we can determine that this current is 0.06 A.
The p.d. across the capacitor decreases and the electric field, and therefore the push on the remaining electrons, is smaller after some electrons have discharged. The light will be weaker and the current [math]\frac{V}{R}[/math] is lower.
Be able to use the equation Q [math]Q = Q_0 e^{-\frac{t}{RC}}[/math] and derive and use related equations for exponential discharge in a resistor-capacitor circuit,[math]I = I_0 e^{-\frac{t}{RC}}, \quad \text{and} \quad V = V_0 e^{-\frac{t}{RC}}[/math] and the corresponding log equations[math]\ln Q = \ln Q_0 – \frac{t}{RC}, \quad \ln I = \ln I_0 – \frac{t}{RC}, \quad \text{and} \quad \ln V = \ln V_0 – \frac{t}{RC}[/math]
⇒ Discharging Capacitors:
Charge, Q:
The charging and discharging of a capacitor follow curving graphs in which the current is constantly changing. It follows that the rate of change of charge and p.d. are also constantly changing.
These graphs are known as exponential curves. The shapes can be produced by plotting mathematical formulae which have power functions in them.
In the case of discharging a capacitor, C, through a resistor, R, the function that describes the charge remaining on the capacitor, Q, at a time, t, is:
[math]Q = Q_0 e^{-\frac{t}{RC}}[/math]
Or corresponding log equation is
[math]\ln Q = \ln Q_0 – \frac{t}{RC}[/math]
Where [math]Q_0[/math] is the initial charge on the capacitor at t = 0, and e is the special mathematical number which is used in the inverse function of natural logarithms ([math]e \approx 2.718[/math]).
Example:
A 0.03 F capacitor is fully charged by a 12 V supply and is then connected to discharge through a 900 resistor. How much charge remains on the capacitor after 20 seconds?
[math]\begin{gather}
\text{Initial charge:} \quad Q_0 = CV \\
Q_0 = (0.03)(12) \\
Q_0 = 0.36 \ \text{C} \\
Q = Q_0 e^{-\frac{t}{RC}} \\
Q = (0.36) e^{-\frac{20}{900 \times 0.03}} \\
Q = (0.36)(0.477) \\
Q = 0.17 \ \text{C}
\end{gather}[/math]
⇒ Voltage, V:
The p.d. across a discharging capacitor will fall as the charge stored falls. By substituting the equation [math]Q = CV[/math] into our exponential decay equation, we can show the formula that describes voltage on a discharging capacitor is in exactly the same form as for the charge itself:
[math]Q = Q_0 e^{-\frac{t}{RC}} \quad \text{and} \quad Q = CV[/math]
Which also means that initially,[math]Q_0 = CV_0[/math]
[math]CV = CV_0 e^{-\frac{t}{RC}}[/math]
from which the capacitance term, C, can be cancelled, leaving:
[math]V = V_0 e^{-\frac{t}{RC}}[/math]
Or corresponding log equation is
[math]\ln V = \ln V_0 – \frac{t}{RC}[/math]
Example:
A 0.03 F capacitor is fully charged by a 12 V supply and is then connected to discharge through a 900 resistor. What is the p.d. on the capacitor after 20 seconds?
Initial voltage is the same as the supply at 12 V.
[math]\begin{gather}
V = V_0 e^{-\frac{t}{RC}} \\
V = (12)(0.477) \\
V = 5.7 \ \text{V}
\end{gather}[/math]
⇒ Current, I:
the discharging current also dies away following an exponential curve. Ohm’s law tells us that
[math]\begin{gather}
V = IR \\
V_0 = I_0 R \\
IR = I_0 R e^{-\frac{t}{RC}}
\end{gather}[/math]
From which the resistance term, R, will cancel on both sides:
[math]I = I_0 e^{-\frac{t}{RC}}[/math]
Example:
A 0.03 F capacitor is fully charged by a 12 V supply and is then connected to discharge through a 900Q resistor. What is the discharge current after 20 seconds?
[math]\begin{gather}
I_0 = \frac{V_0}{R} \\
I_0 = \frac{12}{900} \\
I_0 = 0.013 \ \text{A} \\
I = I_0 e^{-\frac{t}{RC}} \\
I = (0.013)(0.477) \\
I = 6.2 \ \text{mA}
\end{gather}[/math]
105. Understand and use the terms magnetic flux density B, flux [math] \Phi[/math] and flux linkage [math]N \Phi[/math]
⇒ Magnetic Flux density:
Where the flux is perpendicular to the area, a rearrangement of the equation shows why the quantity B is known as the magnetic flux density.
[math]B = \frac{\Phi}{A}[/math]
By sharing the flux over the area, B indicates how close together the magnetic field lines are – how dense they are.
How dense field lines are indicates how strong the field is, so B is also known as the magnetic field strength.
⇒ Flux Linkage:
The interaction between magnetic fields and charged particles, or conductors, allows motors to operate, and electricity to be generated. In most practical applications, magnetic flux is made to interact with a coil of wire, as the effect on a single strand of wire is too small to be useful.
If the single wire is coiled up, then the magnetic field can interact with each turn on the coil, and so any effect is multiplied N times, where N is the number of turns on the coil.
The amount of magnetic flux interacting with a coil of wire is known as the flux linkage, and is the product of the number of turns of wire and the flux in that region:
[math]\begin{align*}
\text{Flux linkage} &= N \Phi \quad \text{(measured in weber-turns)} \\
\text{Remembering that } \Phi &= BA, \text{ we also have:} \\
\text{Flux linkage} &= BAN
\end{align*}[/math]
Example:
Zoran takes a wire and winds it into a coil of ten circular turns with a radius of 5 cm and then holds this in a magnetic field with a strength of 20 mT. What is the flux linkage?
[math]\begin{gather}
\text{Flux linkage} = BAN \\
\text{Area } A = \pi r^2 \\
A = (3.14)(0.05)^2 \\
A = 7.85 \times 10^{-3} \ \text{m}^2 \\
\text{Flux linkage} = (0.02)(7.85 \times 10^{-3})(10) \\
\text{Flux linkage} = 1.6 \times 10^{-3} \ \text{Wb}
\end{gather}[/math]
106. Be able to use the equation [math]F = B q v \sin \theta[/math] and apply Fleming’s left-hand rule to charged particles moving in a magnetic field
⇒ Forces on particles:
The motor effect happens because a charged particle moving at right angles to a magnetic field experiences a force on it at right angles to its direction of motion and also at right angles to the field.
If the charged particle is constrained – like an electron in a current in a wire – then the force will transfer to the wire itself.
If the particle is flying freely, its direction will change and it will travel a circular path whilst in the magnetic field (figure 13).
Figure 13 Charged particles moving in a magnetic field follow a circular path as the motor effect provides a centripetal force.
The direction of the force acting on a charged particle moving in a magnetic field can be found using Fleming’s left-hand rule.
We must be careful with the current direction though: a negative charge moving in one direction is a current flowing in the opposite direction.
The strength of the force on a charged particle moving across a magnetic field is given by the equation:
[math]F = B \times q \times v \sin \theta[/math]
where q is the charge on the particle, v is its velocity, and is the angle between the velocity and the magnetic field lines. A simplified situation that is often considered is that of an electron (charge ‘e’) moving at right angles to the field (sin 90° = 1). This reduces the formula to
[math]F = Bev[/math]
As the force on the charged particle is always at right angles to the direction of its velocity, it acts as a centripetal force, and the particle follows a circular path.
As the charged particle’s velocity is constantly changing along a circular path, it is being accelerated by the magnetic field.
This means that given the right combination of conditions, a moving charged particle could be held by a magnetic field, continuously orbiting a central point.
⇒ Flaming’s Left Hand Rule:
Magnetic fields can affect moving electric charges, as well as magnetic poles. If you place a wire in a magnetic field and pass a current through it, the wire will experience a force on it (figure 14). This is called the motor effect.
The effect is greatest when the wire and the magnetic field are at right angles. In this instance, the force will be at right angles to both, in the third dimension, as shown by Fleming’s left-hand rule in figure 15.
Figure 14 The jumping wire experiment illustrates the motor effect in action.
Figure 15 Fleming’s left-hand rule gives the relative directions of the field, current and movement in the motor effect.
107. Be able to use the equation [math]F = B I l \sin \theta[/math] and apply Fleming’s left-hand rule to current carrying conductors in a magnetic field
⇒ Forces on wires:
The strength of the force, F, on a wire which has a current, I, through it whilst it is in a magnetic field, B, is given by the equation:
[math]F = B \times I \times L \times \sin \theta[/math]
where L is the length of the wire within the field, and is the angle the current makes with the lines of the magnetic field. For simplicity, we will only consider uniform magnetic fields in which the field lines are all parallel.
It is common to set up situations in which the angle is 90° so that sin is a maximum and equals 1. This reduces the formula to
[math]F = B I l \sin \theta[/math]
Figure 16 Investigating the strength of the force on a current-carrying conductor in a magnetic field.
A consequence of the expression [math]F = B I l \sin \theta[/math] is that a motor can be made more powerful, or faster, by:
– increasing the current through the motor (I)
– increasing the number of turns of wire in the motor (L)
– increasing the magnetic field within the motor (B).
The magnetic field strength is usually maximized by making the coil’s core out of soft iron. Some motors use electromagnets to provide the field, and these could be strengthened by increasing the current through them.
108. Understand the factors affecting the e.m.f. induced in a coil when there is relative motion between the coil and a permanent magnet
109. Understand the factors affecting the e.m.f. induced in a coil when there is a change of current in another coil linked with this coil
110. Understand how to use Faraday’s law to determine the magnitude of an induced e.m.f. and be able to use the equation that combines Faraday’s and Lenz’s laws
[math]\varepsilon = -\frac{d(N \Phi)}{dt}[/math]
⇒ Electromagnetic Induction:
The movement of a charged particle in a magnetic field causes it to experience a force.
Newton’s third law of motion reminds us that this force must have a counterpart that acts equally in the opposite direction. This pair of electromagnetic forces is generated whenever there is relative motion between a charge and a magnetic field.
So, a magnetic field moving past a stationary charge will create the same force. The velocity term in the expression [math]F = B q v [/math] actually refers to the relative (perpendicular) velocity between the magnetic field lines and q. (Also remember that if the movement is not at right angles, then we need to work out the component of it that is at right angles by including the sin term: [math]F = B q v \sin \theta[/math])
This means that if we move a magnet near a wire, the electrons in the wire will experience a force to make them move through the wire. This is an e.m.f.; if the wire is in a complete circuit, then the electrons will move, forming an electric current.
Figure 17 Using an electromagnet to induce e.m.f. in a coil.
We can use this principle to generate electricity. Reversing the direction of the magnetic field, or the direction of the relative motion, will reverse the direction of the force on the electrons, reversing the polarity of the e.m.f.. Faraday’s law states that the induced e.m.f. is proportional to the rate of change of flux linkage.
⇒ Electromagnetic Induction Using an Electromagnet:
The induction of e.m.f. was a result of the relative motion between a conductor, or coil, and a permanent magnet.
The magnetic field which interacts with the coil could be produced electrically by another coil. This is the principle of operation of a transformer, and the coil producing the initial magnetic field is referred to as the primary coil.
In figure 17 we have a pair of coils linked together by a soft iron core. Iron is extremely good at carrying magnetism and, in the set-up shown, nearly all the magnetic field generated by the primary coil on the left would interact with the secondary coil on the right.
When the primary switch is first closed, the primary suddenly produces a magnetic field that was not previously there. This means that a changing magnetic field is now within the secondary coil.
This sudden change of flux linkage will generate an e.m.f. in the secondary.
However, once the electromagnetic field is stable there will no longer be any change in flux linkage over time, and so there will be no further induced e.m.f. in the secondary.
The voltmeter needle will kick and then return to zero. If the primary circuit is switched off, the magnetic field it produces will suddenly disappear, and a brief e.m.f. will be induced in the opposite direction to the switch-on voltage. The voltmeter needle will kick in the opposite direction and then return to zero again.
⇒ Calculating Induced E.M.F:
Putting Faraday’s and Lenz’s laws together gives us an expression for calculating an induced e.m.f.:
[math]\varepsilon = -\frac{d(N \Phi)}{dt} \quad \text{or} \quad \varepsilon = -\frac{\Delta (N \Phi)}{\Delta t}[/math]
Faraday’s law told us that the e.m.f. would be proportional to the rate of change of flux linkage, and the minus sign in the equation comes from Lenz’s law, to indicate the opposing direction