Summary:
Provides a detailed overview of electricity. Describes how to find the force caused by a magnetic field on a single charge carrier moving in a wire.
Electricity
Forces on Moving Charges in a Magnetic Field
A current is the result of the motion of charged particles, and a wire carrying a current can experience a force in a magnetic field.
If the force were exerted directly on the charge carriers, even a beam of charge from a charge gun would experience a force due to a magnetic field.
To find the force caused by a magnetic field on a single charge carrier moving in a wire, we must divide the total force on a length L of wire by the number of charge carriers in this length.
If the wire has a cross sectional area A, and there are nu charge carriers per unit volume, the number of charge carriers in length L is nuAL.
Force per charge = force on wire= BfÎ IL= BfÎInumber of carriers nuAL nuA
The number of charge carriers that pass P in time f´t is nuAvf´t. Each carries a charge q, and so¡K
I = charge past P in time f´t =q nuAvf´t=q nuAv
f´t f´t
Using I in the expression for the force per unit charge¡K
F=BfÎI=qvBfÎ
nuA
A charge q moving with speed v perpendicular to a magnetic field of magnitude BfÎ experiences a force of magnitude¡K
F=qv BfÎ
The direction of the force on the moving charge is given by the right hand palm rule.
Notice that the force is perpendicular to the velocity vector of the charge and therefore has no component in the direction of the motion. As a result, it does no work on the charge.
A steady magnetic field does no work on a charge moving through it.
Despite this fact, the force deflects the particle. However, it does not change its kinetic energy, since no work is done.
Particle Motion in a Magnetic Field
Suppose a positively charged particle is moving with velocity v in a magnetic field B. The field is directed into the page, and so v and B are mutually perpendicular.
We know that the particle, which we shall say has charge q, experiences a force whose magnitude is given by F = qv BfÎ.
The right hand palm rule tells you that the force is in the direction as shown in Fig 18.3. The force is perpendicular to v and so v does not change. Viz. A ball on the end of a string.
A force perpendicular to v causes the object to follow a circular path. The force qvB due to the magnetic field furnishes the centripetal force needed for circular motion.
Equating qvB to the centripetal force mv2/r gives
qvB = mv2r = mv
r qB
A particle moving perpendicular to a uniform magnetic field follows a circular path with this radius r.
The Hall Effect
There are very few phenomena that show clearly the sign of the charge carriers. A battery is connected to the ends of a thin uniform conducting ribbon made perhaps from metal. The two symmetric points m and n are at the same potential (i.e no voltage difference between them). When a magnetic field is impressed perpendicular to the ribbon, points m and n differ in potential. Suppose the charges through the ribbon are positive. The right hand palm rule tells us that the charge is forced upward, toward m. Hence point m becomes positively charged and a potential difference appears between m and n (i.e m is positive when the charge carriers are positive). The opposite would apply if the charge was negative (i.e. m would become negatively charged). We have a clear way of determining the sign of the charge carriers in a material.
This is called the Hall Effect.
The Light-Emitting Diode (LED) as an indicator
An LED is a device which glows when it carries a current in one particular direction only. In the diagram below the LED is connected in forward bias. When the LED is connected in the reverse way in a circuit it does not carry a current and the LED is said to be reverse biased. An LED is forward biased when the longer of its two leads (anode) is connected to the higher potential (+) side in a circuit. The shorter lead is called the cathode. Since too great a current will destroy an LED, it can be protected in a circuit by a suitable series resistance which limits the current.
The Potentiometer (POT)
The potentiometer is best thought of as a variable resistor, having a circular rather than linear geometry.
Only two terminals are needed for it to be used as a variable resistor. Another use for the POT is as a voltage divider, in which case all three terminals are used.
Exercise:
Connect the battery terminal potential difference across the outside terminals of the POT. Use a voltmeter to determine the potential difference between the centre terminal and each outside terminal for various positions of the spindle.
Capacitors
This one form of a device that is of considerable practical importance for the storage of electric charge and energy, as it involves the use of two oppositely charged metal plates (viz the Hall Effect). The positive terminal of the battery places a positive charge on one plate, and the negative terminal places an equal-magnitude negative charge on the other plate. These charges attract one another, and so they reside on the inner surfaces of the plates. Such a device is capable of storing charge.
The charge on the positive plate is +q, and the charge on the negative plate is ¡Vq. Because the electric field E between the plates is proportional to q and because the potential difference V between the plates is proportional to E
(remember that V = Ed), we find q is proportional to V. therefore¡K
q = (constant)(V)or q= constant
V
This constant is simply the charge that the capacitor holds for each volt of potential difference between the plates. It is called the capacitance (C) of the capacitor:
Capacitance = C = qcoulombs per volt or farad (F)
V
In a parallel-plate capacitor where each plate has a surface area A and the plates are separated by only a tiny distance d, then capacitance is¡K
C = fÕo Aparallel plates
d
where fÕo (read epsilon sub zero) is a quantity called the permittivity of free space (8.85 x 10 ¡V12 F m-1).
The Capacitor in dc Circuits
Suppose we move from a to b where plate a is the positively charged plate (i.e. higher potential). Since the potential difference V is given by q/C, the potential change in going from a to b is ¡Vq/C. If we are going from b to a the charge would be positive.
Charging and Discharging a Capacitor
Charge the capacitor by connecting points A and B. The green light should glow as charges pass through the 390 fÇ resistor, green LED, and onto the capacitor. The green light will go out when the capacitor is fully charged.
Remove the wires connecting points A and B.
The capacitor is fully charged. We can demonstrate that the capacitor is storing charge by discharging it through the red LED. To do this, connect points A and C. The red light should glow as the stored charge moves through the 680 fÇ resistor and the red LED.
When the switch is closed, the battery tries to send current clockwise around the circuit. Since there is initially no charge on the capacitor, the current i is limited only by the resistor. Therefore, just after the switch is closed (at time t=o) the current is io = Vo/R as shown in (b). As time goes on, however, the capacitor becomes charged and the current decreases. The current must drop to zero when the capacitor¡¦s charge becomes q = CVo.
The curve followed by the current is called an exponential decay curve.
The current drops to a value of 0.3679io in a time equal to the product of RC. We call the product RC the time constant of this circuit. It is the time in seconds required for the current to decrease to about 0.37 times its original value. When the current finally stops, the capacitor¡¦s charge is qo = CVo. The time constant measures the time in seconds it takes the capacitor to become about two-thirds charged. The time constant is a rough measure of the time needed to charge a capacitor. The capacitor discharge current behaves in the same way as the charging current.
AC Quantities; rms Values
Consider an alternating current I that is delivering power to a resistor R. It can be shown mathematically that this current has the same heating effect as a constant dc current of magnitude io /,,©2 (= 0.707 io), where io is the peak, or maximum, current.
This equivalent dc current is called the root-mean-square (or effective) current and is represented by the symbol I.
The root-mean-square (rms) current is therefore given by¡K
I = io /,,©2
Similarly, we define the root-mean-square (rms) voltage V as¡K
V = vo /,,©2
Where vo represents the peak voltage.
This is the complete article, containing 1,454 words
(approx. 5 pages at 300 words per page).
