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Topic 5: Electricity and magnetism
charge comes in two forms + and – charge is a scalar quantity charge is quantized charge is conserved in a closed system charging neutral object ◊ Add electrons and make the object negatively charged. ◊ Remove electrons and make the object positively charged. charge is measured in Coulombs [C] Coulomb is defined as the charge transported by a current of one ampere in one second fundamental amount of charge is known as the electronic (or elementary) charge e =1.6 x 10-19 C
5.1 Charge and field
ratom ≈ 100000 x rnucleus mnucleon ≈ 2000 x melectron
Coulomb’s law: Electric force between TWO POINT charges q1 and q2 separated by distance r is:
𝐹𝐹 = 𝑘𝑘𝑞𝑞1𝑞𝑞2𝑟𝑟2
k = 8.99×109 N m2
C−2
k is Coulomb’s constant.
𝐹𝐹 =1
4𝜋𝜋𝜀𝜀0𝑞𝑞1𝑞𝑞2𝑟𝑟2
ε0 = 8.85×10-12 C2 N-1
m−2
ε0 is permittivity of free space
1/[4πε0] = 1 / [4π×8.85×10-12] = 8.99×109 = k.
Charges that are in a vacuum
Charges immersed in a different medium
𝐹𝐹 =14𝜋𝜋𝜀𝜀
𝑞𝑞1𝑞𝑞2𝑟𝑟2
𝜀𝜀 is permittivity of the medium
Material Permittivity ε /10-12C2 N-1m-2
vacuum(ε0 ) 8.85 air ~ 8.85 paper 34 rubber 62 water 779 graphite 106 diamond 71
q
Q
r
Coulomb’s law – extended distribution • Coulomb’s law works not only for point charges, which have no radii, but for any spherical distribution of charge at any radius.
• Be very clear that r is the distance between the centers of the charges.
Electric field strength (at a point P) is the force per unit positive point/test charge placed at that point. (it is a vector!)
Direction of electric field is the direction of the force on a positive test charge placed at that point.
𝐸𝐸 =𝐹𝑞𝑞
𝐸𝐸 = 𝑁𝑁𝐶𝐶−1
𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑟𝑟𝑒𝑒𝑒𝑒 𝑓𝑓𝑒𝑒𝑒𝑒𝑒𝑒𝑓𝑓 𝑟𝑟 𝑓𝑓𝑒𝑒𝑑𝑑𝑒𝑒𝑑𝑑𝑑𝑑𝑒𝑒𝑒𝑒 𝑓𝑓𝑟𝑟𝑓𝑓𝑓𝑓 𝑝𝑝𝑓𝑓𝑒𝑒𝑑𝑑𝑒𝑒 𝑒𝑒𝑐𝑑𝑑𝑟𝑟𝑐𝑐𝑒𝑒 𝑄𝑄 𝑒𝑒𝑑𝑑:
𝐸𝐸 = 𝑘𝑘𝑄𝑄𝑟𝑟2
=1
4𝜋𝜋𝜀𝜀0𝑄𝑄𝑟𝑟2
𝑟𝑟𝑑𝑑𝑓𝑓𝑒𝑒𝑑𝑑𝑒𝑒𝑒𝑒𝑟𝑟 𝑓𝑓𝑜𝑜𝑒𝑒𝑜𝑜𝑑𝑑𝑟𝑟𝑓𝑓 𝑒𝑒𝑓𝑓 𝑄𝑄 𝑒𝑒𝑑𝑑 +𝑟𝑟𝑑𝑑𝑓𝑓𝑒𝑒𝑑𝑑𝑒𝑒𝑒𝑒𝑟𝑟 𝑒𝑒𝑑𝑑𝑜𝑜𝑑𝑑𝑟𝑟𝑓𝑓 𝑒𝑒𝑓𝑓 𝑄𝑄 𝑒𝑒𝑑𝑑 −
Electric field due to charged conducting sphere whether hollow or solid:
R q
E = k q r2
r
at the surface q R2 E = k
E=0
Mapping fields – Electric field lines Field lines are imaginary
◊ The lines starts on + charges and end on – charges
◊ An arrow is essential to show the direction in which a positive charge would move
◊ Where the field is strong the lines are close together.
◊ The lines never cross.
◊ The lines meet a conducting surface at 90°.
monopol monopol dipol
uniform electric field
El. field at surface of a charged conductor is perpendicular to the conductor’s surface.
• When electric charge is placed in electric field it will experience electric force. Work W is done. • Electric potential difference V (∆V) between two points A and B is the amount of work W done per unit charge in moving a point charge from A to B.
𝑉𝑉 ≡ ∆𝑉𝑉 ≡ 𝑝𝑝.𝑓𝑓. = 𝑣𝑣𝑓𝑓𝑒𝑒𝑒𝑒𝑑𝑑𝑐𝑐𝑒𝑒
• units of V are JC-1 which are volts V.
Q q
A B
Potential difference
The potential difference between two points is one volt if one joule of energy is transferred per coulomb of charge passing between the two points.
𝑉𝑉 =𝑊𝑊𝑞𝑞
electronvolts are almost exclusively used in atomic and nuclear physics.
energy!!!
1 electron-Volt (eV) is the amount of work done when an elementary charge e is moved through a potential difference of 1volt.
1 eV = W = q ∆V = (1.6x10-19 C) (1V)
1 eV = 1.6x10-19 J
Capacitor: uniform electric field (the one that has constant magnitude and direction) is generated between two oppositely charged parallel plates. Edge effect is minimized when the length is long compared to their separation.
If a charge, q, is moved on its own from A to B, through a potential difference, ∆V, the work done on it by electric force is equal to the decrease in its electric potential energy which is converted into kinetic energy:
① Work done by electric force 𝑭𝑭
W = Fd = q Ed = q ∆V = ½ mv2
Positive charge accelerates from higher to lower potential (from positive to negative).
Negative charge accelerates from lower to higher potential. (from negative to positive)
② Work done by external force 𝑭𝑭𝒆𝒆𝒆𝒆𝒆𝒆 on charge against an electric field, from B to A is stored in the charge as the change in electrical potential energy, U. (𝐹𝐹𝑒𝑒𝑒𝑒𝑒𝑒 has to be equal to electric force in magnitude, opposite in direction)
W = Fext d = qEd W = q ∆V
qEd = q ∆V ⇒ 𝑬𝑬 = ∆𝑽𝑽𝒅𝒅
⇒ (E) = NC-1 = Vm-1
Drift speed Imagine a cylindrical conductor that is carrying an electric current I. The cross-sectional area of the conductor is A It contains charge carriers each with charge q. n is charge carriers density We assume that each carrier has a speed v
•Through any time interval ∆t, only the charges ∆Q between the two black cross-sections will provide the current I. •The volume containing the charge ∆Q is V = Av∆t. •Thus ∆Q = nVq = nAv∆tq. •Finally, I = ∆Q / ∆t = nAvq.
v A
v ∆t
∆Q
I = nAvq current vs. drift velocity
Current is the rate at which charge flows past a given cross-section
𝐼𝐼 = 𝑄𝑄𝑒𝑒
1C1A = 1s
Electrical resistance , R is the ratio of the potential difference across the resistor/conductor to the current that flows through it.
𝑅𝑅 = 𝑉𝑉𝐼𝐼
1Ω (ohm) = 1V/1A
The resistance of a conducting wire depends on four main factors:
• length, L • cross-sectional area, A • material/resistivity, ρ • temperature, T:
• if temperature is kept constant: 𝑅𝑅 = 𝜌𝜌 𝐿𝐿𝐴𝐴
In conclusion, we could say that a short fat cold wire makes the best conductor.
If you double the length of a wire, you will
double the resistance of the wire.
If you double the cross sectional area of a wire you will cut its resistance in half.
If you double the radius of a wire you will
cut its resistance in quarter.
The different types of resistors have different schematic symbols.
fixed-value resistor
variable resistor
potentiometer 2 leads
2 leads
3 leads
Electrical Resistance
thermister 2 leads
As temperature increases resistance decreases
light-dependent resistor (LDR)
2 leads
As brightness increases resistance decreases
OHM’S LAW: Current through resistor/conductor is proportional to potential difference on the resistor if the temperature/resistance of a resistor is constant.
or: 𝑉𝑉𝐼𝐼
= 𝑒𝑒𝑓𝑓𝑑𝑑𝑑𝑑𝑒𝑒 I – current through resistor, V – potential difference across R
Ohmic and Non-Ohmic conductors
How does the current varies with potential difference for some typical devices?
devices are non-ohmic if resistance changes
curr
ent
potential difference
curr
ent
potential difference
curr
ent
potential difference
metal at const. temp. filament lamp diode
Devices for which current through them is directly proportional to the potential difference across device are said to be ‘ohmic devices’ or ‘ohmic conductors’ or simply resistors. There are very few devices that are trully ohmic. However, many useful devices obey the law at least over a reasonable range.
𝐼𝐼𝑉𝑉
= 1𝑅𝑅
𝑒𝑒𝑑𝑑 𝑒𝑒𝑓𝑓𝑑𝑑𝑑𝑑𝑒𝑒. → 𝑅𝑅 𝑒𝑒𝑑𝑑 𝑒𝑒𝑓𝑓𝑑𝑑𝑑𝑑𝑒𝑒.
𝐼𝐼 = 𝑉𝑉𝑅𝑅
𝑒𝑒𝑓𝑓 𝑅𝑅 𝑒𝑒𝑑𝑑 𝑒𝑒𝑓𝑓𝑑𝑑𝑑𝑑𝑒𝑒.
When a current is flowing through a load such as a resistor, it dissipates energy in it. In collision with lattice ions electrons’ kinetic energy is transferred to the ions, and as a result the amplitude of vibrations of the ions increases and therefore the temperature of the device increases. KE is transferred to thermal energy.
Electric power is the rate at which energy is supplied to or used by a device.
Electric power is the rate at which electric energy is converted into another form such as mechanical energy, thermal energy, or light.
Power dissipated in a resistor/circuit: P = I V
W = qV → P = qV/t and I = q/t, so P = I V
Resistors in Series: connected in such a way that all components have the same current through them. Burning out of one of the lamp filaments in series or simply opening the switch causes a break. Equivalent resistance is greater that the greatest resistance in series.
𝑅𝑅𝑒𝑒𝑒𝑒= 𝑅𝑅1 + 𝑅𝑅2 + ⋯
Resistors in Parallel: are connected to the same two points of an electric circuit, so all resistors in parallel have the same potential difference across them. The current flowing into the point of splitting is equal to the sum of the currents flowing out at that point: I = I1 + I2 + … A break in any one path does not interrupt the flow of charge in the other paths. Each device operates independently of the other devices. The greater resistance, the smaller current. Equivalent resistance is smaller than the smallest resistance in series.
1 𝑅𝑅𝑒𝑒𝑒𝑒
=1𝑅𝑅1
+1𝑅𝑅2
+ ⋯
Internal resistance, r: some of the power/energy delivered by a cell is used/dissipated in driving the current though the cell itself
Terminal voltage (the actual voltage delivered to the circuit): V = ε – Ir
I = 𝜀𝜀𝑅𝑅𝑒𝑒𝑒𝑒
= 𝜀𝜀𝑅𝑅+𝑟𝑟
= 𝑉𝑉𝑅𝑅
To measure the current, we use an AMMETER.
• Ammeter is in SERIES with resistor R in order that whatever current passes through the resistor also passes through the ammeter. Ram << R, so it doesn’t change the current being measured. • In order to not alter the original properties of the circuit ammeters would ideally have no resistance, to minimize the effect on the current that is being measured (Req = R+ Rammeter ≈ 𝑅𝑅). No energy would be dissipated in ammeter.
To measure the potential difference across resistor, we use a VOLTMETER.
• Voltmeter is in PARALLEL with the resistor we are measuring. Rvoltmeter >> R so that it takes very little current from the device whose potential difference is being measured. • In order to not alter the original properties of the circuit an ideal voltmeter would have infinite resistance (1/ Rvoltmeter ≈ 0) with no current passing through it and no energy would be dissipated in it.
Potential divider circuits •Consider a battery of ε = 6 V. Suppose we have a light bulb that can only use three volts. How do we obtain 3 V from a 6 V battery? •A potential divider is a circuit made of two (or more) series resistors that allows us to tap off any voltage we want that is less than the battery voltage. •The input voltage is the emf of the battery.
• R = R1 + R2.
•The output voltage is the voltage drop across R2.
potential divider R1
R2
• I = VIN / R = VIN / (R1 + R2).
• VOUT = V2 = IR2 ⇒
𝑉𝑉𝑜𝑜𝑜𝑜𝑒𝑒 =𝑅𝑅2
𝑅𝑅1 + 𝑅𝑅2𝑉𝑉𝑖𝑖𝑖𝑖 𝑉𝑉1 =
𝑅𝑅1𝑅𝑅1 + 𝑅𝑅2
𝑉𝑉𝑖𝑖𝑖𝑖
Using a potential divider to give a variable pd
a power supply, an ammeter, a variable resistor and a resistor. When the variable resistor is set to its minimum value, 0 Ω, pd across the resistor is 2 V and a current of 0.2 A in the circuit.
When the variable resistor is set to its maximum value, 10 Ω, pd across the resistor is 1 V and a current of 0.1 A in the circuit.
Therefore the range of pd across the fixed resistor can only vary from 1 V to 2 V. The limited range is a significant limitation in the use of the variable resistor.
The same variable resistor can be used but the set up is different and involves the use of the three terminals on the variable resistor (sometimes called a rheostat.)
variable resistor circuit
potential divider
Terminals of rheostat resistor are connected to the terminals of the cell. The potential at any point along the resistance winding depends on the position of the slider (or wiper) that can be swept across the windings from one end to the other. Typical values for the potentials at various points on the windings are shown for the three blue slider positions.
The component that is under test (again, a resistor in this case) is connected in a secondary circuit between one terminal of the resistance winding and the slider on the rheostat. When the slider is positioned at one end, the full 2 V from the cell is available to the resistor under test. When at the other end, the pd between the ends of the resistor is 0 V (the two leads to the resistor are effectively connected directly to each other at the variable resistor). You should know how to set this arrangement up and also how to draw the circuit and explain its use.
• Potentiometer – rheostat – variable resistor
\ˈkir-ˌkof\
current into a junction has "+" sign current out of a junction has “ − " sign
𝐼𝐼1 + 𝐼𝐼2 + 𝐼𝐼3 = 𝐼𝐼4 + 𝐼𝐼5
Kirchhoff’s first law
OR: the sum of the currents into a junction equals the sum of the currents away from a junction
It is equivalent to a statement of conservation of charge.
−
for any junction ∑I = 0
Kirchhoff’s second law
Equivalent to conservation of energy.
in a complete circuit loop, the sum of the emfs in the loop is equal to the sum of the potential differences in the loop or the sum of all variations of potential in a closed loop equals zero
𝑓𝑓𝑓𝑓𝑟𝑟 𝑑𝑑𝑑𝑑𝑟𝑟 𝑒𝑒𝑒𝑒𝑓𝑓𝑑𝑑𝑒𝑒𝑓𝑓 𝑒𝑒𝑓𝑓𝑓𝑓𝑝𝑝 𝑒𝑒𝑑𝑑 𝑑𝑑 𝑒𝑒𝑒𝑒𝑟𝑟𝑒𝑒𝑜𝑜𝑒𝑒𝑒𝑒 ∑𝜀𝜀 = ∑𝐼𝐼𝑅𝑅
If a circuit has ONLY one cell you are lucky. You use Kirchhoff's laws in simple form (like juniors). Next four slides are the simplest possible way to do it -
If a circuit has MORE than ONE cell you are less lucky. But you are seniors now. You have to use Kirchhoff's laws in purest form. Later.
Find power of the source, current in each resistor, terminal potential, potential drop across each resistor and power dissipated in each resistor.
1. step: find total/equivalent resistance
2. step: find current in main circuit: 𝐼𝐼1 = 𝜀𝜀𝑅𝑅
Req = 120 Ω
I1 = ε ∕ Req = 0.3 A
3. step: to find currents in parallel branches use the Kirchhoff's laws: voltage is the same across all resistors in parallel: 𝐼𝐼2𝑅𝑅2 = 𝐼𝐼3𝑅𝑅3, and 𝐼𝐼1 = 𝐼𝐼2+𝐼𝐼3 one could use 𝜀𝜀 = 𝐼𝐼1r + 𝑉𝑉𝐴𝐴𝐴𝐴 𝑉𝑉𝐴𝐴𝐴𝐴 = 𝐼𝐼2𝑅𝑅2 = 𝐼𝐼3𝑅𝑅3
100𝐼𝐼2 = 50𝐼𝐼3 → 𝐼𝐼3 = 2𝐼𝐼2 0.3 = 𝐼𝐼2+𝐼𝐼3 → 0.3 = 3𝐼𝐼2 → 𝐼𝐼2 = 0.1 A 𝐼𝐼3 = 0.2 A
potential drops V = IR
power dissipated P = IV
80 Ω 0.3x80 = 24 V 0.3x24 = 7.2 W 100 Ω 0.1x100 = 10 V 0.1x10 = 1 W 50 Ω 0.2x50 = 10 V 0.2x10 = 2 W 6.7 Ω 0.3x6.7 = 2 V 0.3x2 = 0.6 W
ε = Σ all potential drops: (Kirchhoff's law) 36 V = 2 V + 24 V + 10 V power dissipated in the circuit = power of the source 0.6 + 2 + 1 + 7.2 = 0.3x36
R1 R2 R3
ε
Resistors in series Three resistors of 330 Ω each are connected to a 6.0 V battery in series
What is the voltage and current on each resistor?
In series the V’s are different if the R’s are different.
R = R1 + R2 + R3 R = 330 + 330 + 330 = 990 Ω
I = V / R = 6 / 990 = 0.0061 A
The current I = 0.0061 A is the same in each resistor.
voltage/potential difference across each resistor: V = I R1 = I R2 = I R3 = (0.0061)(330) = 2.0 V
Resistors in series and parallel
What is the voltage and current on each resistor?
In parallel the I’s are different if the R’s are different.
1/R = 1/R1 + 1/R2 + 1/R3 1/R = 1/330 + 1/330 + 1/330 = 0.00909
The voltage on each resistor is 6.0 V, since the resistors are in parallel. (Each resistor is clearly directly connected to the battery).
I1 = V1 / R1 = 0.018 A
I2 = V2 / R2 = 6 / 330 = 0.018 A
Three resistors of 330 Ω each are connected to a 6.0 V cell in parallel.
R = 110 Ω
I3 = V3 / R3 = 6 / 330 = 0.018 A
Kirchhoff’s rules – solving the circuit (less lucky)
EXCELLENT EXPLANATION of KIRCHOFF’s LAWS – thank you Kiran
Kirchhoff’s rules – solving the circuit
EXAMPLE: Suppose each of the resistors is R = 2.0 Ω, and the emfs are ε1 = 12 V and ε2 = 6.0 V. Find the voltages and the currents of the circuit. I1 – I2 + I3 = 0 (1)
–V1 + –V2 + –V4 + ε1 + – ε2 = 0, – ε2 – V3
– V4 = 0
–V1 + –V2 + –V4 + ε1 + – ε2 = 0 & V=IR ⇒ –2I1 + –2I1 + –2I2 + 12 + –6 = 0 (2) – ε2 – V3
– V4 = 0 & V=IR ⇒ –– 6 – 2I3 – 2I2 = 0 (3)
We now have three equations in I: (1) → I3 = I2 – I1. (2) → 3 = 2I1 + I2. (3) → 3 = -I2 + -I3
I1 = 1.8 A I2 = -0.6 A I3 = -2.4 A
Finally, we can redraw our currents:
resistor voltages: V = IR V1 = 1.8(2) = 3.6 V. V2 = 1.8(2) = 3.6 V. V3 = 2.4(2) = 4.8 V. V4 = 0.6(2) = 1.2 V
PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W” on its package. The potentiometer has a resistance from X to Z of 24 Ω and has linear variation.
R1
R2
(e) The potentiometer is adjusted so that the meter shows 4.0 V. What is the current in the Y-Z portion of the potentiometer?
SOLUTION: V2 = 4.0 V because it is in parallel with the lamp.
I2 = V2 / R2
= 4 / 13.71 = 0.29 A
(f) The potentiometer is adjusted so that the meter shows 4.0 V. What is the current in the ammeter?
SOLUTION: The battery supplies two currents.
The red current is 0.29 A because it is the I2 we just calculated in (e).
The green current is 0.20 A found in (c).
The ammeter has both so I = 0.29 + 0.20 = 0.49 A.
The cells are used until they are exhausted and then thrown away. The original chemicals have completely reacted and been used up, and they cannot be recharged. Examples include AA cells (properly called dry cells) and button mercury cells as used in clocks and other small low current devices.
Primary and secondary cells
primary cells – non rechargeable cells
When the chemical reactions have finished, the cells can be connected to a charger. Then the chemical reaction is reversed and the original chemicals form again. When as much of the re-conversion as is possible has been achieved, the cell is again available as a chemical energy store.
secondary cells – rechargeable cells
To reverse the chemical processes we need to return energy to the cell using electrons as the agents, so that the chemical action can be reversed. When charging, the electrons need to travel in the reverse direction to that of the discharge current and you can imagine that the charger has to force the electrons the “wrong” way through the cell.
of a cell is the quantity used to measure the ability of a cell to release charge: if a cell can supply a constant current of 2 A for 20 hours then it said to have a capacity of 40 amp-hours (40 A h). The implication is that this cell could supply 1 A for 40 hours, or 0.1 A for 400 hours, or 10 A for 4 hours. However, practical cells do not necessarily discharge in such a linear way and this cell may be able to provide a small discharge current of a few milliamps for much
capacity
Internal resistance, emf and terminal voltage of a cell The materials from which the cells are made have electrical resistance in just the same way as the metals in the external circuit. This internal resistance has an important effect on the total resistance and current in the circuit.
model for a real cell
Assumptions: internal resistance is constant (for a practical cell it varies with the state of discharge) emf is constant (which also varies with discharge current).
Kirchhoff’s second law the emf of the cell supplying energy to the circuit = the sum of the pds ε = IR + Ir If the pd across the external resistor is V, then ε = V + Ir or V = ε – Ir
𝐼𝐼
V, which is the pd across the external resistance, is equal to the terminal pd across real cell (in other words between A and B).
The emf is the open circuit pd across the terminals of a power source – in other words, the terminal pd when no current is supplied.
Thus the TERMINAL VOLTAGE (the actual voltage delivered to the circuit) is: V = ε - Ir In the true sense, electromotive force (emf) is the work (energy) per unit charge made available by an electrical source.
Electromotive force, (ε) is the power supplied to the circuit per unit current
P = I V ⟹ V = P/ I electric potential = powercurrent
Energy supplied by the source = ε Q (coming from W = Q ∆V)
Total energy supplied by the source = energy used in the resistors: ε Q = V1 Q + V2 Q + …..
divide it by time ⟹ ε I = V1 I + V2 I + …..
ex: the cell supplies 8.0 kJ of energy when 4 kC of charge moves completely across the circuit with
constant current. Find ε: ε = energy/charge = 2 V
Power supplied by the source will be dissipated in the circuit:
Power of the source = sum of the powers across the resistors: Pout = Σ Pi ⟹ ε I = V1 I + V2 I + …..
Electromotive force (emf) is the work (energy) per unit charge made available by an electrical source.
When voltage sources are connected in series in the same polarity, their emfs and internal resistances are add.
Two voltage sources with identical emfs connected in parallel have a net emf equivalent to one emf source, however, the net internal resistance is less, and therefore produces a higher current.
Emf and internal resistance of the battery (made up of two cells)
COULOMB’S/ NEWTON’S LAW
Electric (Coulomb’s law)/gravitational force (Newton’s law of gravitation) between two POINT charges/masses is proportional to the product of two charges/masses and inversely proportional to the distance between them squared.
Both laws can be applied to the objects that are spherically symmetrical (do not look like potatoes). In that case we consider that the charge/mass is concentrated at the center. Electric /gravitational force is a vector. If two or more point charges/masses act on some charge/mass, the net force on that charge/mass is a vector sum of all individual forces acting on that charge/mass.
The electric/gravitational field at point P is defined as the force per unit charge/mass placed at that point.
qF = E
F = gm
g = G 𝑀𝑀𝑅𝑅2
M – mass of the planet R – radius of the planet gEarth = G 𝑀𝑀
𝑅𝑅2 = 9.80 m s-2
Direction of electric field due to charge Q at some point P is equal to the direction of the force on a positive charge placed at that point.
Direction of gravitational field due to mass M at some point P is toward mass M.
Electric force acting on a charge q placed there is:
𝐹 = 𝑞𝑞 𝐸𝐸
Uniform electric field, E
Gravitational force acting on a mass m placed there is:
𝐹 = 𝑓𝑓 𝑐
commonly called “weight of mass m”
Uniform gravitational field, g
If a positive charge q is released at point A, force F = qE will accelerate it in the direction of the field toward point B. Work done on charge by force F along displacement d is converted into kinetic energy.
W = Fd = qEd = ½ mv2
(remember: const E, const. F so W = Fd)
If a mass q is released at point A, force F = mg will accelerate it in the direction of the field toward point B. Work done on mass by force F along displacement d is converted into kinetic energy.
W = Fd = mgd = ½ mv2
(remember: const E, const. F so W = Fd)
We call the lines along which the magnets align themselves the magnetic field lines.
magnetic field 𝑩𝑩 is measured in Tesla (T).
Magnetic field
Strength of the B-field is proportional to the density of the field lines.
At either pole of the earth the B-field is thus the greatest.
N S
N
S
N S
N S
Magnetic field lines don’t start or stop. There are no magnetic charges (monopoles)
|B| = Fmag
q v 1 T(Tesla) = N·sC·m
Direction at any location is the direction in which the north pole of the compass needle points at that location.
Outside magnet: N → S Inside magnet: S → N (always closed loops)
B OR
OR
View from head of a vector
View from tail of a vector
Sketching conventions for drawing direction of 3-D vector. How do you draw a vector that is directed toward you or away from you
INTO Page away from you
OUT of Page toward you
RHR 2: The direction of the magnetic field produced by electric current is given by the right-hand rule 2: If a wire is grasped in the right hand with the thumb in the direction of current flow, the fingers will curl in the direction of the magnetic field.
Magnetic field B around a wire with current I
B = 𝜇𝜇02𝜋𝜋
𝐼𝐼𝑟𝑟
𝜇𝜇0 = the permeability of free space 4p×10-7 T·m/A I = current [A] r = distance from the center of the conductor n = N/L number of turns of wire per unit length
Magnetic Field B Inside of a Solenoid
B = 𝜇𝜇0 n I
The magnetic field is concentrated into a nearly uniform field in the centre of a long solenoid. The field outside is weak and diverging
Magnetic field caused by a straight line current
RHR 2: The direction of the magnetic field produced by electric current is given by the right-hand rule 2: If a wire is grasped in the right hand with the thumb in the direction of current flow, the fingers will curl in the direction of the magnetic field.
Magnetic field caused by a straight line current
Magnetic field is decreasing away from wire
Most of the time you’ll find this much easier to draw Just keep that in mind!!!!
Lines of B
r B Current I OUT
Determining magnetic field direction – wire loop
Solenoids
A solenoid consists of several current loops stacked together. In the limit of a very long solenoid, the magnetic field inside is very uniform, and outside is almost zero (B ≈ 0 )
I
N S
N S
If we place an iron core inside the solenoid we have an electromagnet. The ferrous core enhances the strength of the B-field
Another way: RHR for solenoids Grasp the solenoid with your right hand in such a way that your fingers curl in the direction of the current. Your extended thumb points in the direction of north pole.
I
I
Force on a charge moving in a B-field Lorentz force (Hendrick Antoon Lorentz, Dutch physicist of the late 19th and early 20th century
Moving charge produces a magnetic field.
Moving charge placed in an external magnetic field will feel a magnetic force Surprise, or isn’t?
Interaction of fields !!! a stationary charge in a magnetic field will feel no magnetic force because the charge will not have its own magnetic field.
Force F felt by a charge q traveling at velocity v through a B-field of strength B is given by
𝐹𝐹 = 𝑞𝑞𝑣𝑣𝑞𝑞 sin 𝜃𝜃 𝜃𝜃 is the angle between 𝑣 and 𝑞𝑞
Force on q due to presence of B
Force on a charge moving in a B-field
Magnetic force acting on a charge q in a magnetic Field B: 𝐹𝐹 = 𝑞𝑞𝑣𝑣𝑞𝑞 𝑑𝑑𝑒𝑒𝑑𝑑 𝜃𝜃
Magnetic force on a wire carrying current I in a magnetic field B: 𝐹𝐹 = 𝑞𝑞𝐼𝐼𝐼𝐼𝑑𝑑𝑒𝑒𝑑𝑑 𝜃𝜃
q = charge [C] v = velocity [m/s] B = magnetic field [T] 𝜃𝜃 is the angle between 𝑣 and 𝑞𝑞
I = current [A] L = length [m] B = magnetic field [T] 𝜃𝜃 = 𝜃𝜃 is the angle between 𝐼𝐼 and 𝑞𝑞
RHR 1: The direction of the magnetic force on a charge/current is given by the right-hand rule 1:
Outstretch fingers in the direction of v (or current I). Curl fingers as if rotating vector v (I ) into vector B. Magnetic force on a positive charge (or I) is in the direction of the thumb. Magnetic force on a negative charge points in opposite direction.
𝐹 is perpendicular to 𝑣 and 𝑞𝑞 and thus to the plane of 𝑣 and 𝑞𝑞 𝐹 is in the opposite direction for a (–) charge.
Charge q in elec. field E and mag. field B
The electric force: Felec = Eq is always parallel to the direction of the electric field. acts on a charged particle independent of the particle’s velocity (even at rest). does the work when moving charge. The work, is converted into kinetic energy which is, in the case of conductors, transferred to thermal energy through collisions with the lattice ions causing increased amplitude of vibrations seen as rise in temperature.
The magnetic force: Fmag = qvB sinθ is always perpendicular to the direction of the magnetic field acts on a charged particle only when the particle is in motion and only if v and B do not point in the same or opposite direction (sin 00 = sin 1800 = 0). Force is perpendicular to the direction of the motion, so the work done by magnetic force is zero. W = ΔKE = 0 Hence change in kinetic energy of the charge is 0, and that means that mag. force cannot change the speed of the charge. Magnetic force can only change direction of the velocity – therefore it acts as centripetal force.
In the presence of magnetic field, the moving charged particle is deflected (dotted lines)
The electric field accelerates charged particles.
B v v F
F
The trajectory of a charge q in a uniform magnetic field B
In the case the charge q is subject to the uniform field B, centripetal force Fc is magnetic force forcing the charge to move in a circle:
Positive charge q in magnetic field B 𝐹𝐹𝑐𝑐 = 𝐹𝐹𝑚𝑚𝑚𝑚𝑚𝑚
𝑓𝑓 𝑣𝑣2
𝑅𝑅= 𝑞𝑞𝑣𝑣𝑞𝑞
𝑅𝑅 =𝑓𝑓𝑣𝑣𝑞𝑞𝑞𝑞
B = magnetic field [T] R =is the radius of the path Fma is magnetic force on the charge directed toward the centre of the circular path m = mass [kg] v = velocity [m/s] q = charge [C]
massive or fast charges – large circles large charges and/or large B – small circles
A charge q traveling at velocity v perpendicular to magnetic field B. Show that r = mv / qB.
Force is perpendicular to v sin θ = 1
Magnetic force (F=qvB) does no work! Speed is constant (W = Δ KE ) Circular motion
Force between wires carrying current
Current-carrying wires create magnetic fields Magnetic fields exert a force on current-carrying wires Current carrying wires exert forces on each other!
Two current-carrying wires exert magnetic forces on one another We already saw that if we put a current carrying wire into a magnetic field it will feel a force....so what will happen when we put two current carrying wires together!?!?! One will create magnetic field that the other will feel a force from, and vice versa! Let us see what is going on.
Force between wires carrying current
I up
B
I up
F Conclusion: Currents in same direction attract!
x
Use RHR #2 to find the direction of the magnetic field at point P
use RHR #1 to find the force on second wire
F
I up
B
I down
F Conclusion: Currents in opposite direction repel!
x
Use RHR #2 to find the direction of the magnetic field at point P
use RHR #1 to find the force on second wire
F
P
P
