Electricity and Electromagnetism

Question 1

Unit for electrical current (I)

Answer 1

The unit for electrical current is amperes (A).

Question 2

Unit for electric charge (q)

Answer 2

The unit for electric charge is coulombs (C).

Question 3

Unit for voltage (V)

Answer 3

The unit for voltage is volts (V).

Question 4

How does applying a voltage to a wire generate an electrical current?

Answer 4

Applying a voltage to a wire produces an electric field in it. This electric field applies a force on the electrons in the wire, causing them to move. The moving electrons are an electrical current.

Question 5

Unit for resistance (R) and equations

Answer 5

The unit for resistance is ohms (Ω).

- Resistance in parallel circuit 
1/Rt = 1/R1 + 1/R2
- Resistance in series circuit
Rt = R1 + R2
- 'Make circuit into series' by calculating resistance in parallel first. Remember to add numbers first if they are on the same line.

Question 6

Unit for power (P)

Answer 6

The unit for power is watts (W).

Question 7

Alternative power equations

Answer 7

P = I ^2 x R
P = V^2 / R

Question 8

How increasing the resistance of one bulb over another affects the brightness of different bulbs in a series circuit?

Answer 8

The brightness of a light bulb depends on the amount of electric power (P) transformed by the bulb into heat and light energy. As the lightbulbs are in a series circuit, an equal amount of current runs through each lightbulb.

The power of each light bulb can be calculated using P = I^2 x R. From this equation we can see that if current is kept constant, as resistance decreases, power will decrease also.

Therefore, the lightbulb with less resistance will use less power, and be dimmer.

Question 9

How increasing the resistance of one bulb over another affects the brightness of different bulbs in a parallel circuit?

Answer 9

The brightness of a light bulb depends on the amount of electric power (P) transformed by the bulb into heat and light energy. As the lightbulbs are in a parallel circuit, an equal amount of voltage is applied to each lightbulb.

The power of each light bulb can be calculated using P = V^2 / R. From this equation we can see that if voltage is kept constant, as resistance decreases, power will increase.

Therefore, the lightbulb with less resistance will use more power, and be brighter.

Question 10

Unit for electrical energy

Answer 10

The unit for electric power is joules (J) or kilowatts per hour (kWh).

Question 11

Conversion from kWh to J

Answer 11

Multiply kWh by 3.6 x 10^6

Question 12

Unit for magnetic field strength (B)

Answer 12

The unit for a magnetic field is teslas (T).

Question 13

Magnetic field lines

Answer 13

• Move from the north to south pole of magnets
• Can’t cross over or have gaps in between them
• Drawn as dots if coming out of the page
• Drawn as crosses if going into the page
• The closer together the magnetic field lines are, the stronger the magnetic force (magnetic field lines are closest together at the poles of the magnet, meaning that the magnetic force is strongest at the poles).

Question 14

Conversion from mT or μT to T

Answer 14

• Divide mT by 1000

- Divide μT by 1,000,000

Question 15

Right hand slap rule for current carrying wire

Answer 15

When a current-carrying wire is placed in a magnetic field, we use the right hand slap rule to find the direction that the magnetic force acts in.

• Point the fingers on your right hand in the same direction as the magnetic field lines
• Point your thumb in the same direction as the conventional current
• The palm of your hand is then facing in the same direction that the magnetic force acts in

Question 16

How is a current-carrying wire affected as it is placed into a magnetic field?

Answer 16

When a current-carrying wire is placed into a magnetic field, it will only experience a magnetic force if the wire is perpendicular to the magnetic field (not parallel).

Question 17

F = BIL

Answer 17

The magnitude of the magnetic force when current and magnetic field are perpendicular

• F is the magnetic force
• B is the magnitude of the magnetic field
• I is the magnitude of the current
• L is the length of wire in the magnetic field.

(Only used to calculate magnitude of magnetic force, right hand slap rule must be used to determine the direction).

Question 18

Right hand slap rule for single charged particle

Answer 18

When a single charged particle moves through a magnetic field, we use the right hand slap rule to find the direction that the magnetic force acts in.

• Point the fingers on your right hand in the same direction as the magnetic field lines
• Point your thumb in the same direction as the particle’s velocity

NOTE - When a positively charged particle travels through a magnetic field, thumb points in the same direction that the particle is travelling.
When a negatively charged particle travels through a magnetic field, thumb points in the opposite direction to the particle’s motion.

• The palm of your hand is then facing in the same direction that the magnetic force acts in

Question 19

How is a charged particle affected as it moves through a magnetic field?

Answer 19

A charged particle experiences a magnetic force when it is placed in a magnetic field, which causes it to accelerate, changing direction.

For example, a positvely charged particle that is moving upwards is placed in a magnetic field facing towards you. At that instant, the magnetic force is directed to your left, so it causes the particle’s motion to curve to the left. Since the particle is now moving in a different direction, the magnetic force must also change direction. By applying the right hand slap rule, we can see that as the partcle curves to the left, the magnetic force is directed on an increasingly steeper diagonal, which causes the particle’s motion to curve even more. Overtime, the curve in the particles motion due to the magnetic force changing its direction will result in the particle moving in a circle. The magnetic force is always directed towards the centre of the circle.

Question 20

Calculating the induced current

Answer 20

1. Calculate the induced voltage (V = BvL)

2. Calculate the induced current (V = IR)

Question 21

Effect of an electric field on a particle entering the field perpendicular to the field lines

Answer 21

If a particle experiences an electrostatic force, the force will make the particle accelerate. If a particle enters the electric field perpendicular to the field lines, it will travel along a parabolic path. This is because whilst the particle’s velocity parallel to the field increases, the particle’s velocity perpendicular to the field remains unchanged.

Question 22

Electron in an electric field.

Answer 22

• When the electron is near the negatively charged plate, it has electric potential energy.
• When the electron is free to move, it is accelerated by the electric field towards the positive plate and the electric potential energy stored inside the particle is transformed into kinetic energy.

NOTE - Only movement parallel to the field lines affects the electric potential energy. Movement perpendicular to the field lines does not affect a particle’s potential energy.

To move a particle in the opposite direction, work must be done to overcome the electrostatic forces on the particle. The work done on the particle is stored as electric potential energy.

The change in a particle’s electric potential energy can be calculated using ΔEp = Eqd

Question 23

A charge moving through a uniform electric field

Answer 23

• The forces acting on the charge are electrical force (upwards) and weight force caused by gravity (downwards)
• When the charge is suspended between the plates, the electrical force is equal and opposite to the weight force (Eq = mg)
• To allow for suspension, the electrical force must be acting upwards, and therefore the type of charge must be opposite to the charge on the top plate.
• If the distance between the plates is decreased while the charge and voltage across the plates remains constant, this will increase the electric field strength (E = V/d) and therefore increase the electric force acting on the charge. This will cause the negative charge to accelerate towards the positive plate, as the forces acting on it are unbalanced (electric force > gravitational force).

Question 24

An electron and proton are placed in fields with the same electric field strength

Answer 24

• The electric force experienced by each particle will be the same as they both have the same sized charge on them and the electric field strength is the same.
• The acceleration of the electron will be greater than the proton because it has a smaller mass (a = F/m)

Question 25

Calculating voltage across components

Answer 25

- In series, the voltage splits between components and the current remains the same - In parallel, the voltage remains the same between components and the current splits equally between the number of electrical pathways (unless the resistances on each path are different, in which case it can be a different current across each pathway) (With component in parallel circuit) 1. Calculate the total resistance of the circuit 2. Calculate the total current in the circuit 3. Calculate the total voltage in the circuit 4. Calculate the voltage across the other component(s) you are not trying to calculate 5. Subtract the answer from the total voltage in the circuit (With component in series circuit) 1. Use V = IR with the resistance of component and the total current

Question 26

Significant figures

Answer 26

- The final answer cannot be any more accurate than the least number of significant figures in the question.

Question 27

Power of components

Answer 27

- The energy per second of each component is the power output - Power relates to voltage and current (P = IV) Parallel If both components are on the same branch... - They both have the same current through them - Voltage is directly proportional to resistance (V = IR), therefore when the current is the same, as resistance increases, voltage increases also. Therefore, the voltage with the highest resistance would have a greater voltage across it and therefore use more energy per second.

Question 28

Affect of adding another bulb in parallel to a circuit on a separate branch

Answer 28

- Adding another bulb in parallel will cause the overall resistance of the circuit to decrease ((1/RT = 1/R1 + 1/R2). - This will cause the overall current in the circuit to increase (I = V/R) - Therefore, the voltage of any bulbs connected in series will increase (V = IR) and the voltage of any bulbs connected in parallel will decrease (Voltage is directly proportional to resistance (V = IR), therefore when the current is the same, as resistance decreases, voltage decreases also). - Therefore, the power used by any bulbs connected in series will have more power, and any bulbs connected in parallel will have less power (P = IV).

Question 29

Rod connected to wire moving across a magnetic field

Answer 29

- The wires used to connect a light are composed of free to move charges, which are moving perpendicular to a magnetic field. From F = Bvq, these charges will experience a magnetic force. This will cause positive charges to experience a force to the top of the wire, and negative charges to the bottom. As work is done to cause this separation of charges, which forms an electric field. This will result in an induced voltage. 1. If one conductor is outside the magnetic field - As one conductor is outside the magnetic field no voltage is induced in the conductor, however, a voltage is induced in the moving conductor as it moves across the magnetic field (V = BvL), so a current would flow (in a clockwise direction). 2. If both conductors are in the same magnetic field - As both conductors are inside the magnetic field, of the same length and with the same speed, the same voltage is induced in both conductors, so no current would flow. (Polarity in both conductors is the same (positive at the top), meaning that no current will flow). 3. If one conductor is not moving, but the other is - As one conductor is not moving, no voltage is induced in the conductor, (V = BvL), however, a voltage is induced in the moving conductor, so a current would flow (in a clockwise direction).

Question 30

Calculations for a rod connected to wire moving across a magnetic field

Answer 30

1. Calculate the induced voltage (V = BvL) 2. Calculate the induced current through the rod (I = V/R) 3. Calculate the force applied to the rod (F = BIL)

Question 31

Electric field in a wire

Answer 31

- The electric force on an electron is F = Eq, and therefore if the charge and electric field strength are constant, the force will be constant also - If another battery is added in series, this will double the voltage, and if the wire length is halved, this will double the electric field strength (E = V/d). Therefore, the electric force on the wire will be 4x as large.

Question 32

Conversion from mV or kV to V

Answer 32

- Divide mV by 1000 | - Multiply kV by 1000

Question 33

Current in electromagnetic loop (swing)

Answer 33

- When the entire loop is inside the magnetic field, the current is 0, as both sides of the loop are cutting across the magnetic field (not parallel) and have the same induced voltage. One voltage pushes the charges clockwise, while the other voltage pushes the charges anti clockwise. This results in a net voltage of 0. - If the speed doubles, the induced voltage doubles (V = Bvl). Assuming the resistance is constant, this will cause the current to double also (I = V/R).

Question 34

Force on a charged particle in a magnetic field

Answer 34

- Magnetic fields need to be perpendicular (90°) to the direction of motion of a charged particle to maximise the force acting upon it. Otherwise, the magnetic force will have little to no effect.

Question 35

Resistance of components in parallel, where A uses more power than B

Answer 35

- Because both components are in parallel, they have the same voltage. Therefore, as the voltage is constant and A uses more power than B, there is more current passing through A (P = IV). As a result, A has less resistance (R = V/I).

Question 36

Unit for electric field strength

Answer 36

Vm^-1

Question 37

Charge of an electron

Answer 37

-1.6 x 10^-19

Question 38

Size of force on rod sliding left to right

Answer 38

- The electric field between the plates is constant, as the distance between each plate and voltage across the plates is constant (E = V/d). Therefore, the force is constant. - The force acts uniformly across the rod, as the charge is uniformly distributed. This means that each part of the rod will experience the same force and no bending will occur (F = Eq).

Question 39

Affect of adding another bulb in parallel to a circuit on the same branch as another component

Answer 39

- Adding another bulb on the same branch as another component (in series) will cause the resistance of the branch to increase, and the overall resistance of the circuit to increase ((1/RT = 1/R1 + 1/R2). - This will cause the overall current in the circuit to decrease (I = V/R) - Therefore, the voltage of any bulbs connected in series will decrease (V = IR) and the voltage of any bulbs connected in parallel will increase (Voltage is directly proportional to resistance (V = IR), therefore when the current is the same, as resistance increases, voltage increases also). - As the bulb and the other component on the same branch are in series, they only get half of what is received by the component on the other branch. - Increased voltage causes more power dissipation, which will cause the component on the other branch to become brighter (P = V^2/R). As a result, it is more likely to blow, as it will receive more than its initial voltage.

Question 40

Force on electromagnetic loop (flying fox)

Answer 40

- The force on the wire connected to the positive terminal is equal and opposite the force on the wire connected to the negative terminal. Therefore the magnetic force has a net force of 0, and will not change the time to complete the flying fox. - This force is so small, even if it was not cancelled out it would be negligible.

Question 41

How to change the velocity of a particle in an electric field

Answer 41

1. Increase the charge (q) - This causes the charge to experience a greater force as F = Eq. This causes it to have greater acceleration (assuming the increase in charge adds negligible mass) and a higher velocity. 2. Increase the electric field strength (E) by decreasing the distance between the plates or by increasing the voltage across the plates - This causes the charge to experience a greater force as F = Eq. This causes it to have greater acceleration and a higher velocity.

Question 42

Why bulb A in parallel will be brighter than bulb B if it has a higher resistance.

Answer 42

- Brightness is determined by power. - As they are in parallel, the voltage across each bulb is the same. However, the current through bulb A is higher, due to having a lower resistance (I = V/R). Therefore, bulb 2 will use more power, as P = IV, and therefore will be brighter.

Question 43

Effect of adding an ammeter in parallel

Answer 43

- Adding an ammeter in parallel will short circuit the connected bulb, as all the current will go through the ammeter instead due to the low resistance of the ammeter (0V across connected bulb). This will cause the connected bulb to go out, and result in the total resistance of the circuit to decrease. - This will cause the overall current in the circuit to increase (I = V/R) - Therefore, the voltage of any bulbs connected in series will increase (V = IR) and the voltage of any bulbs connected in parallel will decrease (Voltage is directly proportional to resistance (V = IR), therefore when the current is the same, as resistance decreases, voltage decreases also). - Therefore, the power used by any bulbs connected in series will have more power, and any bulbs connected in parallel will have less power (P = IV).