# Electromagnetism

## Magnetic Induction

### The Strength of a Magnetic field is known as the Magnetic Induction (B) and is measured in units of Tesla.

### 1 Tesla is the Magnetic Induction of a Magnetic field in which a 1 meter length of wire perpendicular to the field, carrying 1 Ampere of Current experiences 1 Newton of Force.

## A Current Carrying Wire

### As was seen in Higher Physics, a wire with a Current flowing through it generates a Magnetic field around itself, as shown in the diagram below :-

### The Magnetic field lines form concentric circles around the wire, perpendicular to the direction of the Current flow. As there is no north or south pole to give direction to the field lines, the following left hand grip rule (for Electron flow) is used :-

### As can be seen in the above diagram, by placing the thumb pointing in the direction of Current flow, the curl of the fingers shows the direction of the field lines.

## Force on a Current Carrying Wire

### When a wire carrying a Current (which therefore has its own Magnetic field) experiences an external Magnetic field, the wire experiences a Force.

### The Force on the wire depends on several factors :-

### 1. The Magnitude of the Current ( I )

### 2. The Magnetic Induction ( B )

### 3. The Length of the Wire inside the Magnetic Field ( L )

### 4. The Angle between the wire and the Magnetic Field ( θ )

### The diagram below shows wire within a Magnetic field :-

### The following formula is used to calculate the Force applied to the above wire :-

### F = BILsinθ

### Where :-

### F = Force Experienced ( N )

### B = Magnetic Induction ( T )

### I = Current within the conductor ( A )

### L = Length of current carrying conductor within Field ( m )

### θ = Angle between the wire and the Magnetic Field ( ° )

### Note - The sine function in this equation takes into account that it is only the component of the length perpendicular to the field that provides the Force.

### As such, the following applies :-

### Wire Perpendicular to Field - sin 90 = 1, therefore Force at a maximum

### Wire Parallel to Field - sin 0 = 0, therefore no Force experienced

## Example 1 -

### A wire carrying a Current of 6 A has 0.5 m of its length placed within a Magnetic field with a Magnetic Induction of 0.2 T. Calculate the Force acting on the wire if :-

### 1. The Wire is at right angles to the field.

### 2. The Wire is at an angle of 45° to the field.

### 3. The Wire is parallel to the field.

### At a right angle

### F = BILsinθ

### F = 0.2 x 6 x 0.5 x sin 90

### F = 0.2 x 6 x 0.5 x 1

### F = 0.6 N

### At an angle of 45°

### F = BILsinθ

### F = 0.2 x 6 x 0.5 x sin 45

### F = 0.2 x 6 x 0.5 x 0.707

### F = 0.42 N

### Parallel to the Field

### F = BILsinθ

### F = 0.2 x 6 x 0.5 x sin 0

### F = 0.2 x 6 x 0.5 x 0

### F = 0 N

## Fleming's Hand Rules

### In the above diagram, the Force acting on the wire was labelled as acting in a downwards direction. The direction of the force will always act perpendicular to both the Current flow and the Magnetic field.

### The direction of the force can be determined if you know the sign on the charge and the direction of the magnetic field. Magnetic field by definition runs from north to south.

### Electron Current = Fleming's Right Hand Rule

### Conventional Current = Fleming's Left Hand Rule

### How the Rules Work...

### First Finger ( Index Finger ) = ( Magnetic ) Field

### Second Finger ( Middle Finger ) = Current

### Thumb = Direction of Force ( 'Thrust' )

### Note - This means that in order for the Force to act downwards in example 1, the Current type shown must be conventional Current.

### The video below shows a demonstration of a current carrying piece of wire in a magnetic field and its resultant motion. (warning - which version of Current is this?)

## Magnetic Induction at a distance from a Wire

### If the Magnetic Induction around a long Current carrying wire is measured, it can be seen that the Magnetic Induction is proportional to the Current through the wire and inversely proportional to the distance from the wire.

### As such, the following relationship can be derived :-

### Where :-

### B = Magnetic Induction (T)

### I = Current through the Wire (A)

### r = Distance radially from the wire (m)

### μ0 = The Permeability of Free Space (4π x 10-7 N A-2)

## Non Examinable Derivation - Force per unit length between two parallel Wires

### The previous calculation, the field around a single wire was investigated. The single wire generated a Magnetic field around itself, which could be measured using a Hall Probe. If a second Current carrying wire is introduced to the system, then to two separate Magnetic fields will interact, causing the wire to experience a Force. Assuming the wires are placed parallel to each other, the above formula can be used to find the Force between the wires.

### The Magnetic field generated by Wire 1 is given by :-

### By substitution of the Force calculation above, the Force applied to wire two can be derived :-

### As these fields are not based upon a point source, an extended description as a base unit is required - the Force per unit length :-

### The above calculation allow the magnitude of the Force per unit length to be calculated, but does not account for the direction of the Force. The direction in this case is dependent on how the two fields interact with each other :-

### 1. Wires carrying Current in same direction - Force is attractive

### 2. Wires carrying Current in opposite direction - Force is repulsive

### The above diagram shows the fields generated by the two wires as well as the direction of the field lines at the other wire. The Force direction is again given by Fleming's Right Hand Rule.

## Example 2 - 2012 Past Paper Q8

### A long thin horizontal conductor AB carrying a current of 25 A is supported by two fine threads of negligible mass. The tension in each supporting thread is T as shown above.

### Calculate :-

### The Magnetic Induction at a point P, 6·0 mm directly below conductor AB.

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### A second conductor CD carrying Current I is now fixed in a position a distance r directly below AB as shown above. CD is unable to move.

### Question :-

### Show that the force per unit length acting on each conductor can be written as:-

### Force per unit length :-

### The mass per unit length of the conductor AB is 5·70 × 10–3 kg m–1. When the conductors are separated by 6·0 mm, the Current I in conductor CD is gradually increased. Calculate the value of I which reduces the Tension in the supporting threads to zero.

### Current :-

## The Ampere

### As seen previously with Voltage, the definition of values within Electricity require quite a high level of Physics knowledge to understand, and none are as high level at this stage as the definition of a Current of 1 Ampere. The definition of 1 Ampere is based upon above Force between two wires calculation.

### The Ampere is defined as :-

### "The constant Current which, if in two straight parallel conductors of infinite length placed one meter apart in a vacuum, will produce a force between the conductors of 2 x 10-7 newtons per meter."

## Non Examinable Derivation - Motion in a Magnetic Field

### The Force on a wire is not actually applied to the wire itself, but to the charges moving through the wire. Therefore the wire itself is unimportant in this calculation, and the motion of single free charges moving through a Magnetic field can be described.

### The diagram below shows a charged particle with a charge q moving with a constant speed of v perpendicularly to a Magnetic field of Magnetic Induction B :-

### As has been shown above, the Force experienced by the charge can be found using :-

### F = BILsinθ

### The time taken for the charge to travel the distance L and the Current this movement generates can be found using the following formulae and combined :-

### t = L / v and I = q / t

### IL = qv

### By substituting into the above formula and assuming the charge is moving perpendicularly to the Magnetic Field, the following can be derived :-

### F = BILsinθ

### F = BILsin90

### F = BIL

### F = qvB

### Where :-

### F = Force experienced by the Charge ( N )

### q = The Charge of the Charge ( C )

### v = The velocity of the Charge ( ms-1 )

### B = The Magnetic Induction ( T )

### Note - If the charge is not moving perpendicular to the Magnetic field, then obviously the above formula reverts to F = qvB sinθ

### The above formula gives the magnitude of the Force experienced by the charge, but does not show its direction. In order to fully explain motion in a Magnetic field, three different paths for the charged particles must be considered :-

### 1. Charge moving parallel or anti-parallel to the Magnetic field.

### 2. Charge moving perpendicular to the Magnetic field.

### 3. Charge moving at an angle to the Magnetic field.

## Motion Parallel or Anti-Parallel to the Field

### If the charge is moving parallel or anti-parallel to the Magnetic field then the angle θ between the velocity vector and the Magnetic field direction is 0° or 180°. Because of this, the Force experienced is zero. The path is a straight line with a constant speed.

## Non Examinable Derivation - Motion Perpendicularly to the Field

### In the above diagram, the charge has (at the instant shown) a tangential velocity straight upwards of v. The charge in this case has a positive charge, and the Magnetic field acts "into the page".

### The charge is moving perpendicular (sin 90 = 1) to the Magnetic field therefore the Force the charge experiences is equal to F = qvB.

### As the Force experienced (as found by Fleming's Left Hand Rule) acts perpendicularly to the motion, the charge will follow a circular path. In this case, the Force the Magnetic field is providing acts centripetally, and as such follows the calculations in the Radial Acceleration and Centripetal Force section of unit 1.

### By using the Centripetal Force calculations from unit 1, the radius of curvature of the charge can be found :-

### Also, by applying calculations for angular Velocity from the Circular Motion section of unit 1, the Period and Frequency of the rotation can be derived :-

## Motion at an angle to the Field

### If the motion of the charge is not at either of the two extremes discussed above, but instead makes some angle to the Magnetic field, then the components in each case are required.

### The Motion of the charge is a combination of a circular motion within the Magnetic field and an unaltered path, giving helical motion.

### In the above diagram the helical path is caused by two components of the Velocity :-

### v sin θ - Circular motion perpendicular to the Magnetic field B

### v cos θ - Straight line motion in the direction of the Magnetic field B

## Helical Motion and Pitch

### In the above helical motion, the "tightness" of the helix is shown by a quantity known as the Pitch of the helical motion.

### The Pitch of the helical motion is defined as :-

### "The distance traveled in the direction of the Magnetic field B during one Period of Rotation"

### The Pitch is therefore found using the following equation :-

### Where :-

### p = Pitch m)

### v cos θ = The Straight line component of the velocity (ms-1)

### T = The Period of the Rotation (s)

## Example 3 - 2014 Past Paper Q8

### A deuterium ion within an experimental fusion reactor is moving with a velocity of 2·4 × 107 ms–1 perpendicular to a Magnetic field. The maximum diameter of the circular motion, permitted by the design, is 0·50 m. Properties of ions present in the plasma are given in the table below:-

### Calculate :-

### 1. The Magnetic Induction B required to constrain the deuterium ion within the maximum permitted diameter.

### 2. The maximum period of rotation for this deuterium ion.

### Magnetic Induction (B) :-

### Period of Rotation (T) :-

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