SSS 3: ELECTROMAGNETIC FIELD

An electromagnetic field is the field due to the interaction of electric and magnetic forces on a charged body.

An electromagnetic field consists of two closely related components:

(i) Electric Field 

An electric field is a region around a charged body in which another electric charge experiences a force.

It is represented by E and its S.I. unit is newton per coulomb (N C⁻¹) or volt per metre (V m⁻¹).

Force on a Current-carrying conductor in a magnetic field 

The force, F on the wire is proportional to:

(I) Current, (I), flowing in the conductor 

(II) the length, (L), of the conductor in the field.

(III) the magnetic flux density (B)

(IV) the sine of the angle,(θ) between the conductor and the field.

When a current-carrying conductor is placed in a magnetic field, it experiences a force.

The force is given by:

F = BILsin θ

Example:

The force experienced by a current carrying conductor of length 100cm is 2.0N. Calculate the current in the conductor of it is flowing in a direction 45° with the field of 1.0T

Solution:

F=BILsin θ

2= 1 × L × 1 × sin 45°

2= L × sin45°

L= 2/sin45°

L= 2.83A

Note: Greatest force occurs when θ=90°

(ii) Magnetic Field

A magnetic field is a region around a magnet or current-carrying conductor in which a magnetic material or moving charge experiences a magnetic force.

 

The strength of magnetic field is usually measured in terms of a quantity called Magnetic flux density.

MAGNETIC FLUX 

Magnetic flux is the total number of magnetic field lines passing through a given surface area.

It is represented by the symbol (Φ).

S.I. unit of magnetic flux is Weber (Wb)

The formula:

Φ =BAcos θ

Where:

Φ = magnetic flux, measured in weber (Wb)

B = magnetic flux density, measured in tesla (T)

A = area, measured in square metres (m²)

θ = angle between the magnetic field and the normal (perpendicular) to the surface.

    When the magnetic field is perpendicular to the surface (θ=0):

Φ =BA

Example 1: 

A magnetic field of 0.5T passes perpendicularly through a rectangular surface of area 2m². Calculate the magnetic flux.

Solution:

Since the field is perpendicular to the surface, θ=0.

Φ =BAcos θ

Φ =0.5 × 2 × cos 0

Φ =0.5 × 2× 1

Φ = 1.0Wb

Example 2: 

A magnetic field of 0.8T passes parallel to the surface of area 3m² . Calculate the magnetic flux.

Solution:

When the field is parallel to the surface, θ=90° to the normal.

Φ =BAcos θ

Φ =0.8 × 3 × cos90°

Φ =0.8 × 3 × 0

Φ = 0 Wb

Example 3:

A magnetic field of 0.4T passes through an area of 5m². The angle between the magnetic field and the normal to the surface is 60°. Calculate the magnetic flux.

Solution:

Φ =BAcos θ

Φ = 0.4 × 5 × cos60°

Φ = 0.4 × 5 × 0.5

Φ = 1Wb

Magnetic Flux Density 

Magnetic flux density is the magnetic flux passing normally per unit area.

It is represented by B and its S.I. unit is the tesla (T) or Weber per square meter (Wb m⁻²)

1 Tesla = 1 Wb m⁻²

Example 1:

A magnetic flux of 0.8 Wb passes normally through an area of 2 m². Calculate the magnetic flux density.

Solution:

Given:Φ = 0.8 Wb, A = 2 m²

Formula:

B = Φ/A

B = 0.8/2

B = 0.4 T

Example 2:

A magnetic field of 0.5 T passes through a flat surface of area 4 m². The angle between the magnetic field and the normal to the surface is 60°. Calculate the magnetic flux through the surface.

Solution:

Given: B = 0.5 T, A = 4 m², θ = 60°

Φ = BA cos θ

Φ = 0.5 × 4 × cos 60°

Φ = 0.5 × 4 × 0.5

Φ = 1 Wb

FLEMING'S LEFT-HAND RULE

Fleming's Left-Hand Rule states that if the thumb, first finger and second finger of the left hand are held mutually perpendicular to one another, the first finger points in the direction of the magnetic field, the second finger points in the direction of the current, and the thumb points in the direction of the force or motion of the conductor.

      Fleming's Left-Hand Rule is used to determine the direction of the force acting on a current-carrying conductor placed in a magnetic field.

(1) First finger → direction of the magnetic Field (B)

(2) Second finger → direction of the Current(I)

(3) Thumb → direction of Force/Motion(F)

Direction of Magnetic Force

MAGNETIC FORCE ON A CHARGE MOVING IN A MAGNETIC FIELD

When a charged particle moves through a magnetic field, it experiences a force called the magnetic force.

The magnitude of the magnetic force is given by:

F = qvBsin θ

Where:

F= magnetic force in newtons (N)

q = charge of the particle in coulombs (C)

v = velocity of the particle in metres per second (m s⁻¹)

B = magnetic flux density in teslas (T)

θ = angle between the velocity and magnetic field.

(i) When the particle moves perpendicular to the field, θ =90°

(ii) When the particle moves parallel to the field, θ = 0°

Therefore, a charged particle moving parallel to a magnetic field experiences no magnetic force.

The magnetic force is perpendicular to both the velocity of the charged particle and the magnetic field.

For a positive charge, the direction can be determined using the appropriate right-hand rule. For a negative charge, the force acts in the opposite direction.

Example 1:

A proton moves with a velocity of 4 × 10⁶ ms⁻¹  perpendicular to a magnetic field of 0.5 T. Calculate the magnetic force acting on the proton. (Charge of a proton = 1.6 × 10⁻¹⁹ C)

Solution:

Given: q = 1.6 × 10⁻¹⁹ C, v = 4 × 10⁶ ms⁻¹ 

B = 0.5 T, θ = 90°

F = qvB sin θ

Since: sin 90° = 1

Therefore:  F = qvB

F = (1.6 × 10⁻¹⁹) × (4 × 10⁶) × 0.5

F = 3.2 × 10⁻¹³ N

Example 2:

An electron moves with a velocity of 3 × 10⁷ ms⁻¹ at an angle of 30° to a magnetic field of 0.2 T. Calculate the magnetic force acting on the electron. (Charge of an electron = 1.6 × 10⁻¹⁹ C)

Solution:

Given: q = 1.6 × 10⁻¹⁹ C, v = 3 × 10⁷ ms⁻¹

B = 0.2 T, θ = 30°

F = qvB sin θ

F = (1.6 × 10⁻¹⁹) × (3 × 10⁷) × 0.2 × sin 30°

F = (1.6 × 10⁻¹⁹) × (3 × 10⁷) × 0.2 × 0.5

F = 4.8 × 10⁻¹³ N

Note: Since an electron is negatively charged, the direction of its magnetic force is opposite to the direction predicted by the right-hand rule for a positive charge.

Lorentz force 

Lorentz force is the total force experienced by a charged particle moving in the presence of electric and magnetic fields. It is the combined effect of the electric force and the magnetic force acting on the charge.

The Lorentz force is given by:

Lorentz force = Electric force + Magnetic force

F =Eq + qvB

F = q(E + vB)

Where:

• F = total Lorentz force (N)
• q = electric charge of the particle (C)
• E = electric field strength (N/C)
• v = velocity of the charged particle (ms⁻¹)
• B = magnetic flux density (T)

Example:

An electron moves with a velocity of 2 × 10⁶ ms⁻¹ perpendicular to a magnetic field of 0.5 T. If the electric field produces an electric force of 1.6 × 10⁻¹³ N on the electron, calculate the magnetic force and the resultant Lorentz force, assuming the electric and magnetic forces are perpendicular to each other.

(q = 1.6 × 10⁻¹⁹ C)

Solution:

Given: v = 2 × 10⁶ ms⁻¹, B = 0.5 T

θ = 90°, Fₑ = 1.6 × 10⁻¹³ N

Step 1:

Magnetic force, Fᵦ = qvB sinθ

Fᵦ = (1.6 × 10⁻¹⁹) × (2 × 10⁶) × (0.5) × sin90°

Fᵦ = 1.6 × 10⁻¹³ N

Step 2: 

Since Fₑ and Fᵦ are perpendicular:

F = √(Fₑ² + Fᵦ²)

F = √[(1.6 × 10⁻¹³)² + (1.6 × 10⁻¹³)²]

F = √[2 × (1.6 × 10⁻¹³)²]

F ≈ 2.26 × 10⁻¹³ N

Resultant Lorentz force, F ≈ 2.26 × 10⁻¹³ N

Force and Field patterns Between Two Parallel Conductors Carrying Current

When two parallel conductors carry electric currents, each conductor produces a magnetic field which exerts a force on the other conductor. 

Ampere's Law explains the action between two parallel current - carrying wires.

Ampere's Law 

Ampere's Law states that two parallel current - carrying conductors attract each other when the current in them flows in the same direction but repel each other when the current in them flows in opposite direction.

If the currents flow in the same direction, the conductors attract each other.

If the currents flow in opposite directions, the conductors repel each other.

The force , F per unit length between the conductors is proportional to the product of the currents flowing through them and is inversely proportional to their separation (r)

Where:

F = force between the conductors (N)

L = length of the conductors (m)

I₁, I₂ = currents in the conductors (A)

d = separation between the conductors (m)

μ₀ = permeability of free space, μ₀ = 4π × 10⁻⁷ H m⁻¹

For attraction, the currents flow in the same direction; for repulsion, they flow in opposite directions.

Example:

Two long, parallel conductors are separated by a distance of 0.20 m. They carry currents of 5 A and 8 A respectively in the same direction. Calculate the force per unit length between the conductors.

Solution:

F/L = μ₀I₁I₂ / 2πd

where:

μ₀ = 4π × 10⁻⁷ H m⁻¹

Given: I₁ = 5 A, I₂ = 8 A, d = 0.20 m

μ₀ = 4π × 10⁻⁷ H m⁻¹

F/L = (4π × 10⁻⁷ × 5 × 8) / (2π × 0.20)

F/L = (160π × 10⁻⁷) / (0.40π)

F/L = 4.0 × 10⁻⁵ N m⁻¹

Since the currents flow in the same direction, the force is attractive.

Applications of Electromagnetic Fields

1. Power generation: Electromagnetic fields are used in generators to produce electrical energy.

2. Electrical machines: They are used in electric motors, transformers and other electrical devices.

3. Communication: Electromagnetic waves are used to transmit information in radio, television, mobile phones and wireless communication.

4. Medical equipment: Electromagnetic fields are used in medical devices such as MRI scanners and other diagnostic equipment.

5. Transportation: They are used in electric trains, motors and magnetic levitation (Maglev) systems.

6. Industrial heating: Electromagnetic fields are used for induction heating and melting of metals.

7. Security systems: They are used in metal detectors and other security devices.

8. Information technology: Electromagnetic fields are used in data storage, electronic devices and computer systems.

Assignment 

1. Explain the working principle of the following:

(I) D.C motor

(II) Moving coil galvanometer

2. Explain sensitivity and accuracy of galvanometer