Fields and electromagnetic induction

6 visual steps · 3 worked examples · 5 exercises with solutions

First, let’s see the idea.

What determines the direction of induced current in a loop?

Move forward one step at a time. Change the values and observe what happens.

Change creates the responseA = 0.5 m² · N = 4B = 3 T
Drawing description

The coil has 4 turns of 0,5 m². The dot symbols indicate a field of 3 T out of the screen.

Step 1 of 6

Look straight at the coil

Connect the green circles with N and the ⊙ symbols with B.

We have N turns of area A and total resistance 2 Ω. The field is uniform throughout the region and passes perpendicularly through the turns, out of the screen; its initial value is 3 T.

Turns
4
Area per turn
0,5 m²
Field shown
3 T

What if you change this?

-2 T/s2 T/s
0,1 m²1 m²
110

Before moving on

With N=4, A=0.5 m² and α=−1 T/s, determine the emf and its direction.

Compare with the explanation

ε=+2 V; I=+1 A, anticlockwise. The induced field points outwards, opposing the decrease.

You decide when to mark this lesson as studied.

Change creates the response

Slide the drawing to see the details.

Explore the full explanationThe foundations, principles, worked examples and all the exercises, at your own pace.

What you will understand

  • Distinguish electric field, force, potential and potential energy.
  • Interpret magnetic force on moving charges.
  • Calculate flux and induction with consistent angles and directions.

Before you start

You do not need to know everything by heart. The explanation revisits the necessary ideas and shows how they connect. If a foundation is still unclear, you can return to these lessons:

Review vectors and electrical interaction

A vector has magnitude, direction and sense. Adding fields requires adding components, not just magnitudes. Electric charge q can be positive or negative, is measured in coulombs and determines how a body responds to an electric field. Electric field E at a point is defined by force per unit positive test charge, taken small enough not to significantly alter the configuration.

The vector relationship is F=qE. If q>0, force points along the field; if q<0, it points opposite. The unit N/C can also be written V/m. The field exists as a description of the configuration even when no test charge is placed at that point. Field lines are representations: the tangent indicates direction and line density suggests strength, but the lines are not physical threads restraining particles.

Charge fields and superposition

A point charge Q produces a radial field in vacuum with magnitude E=k|Q|/r², for r>0. The constant k is approximately 8,99×10⁹ N·m²/C². The field points outwards if Q is positive and towards the charge if negative. The point model is appropriate when the distribution’s size is small compared with the distance, or in situations of suitable symmetry.

With several charges, total field is the vector sum of individual fields. Between two equal positive charges, fields can oppose and cancel at the midpoint; forces on a test charge also cancel there. However, potential and field are not the same quantity: source potentials add algebraically and can be non-zero even where total field is zero. The distinction becomes clear by separating a scalar quantity from a vector quantity.

Potential is not force

Potential difference ΔV between two points measures the change in electric potential energy per unit charge: ΔU=qΔV. Potential is scalar and depends on an additive reference; differences enter the energy balance. In an electrostatic field, the electric force does work W=−ΔU=−qΔV. A positive charge accelerated only by the field tends to gain kinetic energy as it moves to lower potential.

In a uniform field over displacement Δr, ΔV=−E·Δr. If displacement has magnitude d and makes angle θ with the field, ΔV=−Ed cos θ. Perpendicular to the field, the difference is zero. For negative charge, the energy relationship retains q’s sign: it is not enough to claim that all charges move spontaneously to lower potential. The field points where potential decreases fastest, not towards a compulsory absolute value.

What a magnetic field does

A charge with velocity v in magnetic field B experiences a force of magnitude |q|vB sin θ, where θ is the angle between velocity and field. The direction is perpendicular to both; its sense follows the right-hand rule for positive charge and reverses for negative charge. If the charge is at rest or moves parallel to the field, magnetic force is zero.

Magnetic force is perpendicular to velocity and therefore does no work on the charge in the particle model. It can change direction without changing speed. If v is perpendicular to a uniform field, the trajectory can be circular: equating |q|vB to mv²/r gives r=mv/(|q|B), in the non-relativistic regime and when other forces are negligible. An electric field, by contrast, can do work. This difference prevents treating the fields as simple versions of the same force.

Flux: field and surface orientation

Magnetic flux of a uniform field through a flat surface is Φ=BA cos θ. Area A is measured in m², B in teslas and flux in webers. Angle θ is measured between B and the surface normal, not between B and the plane itself. If the field crosses the plane perpendicularly, θ=0 and flux magnitude is greatest; if parallel to the plane, θ=90° and flux is zero.

The chosen normal fixes the flux sign. Reversing the normal changes the sign without altering the physical situation. For non-uniform fields or curved surfaces, imagine adding small B·ΔA contributions while respecting local orientation. Flux can change because B changes, area changes or the surface rotates. It is not a quantity of material lines crossing the circuit: it is a mathematical measure of the field’s distribution and orientation.

Faraday and Lenz: change that induces

For N turns with the same flux per turn, average induced electromotive force is ε=−NΔΦ/Δt. Instantaneously, ε=−N dΦ/dt. Electromotive force is measured in volts and represents circulation of the induced electric field along the circuit. It can exist in an open circuit; current arises only if there is a closed path and the components’ response permits it.

The negative sign expresses Lenz’s law: the induced response opposes the flux change producing it. If flux into the page increases, induced current tends to create an outward field; if that flux decreases, it tends to reinforce it inwards. Opposing change does not mean always opposing the original field. Energy transferred to the circuit comes from the agent varying the field or moving the system. Induction does not create energy without a source; a generator’s mechanical effort is part of the balance.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Example 1

Field and force with signs

A charge q=−3 μC is in a uniform field E=200 N/C to the right. What is the force?

  1. Convert q=−3×10⁻⁶ C and choose right as positive.
  2. F=qE=(−3×10⁻⁶)·200=−6×10⁻⁴ N.
  3. The sign indicates a leftward force, opposite to the field because the charge is negative.
Result

Magnitude 6×10⁻⁴ N, to the left.

Check the result

The units C·N/C give N. Changing the charge sign would reverse direction without changing magnitude.

Example 2

Potential and energy

A positive charge of 2 μC moves from 120 V to 30 V in an electrostatic field. Calculate ΔU and electrical work.

  1. ΔV=30−120=−90 V, using final minus initial.
  2. ΔU=qΔV=2×10⁻⁶·(−90)=−1,8×10⁻⁴ J.
  3. The electric force does work −ΔU=+1,8×10⁻⁴ J. Without other transfers, this is the kinetic energy gained.
Result

Potential energy decreases by 1,8×10⁻⁴ J and the field does positive work.

Check the result

A difference in volts becomes energy only when multiplied by charge; potential and energy are not interchangeable.

Example 3

A coil with changing flux

A coil of 100 turns and area 0,020 m² has its normal parallel to B. The field increases from 0,10 to 0,40 T in 0,50 s. Calculate the magnitude of average electromotive force.

  1. Since θ=0, cos θ=1. Flux change per turn is AΔB=0,020·0,30=0,0060 Wb.
  2. The magnitude is NΔΦ/Δt=100·0,0060/0,50=1,2 V.
  3. In a closed circuit, polarity must produce a current whose field opposes the increase in the chosen flux.
Result

|average ε|=1,2 V.

Check the result

If the normal made 60° with the field throughout the change, the result would be reduced by the factor cos60°=1/2.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Exercise 1

A field is 500 N/C and q=4 nC. What is the force magnitude?
A hint
Convert nanocoulombs.
See the worked solution
q=4×10⁻⁹ C; F=qE=2×10⁻⁶ N. For positive charge, the direction is along the field.

Exercise 2

In a uniform field of 100 V/m, a displacement of 0,2 m is made along the field. What is ΔV?
A hint
The difference is minus the dot product.
See the worked solution
ΔV=−100·0,2=−20 V. A perpendicular displacement would have ΔV=0 even though the field may be non-zero.

Exercise 3

What is the flux of B=0,5 T through A=0,04 m² if the normal makes 60° with B?
A hint
The angle refers to the normal.
See the worked solution
Φ=0,5·0,04·cos60°=0,01 Wb. If 60° were the angle to the plane, we would need to use 30° to the normal.

Exercise 4

Does constant flux generate induction merely because it is very strong?
A hint
Faraday depends on the rate of flux change.
See the worked solution
No. If flux per turn and the number of turns do not change, dΦ/dt=0 and induced electromotive force is zero in this model.

Exercise 5

A charge moves parallel to B. Does magnetic force change its speed?
A hint
Calculate sinθ.
See the worked solution
With θ=0 or π, sinθ=0, so magnetic force is zero. Even when non-zero, ideal magnetic force is perpendicular to velocity and does not change kinetic energy.

Mistakes worth avoiding

Electric field and potential are synonyms.
Field is a vector and relates to force per charge; potential is a scalar and relates to energy per charge.
The flux angle is always measured to the plane.
In BA cosθ, it is measured to the normal.
Lenz says that current always opposes the existing field.
The response opposes flux change and can reinforce a decreasing field.

The essentials to take with you

  • F=qE retains the charge sign.
  • ΔU=qΔV separates energy and potential.
  • Magnetic force depends on motion and orientation.
  • Induction requires flux change and respects energy conservation.

If you still cannot explain one of these ideas, return to the corresponding section and work through an example again. Recognising the difficulty is part of studying.

References for further study

Text and examples prepared for this learning path. These sources help check the principles and continue studying:

The organisation by school year is a study proposal. Pedagogical review by teachers remains a stage of the project.