BE1-HEM1 Electromagnetics 1
Notes 6:Faraday’s Law
1. Electromotance - emf
So far, we have considered electric and magnetic fields that do not change with respect to time - electro-
static and magnetostatic fields. An electrostatic field is conservative. If a charge goes around a circuit in
a conservative electric field Es , no resultant work is done:-
I
Es .dl = 0
This means that even if a current is initally present in the circuit, it won’t remain flowing if there is any
component in the circuit that converts its energy into other forms - eg to light a bulb or turn a motor,
or to lose energy as heat (eg a resistor). Therefore, for a circuit to be useful, there needs to be, at some
point in the circuit, a non-conservative electric field that imparts energy to the charges in the current.
This is called an electromotance or (emf ). The electromotance is equal to the work done moving 1C
around the circuit - ie the work per unit charge - units are Volts.
Summing the resultant electric field E around a circuit where Es is the conservative electrostatic field
due to the presence of charges, and where Em is any electromotive field in the circuit:-
I I I
E.dl = Es .dl + Em .dl
From above:- I I
E.dl = Em .dl
Therefore, emf E for the circuit can be found as:-
I
E= E.dl
The presence of a constant current means that the energy provided by the electromotance is equal to
the energy lost from the circuit. Batteries and power supplies commonly provide this emf. A battery
chemically separates positive and negative charges thus creating a potential energy difference between its
terminals. When open circuit, the emf moves positive charges to the positive terminal until the potential
difference due to the internal electrostatic field caused by the separation of the positive and negative
charges is equal to the emf.
We now introduce an emf that occurs when there is a change in magnetic flux with respect to time - ie
magnetodynamic conditions.
2. Motional electromotance
v v
B B
F F
As seen previously, the diagram on the left shows the force F on charges flowing as a current along a
wire with drift velocity v. We found in those previous notes that the total force on a charge q moving in
both an electric field and a magnetic field with velocity v is given by the Lorentz Force Law:-
F = q(v × B + E)
1
, BE1-HEM1 Electromagnetics 1:Notes 6:Faraday’s Law
B is pointing into the page in the diagram. Assume for now that E = 0 so that F = q(v × B). Note
that v × B plays the same role as electric field in the force law.
The diagram on the right shows a similar situation, except that the wire has been rotated by 90o and
there is initially no current flowing and the wire is motionless. The wire is then physically moved with
velocity v perpendicular to its length as shown. We might therefore expect a force F on the charges so
that the equation F = q(v × B) is still valid (consider it an identity - cause and effect are not specified).
Moving the wire at constant velocity v would therefore cause a constant force F on the charges, and if
the wire was part of a circuit, a current would flow. There is an induced electromotance on the charges.
As this emf is caused by the movement of the wire, it is called a motional electromotance.
The emf due to the movement of the wire within the magnetic field around a closed circuit is:-
I
E= (v × B).dl
3. Faraday’s Law
Consider the following apparatus:-
B
B
v s dl
v
A closed conducting loop is placed such that one end lies inside the constant and uniform magnetic
field B between the two poles of a horseshoe magnet as shown. The circuit can move without friction
horizontally. If the circuit moves at velocity v a motional emf is generated causing current to flow around
the circuit as follows:- I
E= (v × B).dl
By changing the order of the dot product and cross product:-
I
E =− B.(v × dl)
We integrate around the loop with respect to dl in a clockwise direction, evaluating the cross product
v × dl in the direction of the loop at each point, then dot producting the result with the magnetic field
B:-
- The left hand side of the circuit does not lie in the magnetic field, and so contributes zero.
- For the top and bottom sides, dl is parallel to the velocity v, so the cross product is zero
- For the right hand side - as B is constant along its length, we can sum the cross product v × dl for
each dl along the right hand side for a distance s in the direction shown, and then dot product with B:-
Including all of the above:- I Z s
B.(v × dl) = B. (v × dl)
0
The sum of the cross product over a length s, as v is perpendicular to dl, is a vector of length vs pointing
2
Notes 6:Faraday’s Law
1. Electromotance - emf
So far, we have considered electric and magnetic fields that do not change with respect to time - electro-
static and magnetostatic fields. An electrostatic field is conservative. If a charge goes around a circuit in
a conservative electric field Es , no resultant work is done:-
I
Es .dl = 0
This means that even if a current is initally present in the circuit, it won’t remain flowing if there is any
component in the circuit that converts its energy into other forms - eg to light a bulb or turn a motor,
or to lose energy as heat (eg a resistor). Therefore, for a circuit to be useful, there needs to be, at some
point in the circuit, a non-conservative electric field that imparts energy to the charges in the current.
This is called an electromotance or (emf ). The electromotance is equal to the work done moving 1C
around the circuit - ie the work per unit charge - units are Volts.
Summing the resultant electric field E around a circuit where Es is the conservative electrostatic field
due to the presence of charges, and where Em is any electromotive field in the circuit:-
I I I
E.dl = Es .dl + Em .dl
From above:- I I
E.dl = Em .dl
Therefore, emf E for the circuit can be found as:-
I
E= E.dl
The presence of a constant current means that the energy provided by the electromotance is equal to
the energy lost from the circuit. Batteries and power supplies commonly provide this emf. A battery
chemically separates positive and negative charges thus creating a potential energy difference between its
terminals. When open circuit, the emf moves positive charges to the positive terminal until the potential
difference due to the internal electrostatic field caused by the separation of the positive and negative
charges is equal to the emf.
We now introduce an emf that occurs when there is a change in magnetic flux with respect to time - ie
magnetodynamic conditions.
2. Motional electromotance
v v
B B
F F
As seen previously, the diagram on the left shows the force F on charges flowing as a current along a
wire with drift velocity v. We found in those previous notes that the total force on a charge q moving in
both an electric field and a magnetic field with velocity v is given by the Lorentz Force Law:-
F = q(v × B + E)
1
, BE1-HEM1 Electromagnetics 1:Notes 6:Faraday’s Law
B is pointing into the page in the diagram. Assume for now that E = 0 so that F = q(v × B). Note
that v × B plays the same role as electric field in the force law.
The diagram on the right shows a similar situation, except that the wire has been rotated by 90o and
there is initially no current flowing and the wire is motionless. The wire is then physically moved with
velocity v perpendicular to its length as shown. We might therefore expect a force F on the charges so
that the equation F = q(v × B) is still valid (consider it an identity - cause and effect are not specified).
Moving the wire at constant velocity v would therefore cause a constant force F on the charges, and if
the wire was part of a circuit, a current would flow. There is an induced electromotance on the charges.
As this emf is caused by the movement of the wire, it is called a motional electromotance.
The emf due to the movement of the wire within the magnetic field around a closed circuit is:-
I
E= (v × B).dl
3. Faraday’s Law
Consider the following apparatus:-
B
B
v s dl
v
A closed conducting loop is placed such that one end lies inside the constant and uniform magnetic
field B between the two poles of a horseshoe magnet as shown. The circuit can move without friction
horizontally. If the circuit moves at velocity v a motional emf is generated causing current to flow around
the circuit as follows:- I
E= (v × B).dl
By changing the order of the dot product and cross product:-
I
E =− B.(v × dl)
We integrate around the loop with respect to dl in a clockwise direction, evaluating the cross product
v × dl in the direction of the loop at each point, then dot producting the result with the magnetic field
B:-
- The left hand side of the circuit does not lie in the magnetic field, and so contributes zero.
- For the top and bottom sides, dl is parallel to the velocity v, so the cross product is zero
- For the right hand side - as B is constant along its length, we can sum the cross product v × dl for
each dl along the right hand side for a distance s in the direction shown, and then dot product with B:-
Including all of the above:- I Z s
B.(v × dl) = B. (v × dl)
0
The sum of the cross product over a length s, as v is perpendicular to dl, is a vector of length vs pointing
2
