A.) Force on a current-carrying conductor placed in a magnetic field:
Let us consider a conductor PQ of length dl and area of cross-section A. The conductor is placed in a
uniform magnetic field B making an angle θ with the field as shown in figure.
Let a current I flows through PQ. Hence, the electrons are drifted along QP with drift velocity vd. If n is the
number of free electrons per unit volume in the conductor, then the current is
I = n A vd e
Multiplying both sides by the length of the conductor,
I dl = n A vd e dl ……………….(i)
Since the electrons move under the influence of magnetic field, the magnetic Lorentz force on a moving
electron.
fB = e (vd × B) …………….....(ii)
The negative sign indicates that the charge of the electron is negative.
The number of free electrons in the conductor
N = n A dl …………………......(iii)
The magnetic Lorentz force on all the moving free electrons
FB = N fB = n A dl e (vd × B)
Substituting equations (i) this equation
FB = n A dl e (vd × B)
FB = Idl × B …………………..(iv)
FB = Idl B sinθ …………….(v)
This equation (iv) and (v) gives total magnetic force acting on a current carrying conductor placed in
uniform external magnetic field.
Cases:
(i) If the conductor is placed along the direction of the magnetic field, θ=0∘, therefore force F=0 i.e. force
will be minimum.
(ii) If the conductor is placed perpendicular to the magnetic field, θ=90∘, F = I dl B i.e. conductor will
experiences maximum force.
B.) Lorentz Force:
Consider a charge q moving with velocity v in existence of both electric and magnetic fields. Then we
write:
The force on charge due to the electric field is given by
FE = qE ………………(i)
The force on charge due to the magnetic field is given by
FB = q(v х B) …………………(ii)
Let us consider a conductor PQ of length dl and area of cross-section A. The conductor is placed in a
uniform magnetic field B making an angle θ with the field as shown in figure.
Let a current I flows through PQ. Hence, the electrons are drifted along QP with drift velocity vd. If n is the
number of free electrons per unit volume in the conductor, then the current is
I = n A vd e
Multiplying both sides by the length of the conductor,
I dl = n A vd e dl ……………….(i)
Since the electrons move under the influence of magnetic field, the magnetic Lorentz force on a moving
electron.
fB = e (vd × B) …………….....(ii)
The negative sign indicates that the charge of the electron is negative.
The number of free electrons in the conductor
N = n A dl …………………......(iii)
The magnetic Lorentz force on all the moving free electrons
FB = N fB = n A dl e (vd × B)
Substituting equations (i) this equation
FB = n A dl e (vd × B)
FB = Idl × B …………………..(iv)
FB = Idl B sinθ …………….(v)
This equation (iv) and (v) gives total magnetic force acting on a current carrying conductor placed in
uniform external magnetic field.
Cases:
(i) If the conductor is placed along the direction of the magnetic field, θ=0∘, therefore force F=0 i.e. force
will be minimum.
(ii) If the conductor is placed perpendicular to the magnetic field, θ=90∘, F = I dl B i.e. conductor will
experiences maximum force.
B.) Lorentz Force:
Consider a charge q moving with velocity v in existence of both electric and magnetic fields. Then we
write:
The force on charge due to the electric field is given by
FE = qE ………………(i)
The force on charge due to the magnetic field is given by
FB = q(v х B) …………………(ii)
