Magnetic force on a moving charge: the electric charges moving in a magnetic field experiences a force F such that

(i) The force is proportional to the magnitude of the magnetic field i.e. F ∝ B.

(ii) The force is proportional to the charge q, i.e. F ∝ q.

(iii) The force is proportional to the component of the velocity v in the perpendicular direction of the field B, i.e. F ∝ vsinθ.

Combining the all factors, we get

F ∝ qvBsinθ

F = kqvBsinθ

If k = 1

F = qvBsinθ

In vector form

Special cases:

Case 1: if v = 0, then f = 0, thus a stationary charged particles does not experience any force in a magnetic field.

Case 2: if θ = 0 or 180, the F = 0,

Thus a charged particle moving parallel or antiparallel to a magnetic field does not experience any force in the magnetic field.

Case 3: if θ = 90, then F = qvB,

Thus a charged particle experiences the maximum force when it moves perpendicular to the magnetic field.

Rules for finding the direction of force on a charged particle moving perpendicular to a magnetic field:

(i) Fleming’s left hand rule: Stretch the thumb and the first two fingers of the left hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field, central finger in the direction of the current, then the thumb gives the direction of the force on the charged particle.

(ii) Right hand (palm) rule: Open the right hand place it so that tips of the finger point in the direction of the field B and thumb in the direction of velocity v of the positive charge, then the palm faces towards the force F.

Magnetic field:

We know that

B = F/qvsinθ

If q = 1, v = 1, θ = 90, then B = F

Thus the magnetic field at a point is the force acting on a unit charge moving with a unit velocity at right angles to the direction of the field.

S.l init of magnetic field is Tesla = 1 NA^-1 m^-1

CGS unit is Gauss, 1 Gauss = 10^-4 Tesla

Dimension of magnetic field [MT^-2A^-1]

Lorentz force: The total force experienced by a charged particle moving in a region where both electric and magnetic fields are present, is called Lorentz force. The Lorentz force is given by

Work done by a magnetic force on a charged particle:

The magnetic force

always acts perpendicular to the velocity v or the direction of motion of charge q. therefore

According to Newton’s second law,

(d/dt) (1/2 mv²) = 0    

dk/dt = 0

K = constant

Thus a magnetic force does not change the kinetic energy of the charged particle. This indicate that the speed of the particle does not change. According to work energy theorem, the change in kinetic energy is equal to the work done on the particle by the net force. Hence the work done on the charged particle by the magnetic force is zero.

Motion of a charged particle in a uniform magnetic field:

When a charged particle having charge q and velocity v enters a uniform magnetic field B, it experiences a force

The direction of the force is perpendicular to both v and B. the magnitude of the force is

F = qvB sinθ

Following three cases are possible:

(i) When the initial velocity is parallel to the magnetic field:

Here θ = o⁰, so F = 0, thus the parallel magnetic field does not exert any force on the moving charged particle. The charged particle will continue to move along the line of force.

(ii) When the initial velocity is perpendicular to the magnetic field:

Here = 18o⁰, so F = qvB, a maximum force acts on a particle perpendicular to its velocity. It does not do any work on the particle. It does not charge the kinetic energy or speed of the particle. The magnetic force provides the centripetal force. Let r be the radius of the circular path. Now

Centripetal force = (mv²/r) = magnetic force, qvB

r = (mv/qB)  

thus the radius of the circular orbit is inversely proportional to the specific charge (charge to mass, q/m) and the magnetic field.

Period of revolution, T = circumference/speed

T = 2πr/v = (2π/v).(mv/qB) = 2πm/qB

And frequency of revolution, f = 1/T

= (qB/2πm), called the cyclotron frequency.

(iii) When the initial velocity makes an arbitrary angle with the field direction:

Consider a charged particle q entering a uniform magnetic field B with velocity v inclined at an angle θ with the direction of B as shown in fig.

The velocity v can be resolved in to two rectangular components:

(i) The component vll along the direction of the field i.e. v_ll = vcosθ

The parallel component remains unaffected by the magnetic field.

(ii) The component perpendicular to the direction of the field, i.e. in X Y plane.

Due to this component of velocity, the charged particle experiences a force F = qvB which acts perpendicular to both v and B.

Then r = mvsinθ/qB

Period of revolution, T = circumference/speed

T = 2πr/v = (2π/v).(mvsinθ/qB) = 2πm/qB

And frequency of revolution, f = 1/T

= (qB/2πm), called the cyclotron frequency.

The linear distance travelled by a charged particle in the direction of the magnetic field during its period of revolution is called pitch of the helical path.

Pitch = V_ll * T = v cosθ * (2πm/qB) = (2πmv cosθ/qB)

Motion of a charge in perpendicular magnetic and electric fields:

As shown in fig, the electric field acts in the downward direction and deflects the electrons in the upward direction. The magnetic field acts normally in to the plane of paper and deflects the electrons in the downward direction.

Only those electrons will pass undeflected through the slit S₂ on which the electric and magnetic forces are equal and opposite. The velocity v of the undelected electrons is given by

eE = evB

v=E/B