Motion of charged particle in a magnetic field:

In previous sections we studied about sources of magnetic field. In this section we shall study the effect of magnetic field on a moving charge and on current carrying wire.

In magnetic field force experienced by a charged particle is given by the expression.

$ \displaystyle \vec{F} = q (\vec{v} \times \vec{B}) $

Here, v^{→} = velocity of the particle and B^{→} = magnetic field

On the basis of above expression, we can draw following conclusions :

# Stationary charge (i.e. v^{→} = 0) experiences no magnetic force.

# If v^{→} is parallel or anti parallel to B^{→} then the charged particle experiences no magnetic force.

# Magnetic force is always perpendicular to both v^{→} and B^{→}

# As magnetic force is always perpendicular to v^{→} , it does not deliver any power to the charged particle.

# As magnetic force is always perpendicular to the , it will only cause a change in the direction of v^{→}

On the basis of expression the maximum value of magnetic force is equal to

$ \displaystyle \vec{F} = q (\vec{v} \times \vec{B}) $ ,

which occurs when the charge is projected perpendicular to the uniform magnetic field.

In this case path of charged particle is circular and magnetic force provides the necessary centripetal force

# If radius of the circular path is R then,

$ \displaystyle \frac{m v^2}{R} = q v B $ ,

Where m = mass of the particle .

$ \displaystyle R = \frac{m v}{q B} $

Time taken to complete one revolution is

$ \displaystyle T = \frac{2 \pi R}{v} $

$ \displaystyle T = \frac{2 \pi m}{q B} $

If the charged particle is projected obliquely to the field then its velocity can be resolved into two components, one along the say V_{∥} and the other perpendicular to say v_{⊥} . It experiences a magnetic force and hence has a tendency to move on a circular path.

Due to v_{∥} it experience no force, and hence has a tendency to move on a straight path along the field. So in this case it moves along a helical path.

* Radius of Helix , R = mv_{⊥}/qB

* $ \displaystyle T = \frac{2 \pi m}{q B} $

(Note: T is independent of v )

* The distances moved by the charged particle along the magnetic field during one revolution is called pitch.

Pitch = v_{∥} × T

Pitch = v_{∥} $ \displaystyle \times \frac{2 \pi m}{q B} $

Example:A uniform magnetic field of 30 mT exists in the + X direction. A particle of charge + e and mass 1.67 × 10^{-27} kg is projected through the field in the + Y direction with a speed of 4.8 × 16^{6} m/s.

(a) Find the force on the charged particle in magnitude and direction

(b) Find the force if the particle were negatively charged.

(c) Describe the nature of path followed by the particle in both the cases.

Solution :

(a) F = e vB Sinθ

= (1.6 × 10^{-19}) (4.8 ×10^{6}) (30 × 10^{-3}) sin 90^{0}

= 230.4 × 10^{-16} N.

The direction of the force is in the (-z) direction.

(b) If the particle were negatively charged, the magnitude of the force will be the same but the direction will be along (+z) direction.

(c) As V ⊥ B, the path described is a circle, where radius is given by

R = mv/qB

= (1.67 × 10^{27}) (4.8 × 10^{6})/(1.6 × 10^{-19}) (30 × 10^{-3})

= 1.67 m.

Example : A uniform electric field of magnitude E = 100 kV/m is directed upward. Perpendicular to E and directed into the page, there exists a uniform magnetic field of magnitude B = 0.5 T. A beam of particles of charge + q enters this region. What should be the chosen speed of the particles for which the particles will not be deflected by the crossed electric and magnetic field ?

Solution : The charged particle in the electric field experiences a force qE in the direction of the electric field. This force is directed upward. The purpose of the crossed electric and magnetic field is to cancel each other’s effect on the charged particle. Hence, force ‘ qVB ‘ acts downward.

⇒ qE = qvB

⇒ v = E/B = (100 × 1000)/0.5

= 2 × 10^{5} m/s.

Exercise : An electron moves with velocity v^{→} = ak^{^} in a magnetic field of intensity B^{→} = bi^{^} + cj^{^ }. Find the magnetic force on the electron.