# Deviation of charged particle in uniform magnetic field

Deviation of charged particle in uniform magnetic field:
Case 1:
Suppose a charged particle enters perpendicular to the uniform magnetic field if the magnetic field extends to a distance ‘x’ which is less than or equal to radius of the path. In this case , $\large r = \frac{mv}{qB}$

Angle of deviation ‘θ’ can be determined by using the formula

$\large sin\theta = \frac{x}{r} = \frac{x q B}{m v}$

$\large \theta = sin^{-1}(\frac{x q B}{m v})$

The above relation can be used only when x < r.

Case 2: For x > r : In this case , $\large r = \frac{mv}{qB}$

In this case, deviation θ = 180°.

### Lorentz Force:

(i) When a charge enters a region where both electric and magnetic fields exists simultaneously, force acting on it is called Lorentz force and is given by
$\large \vec{F} = q\vec{E} + q (\vec{v} \times \vec{B})$

### Cyclotron:

(a) The cyclotron is a machine to accelerate charged particles or ions to high energies using both electric and magnetic fields in combination.

(b) Cyclotron uses the fact that the frequency of revolution of the charge particle in a magnetic field is independent of its energy.

(c) Centripetal force is provide by the magnetic force $\large \frac{m v^2}{r} = q v B$

(d) Radius of circular path is $\large r = \frac{m v}{q B}$

(e) Time period of charged particle is $\large T = \frac{2\pi r }{v}$

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

Frequency $\large f = \frac{1}{T} = \frac{q B}{2 \pi m}$ = Cyclotron Frequency

(f) K.E of charged particles is

$\large K.E = \frac{1}{2}mv^2 = \frac{q^2 B^2 r^2}{2 m}$