Question

Derive an expression of radius and angular frequency of circular motion of a charge in uniform magnetic field.

Answer 1

If r is the radius of the circular path of a particle, then a force of m v²/r, acts perpendicular to the path towards the centre of the circle, and is called the centripetal force. If the velocity v is perpendicular to the magnetic field B, the magnetic force is perpendicular to both v and B and acts like a centripetal force. It has a magnitude q v B. Equating the two expressions for centripetal force, m v² /r = q v B, which gives 

r = m v / qB --------- (1) 

for the radius of the circle described by the charged particle. The larger the momentum, the larger is the radius and bigger the circle described. If w is the angular frequency, then v = ω r. So, 

ω = 2πυ = q B/ m ---------- (2) 

which is independent of the velocity or energy . Here n is the frequency of rotation. The independence of n from energy has important application in the design of a cyclotron. 

The time taken for one revolution is T= 2π/ω = 1/υ. If there is a component of the velocity parallel to the magnetic field (denoted by v_||), it will make the particle move along the field and the path of the particle would be a helical one. The distance moved along the magnetic field in one rotation is called pitch p. Using Eq. (2) we have  

p = v_||T = 2πm v_|| / q B --------- (3)

The radius of the circular component of motion is called the radius of the helix.