Question

derive the expression for the radius and frequency of a charge describing in uniform circular motion in a uniform magnetic field??​

Answer 1

Answer:

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. Read more on Sarthaks.com - https://www.sarthaks.com/426619/derive-expression-radius-angular-frequency-circular-motion-charge-uniform-magnetic-field

Answer 2

Let the velocity be V perpendicular to Magnetic field.

The charge particle will experience a force perpendicular to both magnetic field and velocity.And the magnitude of force will be Q(VXB).

As the force is perpendicular to velocity it will Not effect the magnitude of velocity but it will effect the direction of velocity resulting in a circular path.

The centrifugal force generated due to circular path will be equal to force due to magnetic field to maintain the circular motion.

QVB=MV²/R

R=MV/QB--------------------------------(1)

Hence ↑ this is the expression for radius.

Now next question is that what would be the frequency.

Frequency=1/Time period

Time period (T)=2πR/V --------------------(2)

Putting the value of in the equation (2)

T=2πM/QB

Frequency=1/T→QB/2πM

Final answers

Frequency:-

$\frac{QB}{2 \pi m}$

Radius:-

$\frac{mV}{QB}$