📖 generic · CBSE Class 12th English Medium · PHYSICS PART-1 · Page 141table

4.3 M OTION IN A M AGNETIC F IELD · Part 2

Chapter 4: Chapter 4 · PHYSICS PART-1

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 v / r = q v B , which gives r = m v / qB ( . ) 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 ω is the angular frequency, then v = ω r . So, ω = π ν = q B / m [ . (a)] which is independent of the velocity or energy .

Here ν is the frequency of rotation. The independence of ν from energy has important application in the design of a cyclotron (see Section . . ).

The time taken for one revolution is T = π / ω ≡ / ν . 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 (Fig. . ).

The distance moved along the magnetic field in one rotation is called pitch p . Using Eq. [ . (a)], we have p = v || T = π m v || / q B [ .

(b)] The radius of the circular component of motion is called the radius of the helix . FIGURE . Circular motion FIGURE . Helical motion

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