(a) A particle of charge 'q' and mass 'm', moving with velocity '\overrightarrow{v}' is subjected to a uniform magnetic field \overrightarrow{B} perpendicular to its velocity. Show that the particle describes a circular path. Obtain expression for the radius of the circular path of the particle.

(b) Explain, how its path will be affected if the velocity \overrightarrow{v} makes an angle \theta (\neq 90^{\circ}) with the direction of the magnetic field.

 

 

 

 
 
 
 
 

Answers (1)
S safeer

(a)

When a charged particle 'q' and mass 'm' enter into a perpendicular uniform magnetic field \overrightarrow{B} with a velocity \overrightarrow{v}. Then Lorentz magnetic force F = qv \times B is acted on it which acts as a centripetal force. The path followed by charge particle is in circular in shape.

Lorentz magnetic force = Centripetal force

\frac{mv^{2}}{r} =qv \times B

r = \frac{mv^{2}}{qvB}

r = \frac{mv}{qB}      radius of circular path

(b) If the velocity \overrightarrow{v} makes an angle \theta with the direction of magnetic field then we can resolve \overrightarrow{v} into two components. One is perpendicular to the magnetic field and the other one is parallel to the magnetic field. The perpendicular component makes the charged particle into circular motion and the parallel component makes it more into a parallel direction. So the path become helix.

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