contestada

A particle of charge q and rest mass m is moving with velocity v where the magnetic field is B. Here B is perpendicular to v, and there is no electric field. Show that the path of the particle is a curve with radius of curvature R given by R = p/qB, where p is the momentum of the particle, ? mv. (Hint: Note that the force qv × B can only change the direction of the momentum, not the magnitude. By what angle delta ? is the direction of p changed in a short tim

Respuesta :

Answer:

[tex]\Delta \theta = \frac{qB}{m} t[/tex]

Explanation:

As we know that charge is moving in magnetic field then we will have

[tex]F = qvB[/tex]

since this force is perpendicular to the velocity

so this force is considered as centripetal force and hence we will have

[tex]F = F_c[/tex]

[tex]qvB = \frac{mv^2}{R}[/tex]

so here we have

[tex]R = \frac{mv}{qB}[/tex]

[tex]R = \frac{P}{qB}[/tex]

here P = mv (momentum)

So angle changed by the charge is given as

[tex]\Delta \theta = \omega t[/tex]

[tex]\Delta \theta = \frac{v}{R} t[/tex]

[tex]\Delta \theta = \frac{qB}{m} t[/tex]

Otras preguntas