Question

The angle of dip, and the horizontal component of earth’s magnetic field, at a given place, equal α and B_H , respectiely. If a charged particle, of charge q and mass m, were to move with a speed v, horizontally northwards at that place, the magnitude of the acceleration, caused on this charged particle, would equal

(1) \(\frac{qvB_H cotα}{m}\)

(2) \(\frac{qvB_H cosα}{m}\)

(3) \(\frac{qvB_H tan \alpha}{m}\)

(4) \(\frac{qvB_H tan \alpha}{m}\)

Answer 1

 (3) \(\frac{qvB_H tan \alpha}{m}\)

Since the charged particle is moving horizontally north wards, it would experience a force, only due to the vertical component (say BV ) of the earth’s (total) magnetic field (= B say). The angle of dip being α, we can write

The force, due to the earth’s magnetic field, on the charged particle, has a magnitude F, where

F = |q (v x B)| = qvB_v

∴ The acceleration caused, has a magnitude