Question

a charged particle moves in uniform magnetic field perpendicular to it in a circular path of radius r. On passing through a metallic sheet , it loses half of its kinetic energy. Now the radius of curvature is..
a) $\sqrt{2}$r
b) r/$\sqrt{2}$
c) r
d) 2$\sqrt{2}$r

Answer 1

Answer:

A charged particle in uniform magnetic field traces a circular path.. (so centripetal force = force due to field) mv2/R=qvB. => R=mv/qB (R is directly proportional to v) --1

Kinetic energy = 1/2mv2. so K.E directly proportional to square of velocity KE1V22=KE2V12

=>KV22=1/2KV2 => V2=V/rt(2) .

from 1 we can say that

R1V2=R2V1 RV/rt(2)=R2V => R2 = R/rt(2)

hope this helps you

Answer 2

Answer:

R / √2

OPTION B

Explanation:

★ AS Electron passes through metal field it loses energy and HENCE ;

RADIUS OF CURVATURE DECREASES.