IIT JEE , AIEEE Forum - Physics , Mechanics

magiclko

A particle of mass m and charge q is projected in a region where an electric field is existing and is given by E = E₀ x i , (E and i represents vector notation), with a velocity V₀ j from the origin at t=0.Then find the radius of curvature of the particle when its x-coordinate becomes x₀ .

krishna.gopal

Let a=qE/m

Then at any time t

x=0.5at^2

y=v_ot

When x=x_o

t=sqrt(2x_o/a)

y=v₀sqrt(2x_o/a)

v_x=a sqrt(2x_o/a)

v_y = v_o

a_x=a

a_y = 0.

From these three you now velocity and acc of particle at x=x_o

Find magnitude of velocity (v) and component of acc perpendicular to velocity (a_per)

Radius of curvature r = v²/a_per

yahiyafirdous

a_x=qE₀/m

a_y=0

x=(qE₀/2m)t²

y=v₀t

Eliminating t from above equations

y²=(2m/qE₀)v₀²x

Find dy/dx and d²y/dx²

^{Radius of Curvature is given by}

[ 1 + (dy/dx)(dy/dx)]^3/2/[d²y/dx²]

Radius= 2x[1+(v₀²/2x)(m/qE₀)]^3/2Answer

yahiyafirdous

Did I rightly solve this????????

magiclko

i think yes, it's sth like this only, but m nt sure.... will confirm it 2morrow, newayz thanx :)

shakirshafi12

man u havent taken electric field variable
ur formula for radius of curvature is dead right

AAKRITI

yes this is one of my oldest doubts, i know the formula for the curvature radius

but do you know where has it come from/

shakirshafi12

manasi is this q from aits or u have made urself??

aakriti here s how

tanQ=dy/dx

diff wrt s

where

ds=dx^2+dy^2

sec^2Q*dQ/ds=d^2y/dx^2.dx/ds

radius of curvature=ds/dQ

ds=dx^2+dy^2-------1

from 1 ds/dx=sqroot(1+(dy/dx)^2))

we

ds/dQ=(1+y`)^3/2/ y2

as fr the Q i think that as it is released from origin that is x=0;

E=0

at that pt bcoz E=Eo *x i

magiclko

hi shakir,

ofcourse i havnt made it on my own, m not that gud at physics!!!! the question is in fiitjee's GMP...

and the answer given is [ m²V² + 2qE₀mX₀ ]^3/2

q E₀ V_om²

AAKRITI

F= qEox=ma ( a is accn in x dirn ) = mv.dv/dx ( v is vel in x dirn)

so,qEox.dx=mv.dv

integrate, x goes from 0 to xo, v from 0 to v_x

qEx²₀= mv²_x ----1 , dx/dt = v_x

dy/dt = v₀

so dy/dx = v₀/v_x-----2

put the value of v_x from 1 in 2, use the expn of dy/dx in radius of curvature

shakirshafi12

manasi if electric field is a function of x then at origin
E=0 then why will it drift to +ve x direction

shakirshafi12

hey the answer u have given only comes if we take Electric field constant

AAKRITI

this is a very big ambiguity as rightly pointed out

the particle should not be given velocity from a point where its x coordinate is 0

if its x coordinate is not 0 and is l say, then the limits of integration will be l to xo

Ask & Discuss Questions with Community & Experts