Question

A singly charged ion of unknown mass moves in a circle of radius 12.5 cm in a magnetic field of 9.0 T. The ion was accelerated through a potential difference of 7.0 kV before it entered the magnetic field. What is the mass of the ion?
___kg

Answer 1

here,

for the ion radius is given by

r = sqrt(2mV/q)/B

Now ,

0.125 = sqrt(2*m*7000/e)/9

solving for m

m = 1.45 * 10^-23 Kg

Mass of ion is  1.45 * 10^-23 Kg

Answer 2

in this problem when ion is moving in circular path

there are two forces magnetic force and centripetal force

under equllibrium condition both forces are equal

determing the charge quantity

assume that charge as q

q²=4x3.14xepsilonxr²x7x10^3x1.6x10^-19

from this we will get 1.109x10^-20

after this

use the given formula get answer mv²/r =qvb

r=mv/qb

Answer 3

It's a singly charged particle.
It need not be a proton or electron.

(eg: it can be an Na+ ion or an Deuteron, etc)

qvB = mv^2/r --eqn1

(the force due to the magnetic field on a charged particle is q[v(bar)xB(bar))], which provides for the centripetal force for it to revolve in a circle.)

now, the initial speed(v) of the particle is attained by accelerating it in a electric field (assume it being uniform), between points at a potential difference of V.

hence, qV = 1/2 mv^2 --eqn2

[Note that he magnetic force doesnot change the speed of the particle as it is always perpendicular to it.]

manipulating the two eqns, we get,
given
{
V = 7000 volts, [NOT 70000 v]
r = 12.5 x 10^-2 metres,
B = 9 T = 1.2 Weber*m^-2
}

v = 2V/rB

=2*7000/12.5 x 10^-2*1.2

=14000/15*10^-2

=933.3*10^2

= 93333.33 m/s

b)m = (q r^2 * B^2)/(2V)

=(1.6*10^-19*(15*10^-2)^2*9^2/2*7000

= 2.571 * 10 ^ -25 kg..

Answer 4

Kinetic energy

KE=qE=(1/2)mv²

1.6*10^-19*7000=(1/2)*m*v²

v=sqrt[2.24*10^-15/m]

centripetal force=magnetic force

mv²/r =qvB

=>m=qBr/v =(1.6*10^-19)*9*0.125/sqrt[2.24*10^-15/m]

sqrt[2.24*10^-15/m]*m =(1.6*10^-19)*9*0.125

m=1.446*10^-23 kg