An alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 6.40 cm in a uniform magnetic field with b = 1.17 t. calculate (a) its speed, (b) its period of revolution, (c) its kinetic energy, and (d) the potential difference through which it would have to be accelerated to achieve this energy.

(a)

The magnetic force acts as centripetal force, so we can write

where

q is the charge of the particle

v is its speed

B is the magnetic field strength

m is the mass

r is the radius of the circular path

For the alpha particle in the problem,

B = 1.17 T

Re-arranging the equation and solving for v, we find its speed:

(b)

The period of revolution is given by the ratio between the distance travelled in one circle (so, the circumference of the path) and the speed of the particle, so

where

r is the radius of the path

v is the speed

Here we have

So the period of revolution is

(c)

The kinetic energy of a particle is given by

where

m is its mass

v is its speed

For the alpha particle in the problem, we have

So its kinetic energy is

(d)

When accelerated through a potential difference, a particle gains a kinetic energy equal to the change in electric potential energy - so we can write:

where the term on the left is the change in electric potential energy, with

q is the charge of the particle

is the potential difference

Here we have

is the  charge of the alpha particle

is the kinetic energy

Re-arranging the formula, we find

Explanation:

It is given that,

Charge on alpha, q = 2e

Radius of circular path, r = 6.4 cm = 0.064 m

Mass of alpha particle, m = 4 u

Uniform magnetic field, B = 1.17 T

(a) Magnetic force is balanced by the centripetal force as :

(b) Period of revolution,

(c) Kinetic energy,

or

(d) Potential energy = Kinetic energy

V = 134375 volts

Hence, this is the required solution.

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach