A proton starting from rest travels through a potential of 644 X 106V and then moves into a uniform magnetic field, B = 0.02 T, perpendicular to the protons velocity. What is the radius of the proton's resulting orbit in m? Mp 1.67 x 10-27 kg, e = 1.6 x 10-19 C.

Given :

Potential ,

Uniform magnetic field , B = 0.02 T .

Mass of proton , .

Charge on proton , .

Also , velocity of proton is perpendicular to magnetic field .

To Find :

The radius of the proton's resulting orbit .

Solution :

Now , when velocity is perpendicular to magnetic field radius of orbit is given by :

Putting all given values above :

Therefore , radius of orbit is 183.33 m .

Hence , this is the required solution .

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .