A singly charged positive ion has a mass of 2.46 ✕ 10−26 kg. After being accelerated through a potential difference of 270 V, the ion enters a magnetic field of 0.535 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field.

Explanation:

Given data

Mass m=2.46×10⁻²⁶kg

Voltage V=270 V

Magnetic field B=0.5353 T

To find

Radius r

Solution

By using conservation of energy to find speed of charge

Substitute the given values

So the radius can be find as:

Substitute the values

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 .