A proton (mass= 1.67×10-27 kg, charge= 1.6×10-19 C) travelling with speed 1×106 m/s enters a region of space containing a uniform magnetic field of 1.2 T. What is the time t required for the proton to re-emerge into the field-free region? t =

The time required for the proton to re-emerge into the field-free region is 5.47 x 10⁻⁸ s

Explanation:

given;

mass of proton, m = 1.67 × 10⁻²⁷ kg

charge of proton, q = 1.6 × 10⁻¹⁹ C

speed of proton, v = 1 × 10⁶ m/s

uniform magnetic field, B = 1.2 T

Force on the proton due to uniform magnetic field = qvB

Force on the proton in the circular region = ma = mv²/r

The time required for the proton to re-emerge into the field-free region, is the same as the time required for the proton to return to its starting position. During this period, the proton has made one complete revolution.

v = 2πr/t

t = 2πr/v

where;

t is the required for the proton to re-emerge into the field-free region

r is the radius of the circular region

mv²/r = qvB

r = mv²/qvB

r = mv/qB

r = (1.67 × 10⁻²⁷ x 1 × 10⁶) / (1.6 × 10⁻¹⁹ x 1.2)

r = 8.698 x 10⁻³ m

t = 2πr/v

t = (2π x 8.698 x 10⁻³) / (1 × 10⁶)

t = 5.47 x 10⁻⁸ s

Therefore, the time required for the proton to re-emerge into the field-free region is 5.47 x 10⁻⁸ s

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 .