Question

A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34×10−27 kgkg and a charge of 1.60×10−19 CC . The deuteron travels in a circular path with a radius of 6.50 mmmm in a magnetic field with a magnitude of 2.80 TA) Find the speed of the deuteron
B)Find the time required for it to make 12 of a revolution.
C) Through what potential difference would the deuteron have to be accelerated to acquire this speed?

Answer 1

Answer:

a) v = 8.72*10^5

b) t =4.27*10^-9 s

c) V = 7.93 kV

Explanation:

Given

Mass of deuteron, m = 3.34*10^-27 kg

Radius, r = 6.5mm

Magnetic field, B = 2.8T

The force acting on the deuteron is giving by F(B) = q(v * B)

Also, the angle between the magnetic field and velocity is 90°, thus, magnitude of force would be

F(B) =|q|vB

From newton's law, the magnitude of the force has to be equal to the given force, so that F(B) = ma, where

a = v²/r, so that F(B) = mv²/r

Then, mv²/r = |q|vB

[3.34*10^-27 * v²] / 6.5*10^-3 = 1.6*10^-19 * v * 2.8

5.138*10^-25 * v = 4.48*10^-19

v = 871934 m/s

v = 8.72*10^5 m/s

t = d/v = πr/v

t = π * 6.5*10^-3 / 8.72*10^5

t = 2.57*10^-8 s

To get 12 of a revolution, then,

t = 2.57*10^-8 / 6

t = 4.27*10^-9 s

To find the potential difference, the KE must be equal to PE, so

1/2 mv² = |q|V

V = mv²/2|q|

V = 3.34*10^-27 * (8.72*10^5)² / 2 * 1.6*10^-19

V = 2.539*10^-15 / 3.2*10^-19

V = 7934.375 V

V = 7.93 kV