Question

A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34×10⁻²⁷ kg and a charge of 1.60×10⁻¹⁹ C . The deuteron travels in a circular path with a radius of 7.30 mm in a magnetic field with a magnitude of 2.10 T .

A) 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

1.. Fm = Fc

qVB = mV^2/r

V= qBr/m

V = 1.6*10^-19*2.1*7.30*10^-3/(3.34*10^-27) = 7.34*10^5 m/sec

2.

t = d/V = pi*R/V

t = 3.14*7.30*10^-3/(7.34*10^5)= 3.12*10^-8 sec

3. energy conservation:

qV = 0.5mv^2

V = 0.5*3.34*10^-27*7.34^2*10^10/(1.6*10^-19) = 5623.26 V