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 7.50 mm in a magnetic field with a magnitude of 3.00 T .
Part A:Find the speed of the deuteron.
Part B:Find the time required for it to make 12 of a revolution.
Part C:Through what potential difference would the deuteron have to be accelerated to acquire this speed?

Answer 1

Answer:

Explanation:

mass, m = 3.34 x 10^-27 kg

charge, q = 1.6 x 10^-19 C

radius, r = 7.5 mm

magnetic field, B = 3 T

A:

Let the speed is v.

$v=\frac{Bqr}{m}$

By substituting the values

$v=\frac{3\times 1.6\times 10^{-19}\times 7.5\times 10^{-3}}{3.34\times 10^{-27}}$

v = 1.08 x 10^6 m/s

B:

Time period to complete one revolution is given by

$T=\frac{2\pi r}{v}$

$T=\frac{2\times 3.14\times 7.5\times 10^{-3}}{1.08\times 10^{6}}$

T = 4.36 x 10^-8 s

So, time to complete 12 revolutions = 12 x T = 12 x 4.36 x 10^-8  

                                                        = 5.23 x 10^-7 s

C:

Let the potential difference is V

So, kinetic energy = q V

0.5 x 3.34 x 10^-27 x 1.08 x 10^6 x 1.08 x 10^6 = 1.6 x 10^-19 x V

V = 12174.3 volt