91Ó°ÊÓ

The speed of the electrons is\(v = 6 \times {10^7}\;\;\frac{{\rm{m}}}{{\rm{s}}}\)The applied magnetic field is\({\rm{B = 5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{\;T}}{\rm{.}}\)

Required:

Find the charge per mass ratio due to Thomson experiment.

For a circular motion, the centripetal force is given by the following relation:

\({\rm{F = m}}{{\rm{a}}_{\rm{R}}}{\rm{ = }}\frac{{{\rm{m}}{{\rm{v}}^{\rm{2}}}}}{{\rm{r}}}\)

In order to evaluate the force due to the magnetic field that effect on a moving charge, we use the following relation:

\({{\rm{F}}_{\rm{B}}}{\rm{ = qvBsin(\theta )}}\)

Sin the electrons move perpendicular to the magnetic field, then the angle\({\rm{\theta = 9}}{{\rm{0}}^{\rm{^\circ }}}{\rm{.So,sin(\theta ) = 1}}{\rm{.}}\)Substitute from the previous calculation:

\({{\rm{F}}_{\rm{B}}}{\rm{ = qvBsin(\theta ) = }}\frac{{{\rm{m}}{{\rm{v}}^{\rm{2}}}}}{{\rm{r}}}\)

Then the ratio\(\frac{{\rm{q}}}{{\rm{m}}}\)is equal to

\(\begin{array}{l}\frac{{\rm{q}}}{{\rm{m}}}{\rm{ = }}\frac{{\rm{v}}}{{{\rm{rB}}}}\\{\rm{ = }}\frac{{{\rm{6 \times 1}}{{\rm{0}}^{\rm{7}}}{\rm{\;m/s}}}}{{{\rm{6}}{\rm{.8 \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{\;m \times 5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{\;T}}}}\\{\rm{ = 1}}{\rm{.765 \times 1}}{{\rm{0}}^{{\rm{11}}}}{\rm{C/kg}}\end{array}\)

Hence, \(\frac{{\rm{q}}}{{\rm{m}}}{\rm{ = 1}}{\rm{.765 \times 1}}{{\rm{0}}^{{\rm{11}}}}{\rm{C/kg}}\)