Question

An unknown charged particle passes without deflection throughcrossed electric and magnetic fields of strengths 187,500 V/m and0.125T, respectively. The particle passes out of the electricfield, but the magnetic field continues, and the particle makes asemicircle of diameter 25.05 cm. What is the particle'scharge-to-mass ratio? Can you identify the particle?

Answer 1

Explanation:

The given data is as follows.

        Electric field strength (E) = 187,500 V/m

    Magnetic field strength (B) = 0.125 T

       Diameter (d) = 25.05 cm = 0.2505 m    (as 1 m = 100 cm)

    Radius (r) = $\frac{d}{2}$

                    = $\frac{0.2505}{2}$

                    = 0.12525 m

Formula to calculate the magnetic force ($F_{M}$) is as follows.

              $F_{M} = Bqv$ ............ (1)

Electrical force is calculated as follows.

             $F_{E} = qE$ ............ (2)

On both electric and magnetic fields the velocity is perpendicular.

       $F_{M} - F_{E}$ = 0

or,             $F_{M} = F_{E}$

Hence, from equations (1) and (2)

              Bqv = qE

or,            v = $\frac{E}{B}$ ............. (3)

                  = $\frac{187500 V/m}{0.125 T}$

                  = 1,500,000 m/s

As the particle is moving in a semi-circular trajectory and motion of charged particle is given by the electric field as follows.

              $F_{c} = \frac{mv^{2}}{r}$ ........... (4)

where,    $F_{c}$ = centripetal force

             $F_{M} = F_{c}$

Using equation (1) and (4) as follows.

            $F_{M} = F_{c}$

              Bqv = $\frac{mv^{2}}{r}$

                   $\frac{q}{m} = \frac{v}{Br}$

                       = $\frac{15 \times 10^{5}}{0.125 \times 0.12525}$

                       = $958.08 \times 10^{5} C/kg$

Thus, we can conclude that charge-to-mass ratio of the given particle is $958.08 \times 10^{5} C/kg$.