Question

A charged particle is accelerated from rest through a potential difference of magnitude |ΔV|. After exiting the potential difference at an emission point, the particle enters a region of uniform magnetic field. The magnetic field is perpendicular to the particle's velocity, and the particle travels along a complete circular path. The particle's mass is 2.10 ✕ 10−16 kg, its charge is 26.0 nC, and the magnetic field magnitude is 0.600 T. The particle's circular path, as it returns to the emission point, encloses a magnetic flux of 15.0 µWb.
a. What is the speed in (m/s) of the particle when it is in the region of the magnetic field?
b. What is the magnitude of the potential difference through which the particle was accelerated?

Answer 1

Answer:

a

The speed of the particle is  $v = 209485.71 m/s$

b

The potential difference is  $\Delta V = 177.2 \ V$

Explanation:

From the question we are told that

    The mass of the particle is  $m = 2.10 *10^{-16} kg$

     The charge on the particle is  $q = 26.0 nC = 26.0 *10^{-9} C$

    The magnitude of the magnetic field is $B = 0.600 T$

    The magnetic flux is  $\O = 15.0 \mu Wb = 15.0 *10^{-6} Wb$

The magnetic flux is mathematically represented as

              $\O = B *A$

Where A is the the area mathematically represented as  

           $A = \pi r^2$

Substituting this into the equation w have

         $\O = B (\pi r^2 )$

Making r the subject  of the formula  

         $r = \sqrt{\frac{\O}{B \pi} }$

Substituting value  

        $r = \sqrt{\frac{15 *10^{-6}}{3.142 * 0.6} }$

            $r = 2.82 *10^{-3}m$

For the particle to form a circular path the magnetic force the partial experience inside the magnetic must be equal to the centripetal force of the particle and this is mathematically represented as

              $F_q = F_c$

Where $F_q = q B v$

   and   $F_c = \frac{mv^2 }{r}$

Substituting this into the equation above

       $qBv = \frac{mv^2}{r}$

 making v the subject

       $v = \frac{r q B}{m}$

substituting values

         $v = \frac{2.82 * 10^{-3} * 26 *10^{-9} 0.6}{2.10*10^{-16}}$

           $v = 209485.71 m/s$

The potential energy of the particle before entering the magnetic field is equal to the kinetic energy in the magnetic field

   This is mathematically represented as

                    $PE = KE$

     Where $PE = q * \Delta V$            

        and  $KE = \frac{1}{2} mv^2$

Substituting into the equation above  

             $q \Delta V = \frac{1}{2} mv^2$

Making the potential difference the subject

            $\Delta V = \frac{mv^2}{2 q}$

              $\Delta V = \frac{2.16 * 10^ {-16} * (209485.71)^2 }{2 * 26 *10^{-9}}$

              $\Delta V = 177.2 \ V$