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?

a

The speed of the particle is  

b

The potential difference is  

Explanation:

From the question we are told that

    The mass of the particle is  

     The charge on the particle is  

    The magnitude of the magnetic field is

    The magnetic flux is  

The magnetic flux is mathematically represented as

Where A is the the area mathematically represented as  

Substituting this into the equation w have

Making r the subject  of the formula  

Substituting value  

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

Where

   and  

Substituting this into the equation above

 making v the subject

substituting values

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

     Where            

        and  

Substituting into the equation above  

Making the potential difference the subject