A particle with a mass of $2.00\times 10^{-16}\mathrm{kg}$ and a charge of 30.0 nC starts from rest, is accelerated through a potential difference $\Delta V$, and is fired from a small source in a region containing a uniform, constant magnetic field of magnitude 0.600 T. The particle's velocity is perpendicular to the magnetic field lines. The circular orbit of the particle as it returns to the location of the source encloses a magnetic flux of $15.0\mu \mathrm{Wb}$. (a) Calculate the particle's speed.(b) Calculate the potential difference through which the $p$.

Key Concepts: Particle Dynamics, Electromagnetism, Magnetic Flux, Potential Difference Explanation: The motion of the particle can be described using the equations for circular motion and the force on a charged particle in a magnetic field. The potential difference can be found using the work-energy theorem. Step by Step Solution: Step 1. Calculate the radius of the circular orbit using the magnetic flux and the magnetic field. ${\Phi }_B=B\pi r^2$ Step 2. Use the mass and the radius to find the speed of the particle using the equation for circular motion. $v=\sqrt{rB/m}$ Step 3. Calculate the kinetic energy of the particle using the mass and speed. $K=(1/2)m{v}^2$ Step 4. Use the work-energy theorem to find the potential difference $\Delta V$ that accelerates the particle from rest. $K=q\Delta V$ Final Answer: (a) The speed of the particle is 9.00e+6 m/s. (b) The potential difference through which the particle is accelerated is 13.5 V.