Answer:
a
The speed of the particle is ![v = 209485.71 m/s](https://tex.z-dn.net/?f=v%20%3D%20209485.71%20m%2Fs)
b
The potential difference is ![\Delta V = 177.2 \ V](https://tex.z-dn.net/?f=%5CDelta%20%20V%20%20%3D%20%20177.2%20%5C%20V)
Explanation:
From the question we are told that
The mass of the particle is ![m = 2.10 *10^{-16} kg](https://tex.z-dn.net/?f=m%20%3D%202.10%20%2A10%5E%7B-16%7D%20kg)
The charge on the particle is ![q = 26.0 nC = 26.0 *10^{-9} C](https://tex.z-dn.net/?f=q%20%3D%2026.0%20nC%20%3D%2026.0%20%2A10%5E%7B-9%7D%20C)
The magnitude of the magnetic field is ![B = 0.600 T](https://tex.z-dn.net/?f=B%20%3D%200.600%20T)
The magnetic flux is ![\O = 15.0 \mu Wb = 15.0 *10^{-6} Wb](https://tex.z-dn.net/?f=%5CO%20%3D%2015.0%20%5Cmu%20Wb%20%3D%2015.0%20%2A10%5E%7B-6%7D%20Wb)
The magnetic flux is mathematically represented as
![\O = B *A](https://tex.z-dn.net/?f=%5CO%20%3D%20B%20%2AA)
Where A is the the area mathematically represented as
![A = \pi r^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cpi%20r%5E2)
Substituting this into the equation w have
![\O = B (\pi r^2 )](https://tex.z-dn.net/?f=%5CO%20%3D%20%20B%20%28%5Cpi%20r%5E2%20%29)
Making r the subject of the formula
![r = \sqrt{\frac{\O}{B \pi} }](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%7B%5Cfrac%7B%5CO%7D%7BB%20%5Cpi%7D%20%7D)
Substituting value
![r = \sqrt{\frac{15 *10^{-6}}{3.142 * 0.6} }](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%7B%5Cfrac%7B15%20%2A10%5E%7B-6%7D%7D%7B3.142%20%2A%200.6%7D%20%7D)
![r = 2.82 *10^{-3}m](https://tex.z-dn.net/?f=r%20%3D%202.82%20%2A10%5E%7B-3%7Dm)
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](https://tex.z-dn.net/?f=F_q%20%3D%20F_c)
Where ![F_q = q B v](https://tex.z-dn.net/?f=F_q%20%3D%20%20q%20B%20v)
and ![F_c = \frac{mv^2 }{r}](https://tex.z-dn.net/?f=F_c%20%3D%20%20%5Cfrac%7Bmv%5E2%20%7D%7Br%7D)
Substituting this into the equation above
![qBv = \frac{mv^2}{r}](https://tex.z-dn.net/?f=qBv%20%3D%20%5Cfrac%7Bmv%5E2%7D%7Br%7D)
making v the subject
![v = \frac{r q B}{m}](https://tex.z-dn.net/?f=v%20%3D%20%20%5Cfrac%7Br%20q%20B%7D%7Bm%7D)
substituting values
![v = \frac{2.82 * 10^{-3} * 26 *10^{-9} 0.6}{2.10*10^{-16}}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B2.82%20%2A%2010%5E%7B-3%7D%20%2A%2026%20%2A10%5E%7B-9%7D%20%200.6%7D%7B2.10%2A10%5E%7B-16%7D%7D)
![v = 209485.71 m/s](https://tex.z-dn.net/?f=v%20%3D%20209485.71%20m%2Fs)
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](https://tex.z-dn.net/?f=PE%20%3D%20%20KE)
Where
and ![KE = \frac{1}{2} mv^2](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
Substituting into the equation above
![q \Delta V = \frac{1}{2} mv^2](https://tex.z-dn.net/?f=q%20%5CDelta%20V%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
Making the potential difference the subject
![\Delta V = \frac{mv^2}{2 q}](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20%20%5Cfrac%7Bmv%5E2%7D%7B2%20q%7D)
![\Delta V = \frac{2.16 * 10^ {-16} * (209485.71)^2 }{2 * 26 *10^{-9}}](https://tex.z-dn.net/?f=%5CDelta%20%20V%20%20%3D%20%20%5Cfrac%7B2.16%20%2A%2010%5E%20%7B-16%7D%20%2A%20%28209485.71%29%5E2%20%7D%7B2%20%2A%2026%20%2A10%5E%7B-9%7D%7D)
![\Delta V = 177.2 \ V](https://tex.z-dn.net/?f=%5CDelta%20%20V%20%20%3D%20%20177.2%20%5C%20V)