Answer:
a
The speed of the particle is [tex]v = 209485.71 m/s[/tex]
b
The potential difference is [tex]\Delta V = 177.2 \ V[/tex]
Explanation:
From the question we are told that
The mass of the particle is [tex]m = 2.10 *10^{-16} kg[/tex]
The charge on the particle is [tex]q = 26.0 nC = 26.0 *10^{-9} C[/tex]
The magnitude of the magnetic field is [tex]B = 0.600 T[/tex]
The magnetic flux is [tex]\O = 15.0 \mu Wb = 15.0 *10^{-6} Wb[/tex]
The magnetic flux is mathematically represented as
[tex]\O = B *A[/tex]
Where A is the the area mathematically represented as
[tex]A = \pi r^2[/tex]
Substituting this into the equation w have
[tex]\O = B (\pi r^2 )[/tex]
Making r the subject of the formula
[tex]r = \sqrt{\frac{\O}{B \pi} }[/tex]
Substituting value
[tex]r = \sqrt{\frac{15 *10^{-6}}{3.142 * 0.6} }[/tex]
[tex]r = 2.82 *10^{-3}m[/tex]
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
[tex]F_q = F_c[/tex]
Where [tex]F_q = q B v[/tex]
and [tex]F_c = \frac{mv^2 }{r}[/tex]
Substituting this into the equation above
[tex]qBv = \frac{mv^2}{r}[/tex]
making v the subject
[tex]v = \frac{r q B}{m}[/tex]
substituting values
[tex]v = \frac{2.82 * 10^{-3} * 26 *10^{-9} 0.6}{2.10*10^{-16}}[/tex]
[tex]v = 209485.71 m/s[/tex]
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
[tex]PE = KE[/tex]
Where [tex]PE = q * \Delta V[/tex]
and [tex]KE = \frac{1}{2} mv^2[/tex]
Substituting into the equation above
[tex]q \Delta V = \frac{1}{2} mv^2[/tex]
Making the potential difference the subject
[tex]\Delta V = \frac{mv^2}{2 q}[/tex]
[tex]\Delta V = \frac{2.16 * 10^ {-16} * (209485.71)^2 }{2 * 26 *10^{-9}}[/tex]
[tex]\Delta V = 177.2 \ V[/tex]
