Please work all the parts to the question and have legiable handwritting. Assume
ID: 1436110 • Letter: P
Question
Please work all the parts to the question and have legiable handwritting.
Assume that the region to the right of a certain vertical plane contains a vertical magnetic field of magnitude 1.41 mT and that the field is zero in the region to the left of the plane. An electron, originally traveling perpendicular to the boundary plane, passes into the region of the field.
(a) Determine the time interval (t) required for the electron to leave the "field-filled" region, noting that its path is a semicircle.
s
(b) Find the kinetic energy of the electron (K) if the maximum depth of penetration into the field is 2.00 cm.
eV
Explanation / Answer
Begin by relating magnetic force to force:
magnetic force = q*v*B (*where q=charge, v=velocity, B=magnetic field)
force = mass*acceleration = (m*v^2) / r (*where m=mass, r=radius)
set the two equations equal and solve for velocity:
q*v*B = q*v*B
so:
v = (q*B*r) / m
Now we know that velocity is equal to distance divided by time:
v = d/t and so from this we can solve for t:
t = d/v (where d=distance of a semicircle, which is pi*r )
Subsituting in the variables for d and v we get:
t = (pi*r) / ((q*B*r) / m)
after simplifiying:
t = (pi*m) / (q*B)
where m = 9.109*10^-31 kg
q = 1.602*10^-19 C
B = 1.41 mT = 1.41*10^-3 T
plugging these values back in for t we get:
t = (pi *(9.109*10^-31 kg)) / ((1.602*10^-19 C) * (1.41*10^-3 T)) = 1.2668 *10^ -8 s
t = 1.2668 *10^ -8 s
Part B:
kinetic energy = (1/2)*m*v^2
we can substitute the equation for v and get:
K = (1/2)*m*((q*B*r) / m)^2
further pluggin in the values we get:
K = (1/2)*(9.109*10^-31 kg)*(((1.602*10^-19 C)*(1.41*10^-3 T)*(0.02 m)) / (9.109*10^-31 kg))^2
solving we get:
K =1.120*10^-17 V however, we want our answer in eV, (electron Volts), and so we divide our answer by the charge of an electron:
K =1.120*10^-17 V / (1.602*10^-19 C) = 6.99 eV
K = 6.99 eV