# Determine the time interval required for the electron to leave the "field-filled" region, noting that the electron's path is a semicircle.

Assume the region to the right of a certain plane contains a uniform magnetic field of magnitude b = 1.04 mT and the field is zero in the region to the left of the plane as shown in the figure below. An electron, originally traveling perpendicular to the boundary plane, passes into the region of the field.

(a) Determine the time interval required for the electron to leave the "field-filled" region, noting that the electron's path is a semicircle.
s

(b) Assuming the maximum depth of penetration into the field is 2.18 cm, find the kinetic energy of the electron.
eV
asked Feb 10, 2012 in Physics

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 = 1.04 mT = 1.04*10^-3 T

plugging these values back in for t we get:

t = (pi *(9.109*10^-31 kg)) / ((1.602*10^-19 C) * (1.04*10^-3 T))  = 1.7176 *10^ -8 s

t = 1.7176 *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)*(1.04*10^-3 T)*(0.0218 m)) / (9.109*10^-31 kg))^2

solving we get:

K = 7.24*10^-10 V  however, we want our answer in eV, (electron Volts), and so we divide our answer by the charge of an electron:

K = 7.24*10^-10 V / (1.602*10^-19 C)  = 45.2 eV

K = 45.2 eV
answered Feb 19, 2012 by ~Expert~ (3,856 points)
selected Feb 19, 2012 by Joey33

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