Question

An electron has a kinetic energy of 2.17E-17 J. It moves on a circular path that is perpendicular to a uniform magnetic field of magnitude 5.13E-5 T. Determine the radius of the path.

Answer 1

Given,
Kinetic energy = 7.20*10^ -17 Joules
Magnetic field = B = 5.20*10^ -5 Tesla
We will use the equation for Newton's second law:
(q)(v)(B) = m(v^2 / r)
In which "q" is the charge of the electron and "m" is the mass of the electron. The charge and mass of the electron have been well studied, and they should be in your physics textbook:
q = -e [because electrons are negatively charged] = -(1.60*10^ -19 Coulombs)
m = 9.11*10^ -31 kilograms
We like "Joules," "Teslas," "kilograms" "Coulombs"
Now, we are missing a variable "velocity"
We can find it with the formula you learned some time ago:
KE = (1/2)(m)(v^2)
[ KE = (1/2)(m)(v^2)]*(2 / m)
2(KE) / m = v^2
√[2(KE) / m] = √v^2
√[2(KE) / m] = velocity
substitute in values:
√[2(7.20*10^ -17) / 9.11*10^ -31]
√1.581*10^-14
1.257*10^7 meters / second = velocity ...in condensed scientific notation.
That's ridiculously fast. However, electrons cover distance so slowly in wires because they are constantly "colliding" with their neighbors.
Now, back to that original equation:
(q)(v)(B) = m(v^2 / r)
When you isolate "r" using some division and multiplication, you get this:
r = [(m)(v)] / [q(B)]
r = (9.11*10^ -31)(1.257*10^7) / (1.60*10^ -19)(5.20*10^ - 5)
r = 1.376 meters...........if I did that final calculation correctly.
I put a plus sign on the "q" because the radius cannot possibly be negative, so maybe I was not thinking right.