An electron has a kinetic energy of 7.00 10-17 J. It moves on a circular path that is perpendicular to a uniform magnetic field of magnitude 4.40 10-5 T. Determine the radius of the path?
you know the equation for force in a magnetic field for a moving particle. Find force.
then set that force equal to centripetal force, mv^2/r, and solve for r.
To determine the radius of the path, we can use the formula for the force on a charged particle moving through a magnetic field:
F = qvB
Where:
F is the magnetic force acting on the electron,
q is the charge of the electron, which is equal to the elementary charge (e) and has a value of 1.6 x 10^(-19) C,
v is the velocity of the electron, and
B is the magnitude of the magnetic field.
The magnetic force (F) acting on the electron provides the necessary centripetal force, so we can equate them:
F = mv^2 / r
Where:
m is the mass of the electron, which is approximately 9.11 x 10^(-31) kg,
v is the velocity of the electron, and
r is the radius of the circular path.
Since the electron's kinetic energy is given, we can relate it to its velocity using the formula:
KE = (1/2)mv^2
Given KE = 7.00 x 10^(-17) J, we can solve for v:
7.00 x 10^(-17) J = (1/2)(9.11 x 10^(-31) kg)v^2
Simplifying the equation gives:
v^2 = (2)(7.00 x 10^(-17) J) / (9.11 x 10^(-31) kg)
Now, we can solve for v by taking the square root of both sides:
v = √[(2)(7.00 x 10^(-17) J) / (9.11 x 10^(-31) kg)]
Substituting the given values, we can calculate v.
Next, we can calculate the magnetic force (F) using the formula:
F = qvB
Once we have the magnetic force, we can equate it to the centripetal force:
F = mv^2 / r
Simplifying the equation gives:
mv^2 / r = qvB
We can now rearrange the equation to solve for the radius (r):
r = mv / (qB)
Substituting the known values of m, v, q, and B, we can calculate the radius (r).