A proton with a kinetic energy of 5.00E-16 J moves perpendicular to a magnetic field of 0.289 T. What is the radius of its circular path?
To find the radius of the circular path of a proton moving perpendicular to a magnetic field, we need to apply the formula for the radius of the circular path of a charged particle in a magnetic field.
The formula is:
r = (mv) / (qB)
Where:
r is the radius of the circular path
m is the mass of the charged particle (in this case, the proton) - approximately 1.67E-27 kg
v is the velocity of the proton
q is the charge of the proton - approximately 1.6E-19 C
B is the magnetic field strength
Given that the kinetic energy (K.E.) of the proton is 5.00E-16 J, we need to convert it to the velocity (v) of the proton.
The kinetic energy of a particle can be expressed as:
K.E. = (1/2)mv^2
Rearranging the formula, we can solve for v:
v = √(2K.E. / m)
Substituting the values into the formula, we get:
v = √(2 * 5.00E-16 J / 1.67E-27 kg)
v ≈ 2.24E7 m/s
Now, we have all the values we need to calculate the radius (r). Let's substitute them into the formula:
r = (m * v) / (q * B)
r = (1.67E-27 kg * 2.24E7 m/s) / (1.6E-19 C * 0.289 T)
Calculating this expression, we find:
r ≈ 1.95E-3 meters
Therefore, the radius of the proton's circular path is approximately 1.95E-3 meters.