Initially at rest, a small copper sphere with a mass of 3.00 - 10-6 kg and a cha
ID: 1517691 • Letter: I
Question
Initially at rest, a small copper sphere with a mass of 3.00 - 10-6 kg and a charge of 5.00. 10-4C is accelerated through a 7000. -V potential difference before entering a magnetic field of magnitude 4.00 T, directed perpendicular to its velocity. What is the radius of curvature of the sphere's motion in the magnetic field? An accelerator produces alphas using an electric potential difference Vo from 1.50 MV to 9.00 MV as shown below: What is the kinetic energy of the alphas as they leave the accelerator if the electric potential difference V_o on the accelerator is 2.50 MV and they initially start from rest? An analyzing magnet supplies a uniform magnetic field which is perpendicular to the path of the alphas. Calculate the magnitude of the magnetic field required to select a 4.00 MeV alpha for a radius of curvature of 45.0 cm.Explanation / Answer
(6)
m = 3.0 * 10^-6 Kg
q = 5.0 * 10^-4 C
V = 7000 V
B = 4.0 T
So,
Kinetic Energy gained by sphere, = q*V
1/2*m*v^2 = q*V
1/2*3.0 * 10^-6 * v^2 = 5.0 * 10^-4 * 7000
v = 1527.5 m/s
Now in Magnetic Field,
Magnetic Force on Sphere = Centripetal Force
q*v*B = m*v^2/r
q*B = m*v/ r
r = (m*v) / (q*B)
r = ( 3.0 * 10^-6 * 1527.5) / (5.0 * 10^-4 * 4.0)
r = 2.29 m