Question

A. An α-particle has a charge of +2e and a mass of 6.64 × 10^-27 kg. It is accelerated from rest through a potential difference that has a value of 1.67 × 10⁶ V and then enters a uniform magnetic field whose magnitude is 2.49 T. The α-particle moves perpendicular to the magnetic field at all times. What is (a) the speed of the α-particle, (b) the magnitude of the magnetic force on it, and (c) the radius of its circular path?

B. A particle that has an 6.9-μC charge moves with a velocity of magnitude 3 × 10⁵ m/s along the +x axis. It experiences no magnetic force, although there is a magnetic field present. The maximum possible magnetic force that the charge with the given speed could experience has a magnitude of 0.410 N. Find the magnitude and direction of the magnetic field. Note that there are two possible answers for the direction of the field.

Three answers: Magnitude B and direction of magnetic field = Number, Unit, direction

C.

A charged particle enters a uniform magnetic field and follows the circular path shown in the drawing. The particle's speed is 171 m/s, the magnitude of the magnetic field is 0.103 T, and the radius of the path is 861 m. Determine the mass of the particle, given that its charge has a magnitude of 6.46 × 10^-4 C.

it (out of paper) 4

D. Two isotopes of carbon, carbon-12 and carbon-13, have masses of 19.93 × 10^-27 kg and 21.59 × 10^-27 kg, respectively. These two isotopes are singly ionized (+e) and each is given a speed of 3.111 × 10⁵ m/s. The ions then enter the bending region of a mass spectrometer where the magnetic field is 0.6165 T. Determine the spatial separation between the two isotopes after they have traveled through a half-circle. Use 1.602 × 10^-19 C as the elementary charge and give your answer to four significant figures.

Answer 1

(A) charge=2×1.​​​​​​b))6×10^-19= 3.2×10^-19C

Mass=6.64×10^-27 kg

Potential difference=1.67×10⁶ volt,. B=2.49 T

(a) qV = (1/2)× mv^{​​​​​2}​

v= (2qV/m)^1/2

v=(2×3.2×10^-19/6.64×10^-27)^1/2

v=9.816×10⁷ m/s

(b) Force= qvB

F=3.2×10^-19×9.816×10⁷×2.49

F=78.2138×10^-12 N

(c) R= mv/qB

R= (6.64×10^-27×9.816×10⁷)/3.2×10^-19×2.49

R=8.18×10

R=81.8 m

Thank you