An Alpha particle is accelerated from rest through a potential difference of 800V. The particle then enters a region of uniform magnetic filed B = 0.5T, which is perpendicular to the direction of motion. Find the radius of the new trajectory. Answer to three decimals. m? = 6.68 × 10?²? kg & q? = +2e

          An Alpha particle is accelerated from rest through a potential difference of 800V. The particle then enters a region of uniform magnetic filed B = 0.5T, which is perpendicular to the direction of motion. Find the radius of the new trajectory. Answer to three decimals.

m? = 6.68 × 10?²? kg & q? = +2e

An Alpha particle is accelerated from rest through a potential difference of 800V. The particle then enters a region of uniform magnetic filed B = 0.5T, which is perpendicular to the direction of motion. Find the radius of the new trajectory. Answer to three decimals.

m? = 6.68 × 10?²? kg q? = +2e

Added by Hugo W.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Hunza Gilgit

06/15/2022

Step 1

563 \times 10^{-16} \] \[ v = \sqrt{\frac{2 \times 2.563 \times 10^{-16}}{6.68 \times 10^{-27}}} \] \[ v = 2.8 \times 10^5 \, m/s \] Show more…

Show all steps

Thanks for your feedback!

An alpha particle is accelerated from rest through a potential difference of 800V. The particle then enters a region of uniform magnetic field B = 0.5T, which is perpendicular to the direction of motion. Find the radius of the new trajectory. Answer to three decimals: ma = 6.68 x 10^-27 kg & Qa = +2e