Question

In a particle accelerator, a magnetic force keeps a lithium nucleus (mass 6.02 u) traveling in a circular path with a radius of 0.450 m. The lithium nucleus moves at a speed of 11.00% of the speed of light. What is the magnitude of the magnetic force on the lithium nucleus (in N)?

Answer 1

Answer:

F = 2.42 × 10⁻¹¹N

Explanation:

given,

mass of lithium nucleus = 6.02 u

radius of the circular path = 0.450 m

speed of the lithium nucleus = 11% of speed of light

                                                = 0.11 × 3 × 10⁸

                                                = 33 × 10⁶ m/s

magnetic force on the lithium nucleus =

                = $\dfrac{mv^2}{r}$

                = $\dfrac{6.02\times 1.66\times 10^{-27}\times (33\times 10^6)^2}{0.45}$

              F = 2.42 × 10⁻¹¹N

so, magnetic force of the lithium nucleus is 2.42 × 10⁻¹¹N