In a particle accelerator, a magnetic force keeps a lithium nucleus (mass 6.02 u) traveling in a circular path with a radius of 0.536 m. The lithium nucleus moves at a speed of 8.00% of the speed of light. What is the magnitude of the magnetic force on the lithium nucleus (in N)?

Respuesta :

Answer:

The value  is   [tex]F_B  = 1.074 *10^{-11} \  N[/tex]

Explanation:

From the question we are told that

   The mass of the  lithium nucleus  is  [tex]m =  6.02 \  u = 6.02 * 1.66*10^{-27} =  9.9932*10^{-27}\ kg   [/tex]

    The radius is  [tex]r =  0.536 \ m[/tex]

    The speed of the lithium nucleus is [tex]v  = 0.08 c = 0.08 * 3.0*10^{8} = 2.40*10^{7} \  m/s [/tex]

   Given that the lithium nucleus is travelling in a circular path, the magnetic force will be equivalent to a centripetal force so

       [tex]F_B  =  F_C[/tex]

And  [tex]F_C[tex] is centripetal  force mathematically represented as

        [tex]F_C = \frac{m *  v^2}{r}[/tex]

So

        [tex]F_B = F_C = \frac{m *  v^2}{r}[/tex]

=>    [tex]F_B = F_C = \frac{9.9932*10^{-27} * [2.40*10^{7}]^2}{ 0.536}[/tex]

=>   [tex]F_B  = 1.074 *10^{-11} \  N[/tex]