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 0.536 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)?

Answer 1

The value is
`F_B  = 1.074 *10^(-11) \  N`

Step-by-step explanation:

From the question we are told that

The mass of the lithium nucleus is
`m =  6.02 \  u = 6.02 * 1.66*10^(-27) =  9.9932*10^(-27)\ kg`

The radius is
`r =  0.536 \ m`

The speed of the lithium nucleus is
`v  = 0.08 c = 0.08 * 3.0*10^(8) = 2.40*10^(7) \  m/s`

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

`F_B  =  F_C`

And
`F_C[tex] is centripetal  force mathematically represented as </p><p>         [tex]F_C = (m *  v^2)/(r)`

So

`F_B = F_C = (m *  v^2)/(r)`

=>
`F_B = F_C = (9.9932*10^(-27) * [2.40*10^(7)]^2)/( 0.536)`

=>
`F_B  = 1.074 *10^(-11) \  N`