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

Answer:

The value  is   $F_B = 1.074 *10^{-11} \ N$

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  $F_C[tex] is centripetal force mathematically represented as [tex]F_C = \frac{m * v^2}{r}$

So

        $F_B = F_C = \frac{m * v^2}{r}$

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

=>   $F_B = 1.074 *10^{-11} \ N$