Question

At a certain location, the horizontal component of the earth's magnetic field is 2.5 × 10−5 T, due north. A proton moves eastward with just the right speed for the magnetic force on it to balance its weight. Find the speed of the proton.

Answer 1

Answer:

  v = 4.10 10⁻³ m / s

Explanation:

For this exercise we will use Newton's second law where the force is magnetic

       F -W = m a

As the field is directed to the north and the proton to the east, using the rule of the right hand the force is vertical upwards, the force balances the weight the acceleration is zero

      F = W

      q v B = m g

Let's calculate the speed

      v = m g / (q B)

      v = 1,673 10⁻²⁷ 9.8 / (1.6 10⁻¹⁹ 2.5 10⁻⁵)

       v = 4.10 10⁻³ m / s