Question

Suppose the Earth's magnetic field at the equator has magnitude 0.50×10−4T and a northerly direction at all points. Part A Estimate the speed a singly ionized uranium ion (m=238u,we) would need to circle the Earth 5.0 km above the equator. Express your answer using two significant figures. vv = nothing m/s

Answer 1

Answer:

v = 1.3 10⁸  m / s

Explanation:

The magnetic force is given by

       F = q v x B

Where blacks indicate vectors

The magnetic field has a direction to the north, the fixed uranium atom on Earth has a direction to the east, so the field and velocity are perpendicular, the expression of the force remains

         F = q v B

If we use Newton's second law

       F = ma

The acceleration is centripetal

      F = m v² / r

The distance is the radius of the earth plus the distance from the surface

      r = Re + 50.103

      r = 6.37 10⁶ + 5 10³

      r = 6.3705 10⁶ m

      m v² / r = q v B

      v = q/m   B r

uranium mass

     m = 238 (1.66 10⁻²⁷) kg = 395 10⁻²⁷ kg

Let's calculate

      v = 1.6 10⁻¹⁹ 0.5 10⁻⁴ 6.3705 10⁶ /395 10⁻²⁷

      v = 1.29 10⁸  m / s