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

v = 1.3 10⁸ m / s

Step-by-step 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