Question

A singly charged positive ion has a mass of 2.46 ✕ 10−26 kg. After being accelerated through a potential difference of 270 V, the ion enters a magnetic field of 0.535 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field.

Answer 1

`r=0.017m`

Step-by-step explanation:

Given data

Mass m=2.46×10⁻²⁶kg

Voltage V=270 V

Magnetic field B=0.5353 T

To find

Radius r

Solution

By using conservation of energy to find speed of charge

`(1/2)mv^2=qV\\v^2=2qV/m\\v=\sqrt{(2qV)/(m) }\\`

Substitute the given values

`v=\sqrt{(2(1.6*10^(-19C))(270V))/((2.46*10^(-26)kg)) }\\v=5.93*10^(4)m/s`

So the radius can be find as:

`r=(mv)/(qB)\\`

Substitute the values

`r=((2.46*10^(-26)kg)(5.93*10^(4)m/s))/((1.6*10^(-19)C)(0.535T)) \\r=0.017m`