Answered

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 :

Khoso123

Answer:

[tex]r=0.017m[/tex]

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

[tex](1/2)mv^2=qV\\v^2=2qV/m\\v=\sqrt{\frac{2qV}{m} }\\[/tex]

Substitute the given values

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

So the radius can be find as:

[tex]r=\frac{mv}{qB}\\[/tex]

Substitute the values

[tex]r=\frac{(2.46*10^{-26}kg)(5.93*10^{4}m/s)}{(1.6*10^{-19}C)(0.535T)} \\r=0.017m[/tex]

Other Questions