A charged particle is injected at 211 m/s into a 0.0633‑T uniform magnetic magnetic field perpendicularly to the field. The diameter of its orbit is measured and found to be 0.0389 m. What is the charge–to–mass ratio of this particle?

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cjmejiab cjmejiab · Answer 1 · 2019-11-12T06:40:18+01:00

Answer:

[tex]85.68*10^3C/kg[/tex]

Explanation:

For this problem we need the concept about Force in a Magnetic field,

For definition we know that

[tex]F=m\frac{v^2}{r}[/tex]

Where v is the velocity, m the mass and r the radius or distance between the two points.

We know as well that

[tex]F = qvB[/tex]

where q is the charge of a proton

v the velocity and B the magnetic field, then matching the two equation,

[tex]qvB=m\frac{v^2}{r}[/tex]

Re-arrange for q/m (charge to mass ratio)

[tex]\frac{q}{m} = \frac{v}{Br}[/tex]

Our values are,

[tex]v=211m/s[/tex]

[tex]B= 0.0633T[/tex]

[tex]r=0.0389m[/tex]

Substituting,

[tex]\frac{q}{m} = \frac{211}{0.0633*0.0389}[/tex]

[tex]\frac{q}{m} = 85689.8 C/kg = 85.68*10^3C/kg[/tex]