Question

An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths $187,500\mathrm{V}/\mathrm{m}$and $0.1250\mathrm{T}$, respectively. The particle passes out of the electric field, but the magnetic field continues, and the particle makes a semicircle of diameter $25.05\mathrm{cm}$. What is the particle's charge-to-mass ratio? Can you identify the particle?

Short answer

The particle's charge-to-mass ratio is found to be $\frac{q}{m}=\frac{v}{Br}=\frac{1.5\times {10}^6}{0.1250\times 0.12525}=9.58\times {10}^7\mathrm{C}/\mathrm{kg}$

Step-by-step solution

Given Information

The velocity of the particle is given by
$v=\frac{E}{B}=\frac{187500}{0.1250}=1.5\times {10}^6\mathrm{m}/\mathrm{s}$

Calculation

Radlus of curvature is $r=25.05/2=12.525\mathrm{rmcm}=0.12525\mathrm{m}$.

Now charge-to-mass ratio is given by $\frac{q}{m}=\frac{v}{Br}=\frac{1.5\times {10}^6}{0.1250\times 0.12525}=9.58\times {10}^7\mathrm{C}/\mathrm{kg}$