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Chapter 37: Q. 35 (page 1083)

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

Short Answer

Expert verified

The particle's charge-to-mass ratio is found to be qm=vBr=1.5脳1060.1250脳0.12525=9.58脳107C/kg

Step by step solution

01

Given Information

The velocity of the particle is given by
v=EB=1875000.1250=1.5脳106m/s

02

Calculation

Radlus of curvature is r=25.05/2=12.525rmcm=0.12525m.

Now charge-to-mass ratio is given by qm=vBr=1.5脳1060.1250脳0.12525=9.58脳107C/kg

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