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An electron of energy 1800 eV describes a circular path in magnetic field of flux density 0.4 T. The radius of path is `(q = 1.6 xx 10^(-19) C, m_(e)=9.1 xx 10^(-31) kg)`A. `2.58xx10^(-4)m`B. `3.58xx10^(-4)m`C. `2.58xx10^(-3)m`D. `3.58xx10^(-4)m` |
Answer» Correct Answer - B `E=(1)/(2)mv^(2) rArr v=sqrt((2E)/(m))` `r=(mv)/(Be)=(m)/(Be)sqrt((2E)/(m))=(sqrt(2mE))/(Be)` `r=(sqrt(2xx1800xx1.6xx10^(-19)xx9.1xx10^(-31)))/(1.6xx10^(-19)xx0.4)=3.58xx10^(-4)m` |
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