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A beam of light travelling along x-axis is described by the magnetic field, `E_y=(600 Vm^-1) sin omega (t-x//c)` Calculating the maximum electric and magnetic forces on a charge q=2e, moving along y-axis with a speed of `3xx10^7m//s`, where `e=1.6xx10^(-19)C`. |
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Answer» Correct Answer - `1.92xx10^-16N; 1.92xx10^(-17)N` `B_0=(E_0)/c=600/(3xx10^8)=2xx10^-6T`, which is along Z-axis, Thus maximum electric force, `F_e=qE_0=2eE_0=2xx1.6xx10^(-19)xx600` `1.92xx10^(-16)N` Maximum magnetic force, `F_m=qvB_0=2evB_0` `=2xx1.6xx10^(-19)xx3xx10^7xx2xx10^-6` `=1.92xx10^(-17)N` |
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