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A light beam travelling in the x- direction is described by the electric field `E_y(300V (m^-1) sin omega (t-(x//c))`. An electron is constrained to move along the y-direction with a speed`(2.0 xx (10^7) m(s^-1))`. Find the maximum electric force and the maximum magnetic force on the electron. |
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Answer» Maximum Electric Force maximum electric field, `E_0=300 V//m` maximum electricfoce `F=qE_0` `= (1.6xx10^-19)(300)` `= 4.8xx10^-17 N` Maximum Magnetic Force From the equation, ` c=E_0/B_0` Maximum magnetic field, or `B_0 = 300/(3.0xx10^8)=10^-6T` `:.` maximum magnetic force = `B_0qupsilon` Substituting the values,we have maximum magnetic force `(10^-6)(1.6xx10^-19)(2.0xx10^7)` ` 3.2xx 10^-18 N`. |
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