A beam of light travelling along $$x-axis$$ is described by the electric field $$EY=(600Vm$$−$$1)$$$$sin$$ω(t$$−$$xc), then the maximum magnetic force on a charge $$q=2e$$ moving along $$y-axis$$ with a speed of $$3$$×$$107m/s$$ is (where$$ $$e=1.6$$×$$10$$−$$19$$ $$C)

Question

A beam of light travelling along $$x-axis$$ is described by the electric field $$EY=(600Vm$$−$$1)$$$$sin$$ω(t$$−$$xc),  then the maximum magnetic force on a charge $$q=2e$$ moving along $$y-axis$$ with a speed of $$3$$×$$107m/s$$ is (where$$ $$e=1.6$$×$$10$$−$$19$$ $$C)

Infinity Learn · Accepted Answer

Amplitude of magnetic field is  $$B0=E0C=6003$$×$$108=2$$×$$10$$−$$6T($$ from the given equation,  $$E0=600V/m)Maximum$$ magnetic force is  $$F=qvB0=2evB0$$ ⇒$$F=2$$×$$1.6$$×$$10$$−$$19$$×$$3$$×$$107$$×$$2$$×$$10$$−$$6$$ ⇒$$F=1.92$$×$$10$$−$$17N$$

A beam of light travelling along x-axis is described by the electric field $E_{Y} = (600 V m^{- 1}) \sin ω (t - \frac{x}{c})$ , then the maximum magnetic force on a charge $q = 2 e$ moving along y-axis with a speed of $3 \times 10^{7} m / s$ is (where $e = 1.6 \times 10^{- 19} C$ )

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A beam of light travelling along x-axis is described by the electric field EY=(600Vm−1)sinω(t−xc), then the maximum magnetic force on a charge q=2e moving along y-axis with a speed of 3×107m/s is (where e=1.6×10−19 C)

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A beam of light travelling along x-axis is described by the electric field $E_{Y} = (600 V m^{- 1}) \sin ω (t - \frac{x}{c})$ , then the maximum magnetic force on a charge $q = 2 e$ moving along y-axis with a speed of $3 \times 10^{7} m / s$ is (where $e = 1.6 \times 10^{- 19} C$ )