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Question

(C) The velocity component perpendicular to the magnetic field is $v_{\perp} = v \sin(	heta) = 10^7 	imes \sin(30^{\circ}) = 10^7 	imes 0.5 = 5 	i

Vedclass · Accepted Answer

$2.5 	imes 10^{-6} \,s$

Vedclass · Answer

(C) The velocity component perpendicular to the magnetic field is $v_{\perp} = v \sin(	heta) = 10^7 	imes \sin(30^{\circ}) = 10^7 	imes 0.5 = 5 	imes 10^6 \,m/s$.The radius of the circular path is given by $r = \frac{mv_{\perp}}{qB}$,where $m = 9.1 	imes 10^{-31} \,kg$ and $q = 1.6 	imes 10^{-19} \,C$.The time period $T$ for one revolution is given by $T = \frac{2\pi r}{v_{\perp}}$.Substituting the values: $T = \frac{2 	imes 3.14 	imes 2}{5 	imes 10^6} = \frac{12.56}{5 	imes 10^6} = 2.512 	imes 10^{-6} \,s$.Rounding to the nearest option,we get $T \approx 2.5 	imes 10^{-6} \,s$.

$A$ magnetic field is applied on an electron moving with a velocity of $10^7 \,m/s$ at an angle of $30^{\circ}$ to the magnetic field. The time period of revolution of the electron in a circular path of radius $2 \,m$ is . . . . . .

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