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

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

Class 12 Physics Moving Charges and Magnetism Motion of Charged Particle In Magnetic Field

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$A$ 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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A
$5.5 \times 10^{-6} \,s$
B
$7.0 \times 10^{-7} \,s$
C
$2.5 \times 10^{-6} \,s$
D
$3.5 \times 10^{-7} \,s$

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Moving Charges and Magnetism

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Motion of Charged Particle In Magnetic Field

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$A$ 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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