(B) The magnetic force on a charged particle is given by $F = q(v \times B) = qvB \sin \theta$. For the proton,the velocity is perpendicular to the ma

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Proton will move in a circle with constant speed and there will be no effect on the motion of electron.

(B) The magnetic force on a charged particle is given by $F = q(v \times B) = qvB \sin \theta$. For the proton,the velocity is perpendicular to the magnetic field,so $\theta = 90^\circ$. The force $F = qvB \sin 90^\circ = qvB$ acts as a centripetal force,causing the proton to move in a circular path with constant speed. For the electron,the velocity is parallel to the magnetic field lines,so $\theta = 0^\circ$. The force $F = qvB \sin 0^\circ = 0$. Since no force acts on the electron,its motion remains unaffected (it continues to move in a straight line).

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

Easy

If a proton is projected in a direction perpendicular to a uniform magnetic field with velocity $v$ and an electron is projected along the lines of force,what will happen to the proton and electron?

A
The electron will travel along a circle with constant speed and the proton will move along a straight line.
B
Proton will move in a circle with constant speed and there will be no effect on the motion of electron.
C
There will not be any effect on the motion of electron and proton.
D
The electron and proton both will follow the path of a parabola.

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Moving Charges and Magnetism

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Motion of Charged Particle In Magnetic Field

$A$ $2 \, MeV$ proton is moving perpendicular to a uniform magnetic field of $2.5 \, T$. The force on the proton is

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$A$ monoenergetic $(18 \;keV)$ electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of $0.04 \;G$ normal to the initial direction. Estimate the up or down deflection of the beam over a distance of $30 \;cm$ $(m_{e} = 9.11 \times 10^{-31} \;kg)$.

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$A$ proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of $2 \times 10^5 \text{ ms}^{-1}$. When the electric field is switched off,the proton moves along a circular path of radius $2 \text{ cm}$. The magnitude of the electric field is $x \times 10^4 \text{ N/C}$. The value of $x$ is . . . . . . . (Take the mass of the proton $= 1.6 \times 10^{-27} \text{ kg}$ and charge $e = 1.6 \times 10^{-19} \text{ C}$)

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$A$ proton accelerated by a potential difference of $500 \ kV$ moves through a transverse magnetic field of $0.5 \ T$ as shown in the figure. The angle $\theta$,through which the proton deviates from the initial direction of its motion,is (mass of proton $= 1.6 \times 10^{-27} \ kg$):- (in $^{\circ}$)

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$A$ proton of velocity $v = (3 \hat{i} + 2 \hat{j}) \ m/s$ enters a magnetic field of induction $B = (2 \hat{j} + 3 \hat{k}) \ T$. The acceleration produced in the proton in $m/s^2$ is (Specific charge of proton $= 0.96 \times 10^8 \ C/kg$)

EasyAP EAMCET 2002

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If a proton is projected in a direction perpendicular to a uniform magnetic field with velocity $v$ and an electron is projected along the lines of force,what will happen to the proton and electron?

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$A$ $2 \, MeV$ proton is moving perpendicular to a uniform magnetic field of $2.5 \, T$. The force on the proton is

$A$ monoenergetic $(18 \;keV)$ electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of $0.04 \;G$ normal to the initial direction. Estimate the up or down deflection of the beam over a distance of $30 \;cm$ $(m_{e} = 9.11 \times 10^{-31} \;kg)$.

$A$ proton of velocity $v = (3 \hat{i} + 2 \hat{j}) \ m/s$ enters a magnetic field of induction $B = (2 \hat{j} + 3 \hat{k}) \ T$. The acceleration produced in the proton in $m/s^2$ is (Specific charge of proton $= 0.96 \times 10^8 \ C/kg$)

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