An alpha particle moving with a certain speed | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The force on a charged particle moving in a magnetic field is given by the Lorentz force formula: $\vec{F} = q(\vec{v} 	imes \vec{B})$.Here,the v

Vedclass · Accepted Answer

a horizontal circular path with the same speed

Vedclass · Answer

(B) The force on a charged particle moving in a magnetic field is given by the Lorentz force formula: $\vec{F} = q(\vec{v} 	imes \vec{B})$.Here,the velocity $\vec{v}$ is towards the east (let this be the $+x$ direction) and the magnetic field $\vec{B}$ is directed vertically up (let this be the $+z$ direction).The direction of the force is given by the cross product of $\vec{v}$ and $\vec{B}$. Using the right-hand rule,$\hat{i} 	imes \hat{k} = -\hat{j}$.Thus,the force acts towards the south ($-y$ direction).Since the force is always perpendicular to the velocity,the particle moves in a circular path. Because the force is horizontal (in the $xy$-plane),the particle will move in a horizontal circular path.Furthermore,since the magnetic force does no work on the charged particle,its speed remains constant.

An alpha particle moving with a certain speed towards the east enters a uniform magnetic field directed vertically up. The alpha particle will then move in:

Similar Questions

$A$ particle of mass $m$ and carrying a charge $q$ enters with a velocity $v$ perpendicular to a uniform magnetic field $B$. The time period of rotation of the particle:

$A$ charged particle is released from rest in a region of uniform electric and magnetic fields,which are parallel to each other. The locus of the particle will be

$A$ stream of electrons is projected horizontally to the right. $A$ straight conductor carrying a current is supported parallel to the electron stream and above it. If the current in the conductor is from left to right,then what will be the effect on the electron stream?

An electron and a proton of equal linear momentum enter in a direction perpendicular to a uniform magnetic field. If the radii of their circular paths are $r_e$ and $r_p$ respectively,then $\frac{r_e}{r_p}$ is equal to - (mass of electron $= m_e$,mass of proton $= m_p$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module