The distance moved by a charged particle alon | vedclass

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

Question

(A) When a charged particle moves in a magnetic field with a velocity component $v_{\perp}$ perpendicular to the field and $v_{\parallel}$ parallel to

Vedclass · Accepted Answer

$\frac{2 \pi m v}{q B}$

Vedclass · Answer

(A) When a charged particle moves in a magnetic field with a velocity component $v_{\perp}$ perpendicular to the field and $v_{\parallel}$ parallel to the field,it follows a helical path.The time period $T$ for one complete rotation is determined by the perpendicular component of velocity:$T = \frac{2 \pi m}{q B}$The distance moved along the magnetic field in one rotation is called the pitch $(p)$.$p = v_{\parallel} 	imes T$Substituting the value of $T$:$p = v_{\parallel} 	imes \frac{2 \pi m}{q B}$If the total velocity $v$ is considered as the parallel component,the distance is $\frac{2 \pi m v}{q B}$.

The distance moved by a charged particle along the magnetic field (the component of velocity is parallel to the magnetic field) in one rotation is given by ($m$ - mass of the particle,$v$ - velocity of the particle,$q$ - charge of the particle,$B$ - magnetic field).

Similar Questions

$A$ beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched off,and the same magnetic field is maintained,the electrons move

An electron is projected along the axis of a circular conductor carrying current $I$. The electron will experience:

$A$ proton of mass $m$ and charge $+e$ is moving in a circular orbit in a magnetic field with energy $1\, MeV$. What should be the energy of $\alpha$-particle (mass = $4m$ and charge = $+2e$) so that it can revolve in the path of the same radius?

An electron and a proton having same momenta enter perpendicular into a magnetic field then

Vedclass Test Series

Exam Paper Generator

Online Exam Module