Statement-1: The path of a charged particle m | vedclass

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

Question

(D) The magnetic force on a charged particle is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$,which has a magnitude $F = qvB \sin 	heta$,where $	he

Vedclass · Accepted Answer

Statement-$1$ is true,Statement-$2$ is true; Statement-$2$ is the correct explanation of Statement-$1$.

Vedclass · Answer

(D) The magnetic force on a charged particle is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$,which has a magnitude $F = qvB \sin 	heta$,where $	heta$ is the angle between the velocity vector $\vec{v}$ and the magnetic field vector $\vec{B}$.If $	heta = 0^\circ$ or $180^\circ$,the magnetic force $F$ becomes $0$. In this case,the particle continues to move in a straight line with constant velocity.Thus,Statement-$1$ is true.The path of the charged particle is determined by the angle $	heta$ between the velocity $\vec{v}$ and the magnetic field $\vec{B}$. If $	heta = 0^\circ$ or $180^\circ$,the path is a straight line; if $	heta = 90^\circ$,the path is circular; and for other angles,the path is helical.Therefore,Statement-$2$ is true and it correctly explains why the path can be a straight line as mentioned in Statement-$1$.

Statement-$1$: The path of a charged particle may be a straight line in a uniform magnetic field.
Statement-$2$: The path of a charged particle is decided by the angle between its velocity and the magnetic field acting on it.

Similar Questions

An ionized gas contains both positive and negative ions. If it is subjected simultaneously to an electric field along the $+x$ direction and a magnetic field along the $+z$ direction,then

An electron,a proton,a deuteron,and an $\alpha -$ particle enter a uniform magnetic field normally with the same velocity. Which particle will have the maximum rotational frequency?

$A$ particle of charge $q$ and mass $m$ is moving with a velocity $-v \hat{i} (v \neq 0)$ towards a large screen placed in the $Y-Z$ plane at a distance $d$. If there is a magnetic field $\vec{B} = B_{0} \hat{k}$,the minimum value of $v$ for which the particle will not hit the screen is

$A$ positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. The particle will

An electron moves at a right angle to a magnetic field of $1.5 \times 10^{-2} \, T$ with a speed of $6 \times 10^7 \, m/s$. If the specific charge of the electron is $1.7 \times 10^{11} \, C/kg$,the radius of the circular path will be.....$cm$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Statement-$1$: The path of a charged particle may be a straight line in a uniform magnetic field.Statement-$2$: The path of a charged particle is decided by the angle between its velocity and the magnetic field acting on it.