An electron is moving along the positive X-ax | vedclass

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

Question

(A) The force on a moving charge in a magnetic field is given by the Lorentz force formula: $\vec{F} = q(\vec{v} 	imes \vec{B})$.For the electron to

Vedclass · Accepted Answer

$Y$-axis

Vedclass · Answer

(A) The force on a moving charge in a magnetic field is given by the Lorentz force formula: $\vec{F} = q(\vec{v} 	imes \vec{B})$.For the electron to reverse its direction,it must follow a semi-circular path in a plane perpendicular to the magnetic field.If the electron is moving along the positive $X$-axis (velocity $\vec{v} = v\hat{i}$),and we apply a magnetic field along the $Y$-axis $(\vec{B} = B\hat{j})$,the force will be $\vec{F} = -e(v\hat{i} 	imes B\hat{j}) = -evB\hat{k}$.This force acts in the $Z$-direction,causing the electron to move in a semi-circle in the $X-Z$ plane.After completing a semi-circle,the velocity vector will point in the negative $X$-direction $(-v\hat{i})$,effectively reversing the electron's direction.Thus,applying the magnetic field along the $Y$-axis (or $Z$-axis) will achieve the desired result.

An electron is moving along the positive $X$-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative $X$-axis. This can be done by applying the magnetic field along

Similar Questions

The direction of the magnetic force on the electron as shown in the diagram is along:

$A$ charged particle having charge $q$ is moving perpendicularly to a uniform magnetic field with linear speed $v$ on a circular path of radius $R$. The periodic time of revolution of the particle . . . . . . .

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

Bohr model is applied to a particle of mass $m$ and charge $q$ moving in a plane under the influence of a transverse magnetic field $B$. The energy of the charged particle in the $n^{\text{th}}$ level will be $[h = \text{Planck's constant}]$

$A$ proton and a deuteron,both having the same kinetic energy,enter perpendicularly into a uniform magnetic field $B$. For the motion of the proton and deuteron on circular paths of radii ${R_p}$ and ${R_d}$ respectively,the correct statement is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module