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### Related Questions

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An electron is moving along $+x$ direction with a velocity of $6 \times 10^{6} \mathrm{~ms}^{-1}$. It enters a region of uniform electric field of $300 \mathrm{~V} / \mathrm{cm}$ pointing along $+y$ direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the $x$ direction will be:

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 $\mathrm{Y}-\mathrm{Z}$ 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 charged particle carrying charge $1 \mu \mathrm{C}$ is moving with velocity $(2 \hat{i}+3 \hat{j}+4 \hat{k}) \mathrm{ms}^{-1}$. If an external magnetic field of $(5 \hat{i}+3 \hat{j}-6 \hat{k}) \times 10^{-3} \mathrm{~T}$ exists in the region where the particle is moving then the force on the particle is $\vec{F} \times 10^{-9} \mathrm{~N}$. The vector $\vec{F}$ is :

A beam of protons with speed $4 \times 10^{5} \mathrm{~ms}^{-1}$ enters a uniform magnetic field of $0.3 \mathrm{~T}$ at an angle of $60^{\circ}$ to the magnetic field. The pitch of the resulting helical path of protons is close to : (Mass of the proton $=1.67 \times 10^{-27} \mathrm{~kg}$, charge of the proton $=1.69 \times 10^{-19} \mathrm{C}$ )

The figure shows a region of length ‘ $l$ ‘ with a uniform magnetic field of $0.3 \mathrm{~T}$ in it and a proton entering the region with velocity $4 \times 10^{5} \mathrm{~ms}^{-1}$ making an angle $60^{\circ}$ with the field. If the proton completes 10 revolution by the time it cross the region shown, ‘ $l$ ‘ is close to (mass of proton $=1.67 \times 10^{-27} \mathrm{~kg}$, charge of the proton $=1.6 \times 10^{-19} \mathrm{C}$ )

Proton with kinetic energy of $1 \mathrm{MeV}$ moves from south to north. It gets an acceleration of $10^{12} \mathrm{~m} / \mathrm{s}^{2}$ by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is $1.6 \times 10^{-27} \mathrm{~kg}$ )

A particle having the same charge as of electron moves in a circular path of radius $0.5 \mathrm{~cm}$ under the influence of a magnetic field of $0.5 \mathrm{~T}$. If an electric field of $100 \mathrm{~V} / \mathrm{m}$ makes it to move in a straight path then the mass of the particle is (Given charge of electron $=1.6 \times 10^{-19} \mathrm{C}$ )

An electron, moving along the $x$-axis with an initial energy of $100 \mathrm{eV}$, enters a region of magnetic field $\vec{B}=\left(1.5 \times 10^{-3} \mathrm{~T}\right) \hat{k}$ at $\mathrm{S}$ (see figure). The field extends between $x=0$ and $x=$ $2 \mathrm{~cm}$. The electron is detected at the point Q on a screen placed $8 \mathrm{~cm}$ away from the point $\mathrm{S}$. The distance $d$ between $\mathrm{P}$ and $\mathrm{Q}$ (on the screen) is :

(Electron’s charge $=1.6 \times 10^{-19} \mathrm{C}$, mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ )

A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbits in a plane due to magnetic field perpendicular to the plane. Let $\mathrm{r}_{\mathrm{p}}, \mathrm{r}_{\mathrm{e}}$ and $\mathrm{r}_{\mathrm{He}}$ be their respective radii, then,