Motion in a Magnetic Field

A particle having charge of $1 \mathrm{C}$, mass $1 \mathrm{kg}$ and speed $1 \mathrm{m} {\mathrm{s}}^{-1}$ enters a uniform magnetic field, having magnetic induction of $1 \mathrm{T}$, at an angle $\mathrm{\theta }=\mathrm{30}°$  between velocity vector and magnetic induction. The pitch of its helical path is (in meters)

An electron gun G emits electron of energy $2 \mathrm{keV}$ travelling in the (+)ve x-direction. The electrons are required to hit the spot S where $GS=0.1 \mathrm{m}$ & the line GS makes an angle of ${\mathrm{60}}^{\mathrm{o}}$ with the x-axis, as shown in the fig. A uniform magnetic field $\overrightarrow{B}$ parallel to GS exists in the region outsides to electron gun. Find the minimum value of B needed to make the electron hit S.

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

In the cyclotron, as radius of the circular path of the charged particle increases ($\omega =$ angular velocity, $v=$ linear velocity)

In the cyclotron,as radius of the circular path of the charged particle increases($\omega$=angular velocity, $v$=linear velocity).

The expression of radius of the helical motion in magnetic field is 

Write the expression of radius of the helical motion in magnetic field.

In the cyclotron, as radius of the circular path of the charged particle increases ($\omega =$ angular velocity, $v=$ linear velocity)

A charged particle enters into a magnetic field whose direction is parallel to the velocity of the particle. Choose the incorrect statement regarding the path of the charged particle:

A particle of charge $e$ and mass $m$ moves with a velocity $v$ in a magnetic field $B$ applied perpendicular to the motion of the particle. The radius $r$ of its path in the field is

A proton, a deuteron and an $\alpha$-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If $r_p, r_d$ and $r_{\alpha }$ denote, respectively, the radii of the trajectories of these particles, then

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

A particle of charge $q$ and mass$m$, moving with a velocity $v$ along the $x$ -axis, enters the region $\left(x>0\right)$ with uniform magnetic field$B$ along the $\hat{\mathrm{k}}$ direction. The particle will penetrate in this region in the $x$ -direction up to a distance $d$ equal to

If a very high magnetic field is applied to a stationary charge then, the charge experiences no force.

A charge enters a uniform magnetic field at an angle $(0<\theta <{90}^{\circ })$. What will be the path of the charge? Also find its pitch?

A proton, a deuteron and an $\mathrm{\alpha }$ -particle with same kinetic energy enter perpendicularly in a uniform magnetic field. Then, the ratio of radii of their circular path is:

An electron of energy $1800\mathrm{ }\mathrm{e}\mathrm{V}\mathrm{ }$describes a circular path in the magnetic field of flux density $0.4\mathrm{ }\mathrm{T}$. The radius of the path is: $\left(q=1.6\times {10}^{-19}\mathrm{ }\mathrm{C},\mathrm{ }{m}_{\mathrm{e}}=9.1\times {10}^{-31}\mathrm{ }\mathrm{k}\mathrm{g}\right)$

An electron passes undeflected through perpendicular electric and magnetic fields of intensity  $3.4\times {10}^3 \mathrm{V} {\mathrm{m}}^{-1}$ and $2\times {10}^{-3} \mathrm{Wb} {\mathrm{m}}^{-2}$, respectively. Then its velocity is

An electron whose e/m is $1.76\times {10}^{-11}$ c/k enter a region where there is a uniform magnetic field of induction $2\times {10}^{-3}$ tesla with a velocity of $3\times {10}^6$ m/sec in a direction making an angle of ${45}^{\mathrm{o}}$ with the field. The pitch of its helical path in the region is

Assertion: An electric field is preferred in comparison to magnetic field for detecting the electron beam in a television picture tube.

Reason: Electric field require high voltage.