Force Acting on a Charged Particle in Magnetic Field

The magnetic field due to a straight conductor of uniform cross section of radius $a$ and carrying a steady current is represented by

 An uniform magnetic field, $\overrightarrow{\mathrm{B}}=B_0\widehat{ \mathrm{j}}$ exist in a space. A particle of mass $m$ and charge $q$ is projected towards negative $x-axis$ with speed $v$ from the a point $(d,0,0)$.The maximum value of $v$ for which the particle does not hit $y-z plane$ is

A charged particle moves in a magnetic field $\overrightarrow{\mathrm{B}}=10\hat{\mathrm{i}}$ with initial velocity $\overrightarrow{\mathrm{u}}=5\hat{\mathrm{i}}+4\hat{\mathrm{j}}.$ The path of the particle will be 

A proton is projected with a speed of  $2\times {10}^6 \mathrm{m} {\mathrm{s}}^{-1}$ at an angle ${60}^{\circ }$ towards the $x-\mathrm{axis}$. If a uniform magnetic field of  $0.1 \mathrm{T}$ is applied along the $y-\mathrm{axis}$, the path of the proton is

An $\alpha$-particle of $1 \mathrm{MeV}$ energy moves along a circular path in a uniform magnetic field. The kinetic energy of a proton in the same magnetic field for a circular path of double radius is

An electron is projected in vertically upward direction. If a uniform magnetic field acts from south to north, then it will be deflected

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 charged particle moves through a magnetic field perpendicular to its direction. Then 

The charges $1,2,3$ are moving in a uniform transverse magnetic field. Then 

A particle of charge $q$ and mass $m$ enters normally (at point $P$) in a region of magnetic field with speed $v$. It comes out normally from $Q$ after time $T$ as shown in figure. The magnetic field $B$ is present only in the region of radius $R$ and is uniform as shown. Initial and final velocities are along radial direction and they are perpendicular to each other. For this to happen, which expression is correct?

Two particles $X$ and $Y$ are having equal charges. After being accelerated through the same potential difference, they enter a region of uniform magnetic field and describe circular paths of radii $R_1$ and $R_2$_, respectively. The ratio of the mass of $X$ to that of $Y$ is,

In a region, a uniform magnetic field acts in horizontal plane towards north. If cosmic particles ($80\%$ protons) are falling vertically downwards, then they are deflected towards

An electric charge $+q$ moves with velocity $\overrightarrow{\mathrm{v}}=3\hat{\mathrm{i}}+4\hat{\mathrm{j}}+\hat{\mathrm{k}}$, in an electromagnetic field given by: $\overrightarrow{\mathrm{E}}=3\hat{\mathrm{i}}+\hat{\mathrm{j}}+2\hat{\mathrm{k}}$ and  $\overrightarrow{\mathrm{B}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-3\hat{\mathrm{k}}$. The $y-$component of the force experienced by $+q$ is

An electron is moving along $+x$ direction. To get it moving in an anticlockwise circular path in $x-y$ plane, a magnetic field is applied along

A particle of mass $0.5 \mathrm{g}$ and charge $2.5\times {10}^{-8} \mathrm{C}$is moving with velocity $6\times {10}^4 \mathrm{m} {\mathrm{s}}^{-1}$. What should be the minimum value of magnetic field acting on it so that the particle is able to move in a straight line? ($g=9.8 \mathrm{m} {\mathrm{s}}^{-2}$) 

$A$ and $B$ are two concentric circular conductors of centre $O$ and carrying current $i_1$ and $i_2$, respectively as shown in the figure. If the ratio of their radii is $1:2$ and the ratio of the flux densities at $O$ due to $A$ and $B$ is $1:3$, then the value of $\frac{i_1}{i_2}$ will be