A particle of mass $m$ and charge $q$ is thrown from origin at $t = 0$ with velocity $2\hat{i}$ + $3\hat{j}$ + $4\hat{k}$ units in a region with uniform magnetic field $\vec B$ = $2\hat{i}$ units. After time $t =\frac{{\pi m}}{{qB}}$ , an electric field is switched on such that particle moves on a straight line with constant speed. $\vec E$ may be
$5\hat k - \,10\hat j \ units$
$-6\hat k - \,9\hat j \ units$
$-6\hat k + \,8\hat j \ units$
$6\hat k + \,8\hat j \ units$
A electron experiences a force $\left( {4.0\,\hat i + 3.0\,\hat j} \right)\times 10^{-13} N$ in a uniform magnetic field when its velocity is $2.5\,\hat k \times \,{10^7} ms^{-1}$. When the velocity is redirected and becomes $\left( {1.5\,\hat i - 2.0\,\hat j} \right) \times {10^7}$, the magnetic force of the electron is zero. The magnetic field $\vec B$ is :
A proton of energy $8\, eV$ is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be.....$eV$
A proton and an $\alpha -$ particle (with their masses in the ratio of $1 : 4$ and charges in the ratio of $1:2$ are accelerated from rest through a potential difference $V$. If a uniform magnetic field $(B)$ is set up perpendicular to their velocities, the ratio of the radii $r_p : r_{\alpha }$ of the circular paths described by them will be
A charged particle carrying charge $1\,\mu C$ is moving with velocity $(2 \hat{ i }+3 \hat{ j }+4 \hat{ k })\, ms ^{-1} .$ If an external magnetic field of $(5 \hat{ i }+3 \hat{ j }-6 \hat{ k }) \times 10^{-3}\, T$ exists in the region where the particle is moving then the force on the particle is $\overline{ F } \times 10^{-9} N$. The vector $\overrightarrow{ F }$ is :
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