$A$ negatively charged particle projected towards east is deflected towards north by a magnetic field. The direction of the magnetic field is

$A$ particle of charge per unit mass $\alpha$ is released from the origin with a velocity $\vec{v} = v_0 \hat{i}$ in a uniform magnetic field $\vec{B} = -B_0 \hat{k}$. If the particle passes through $(0, y, 0)$,then $y$ is equal to

$A$ particle of charge $16 \times 10^{-16} \, C$ moving with velocity $10 \, ms^{-1}$ along the $x$-axis enters a region where a magnetic field of induction $\vec{B}$ is along the $y$-axis and an electric field of magnitude $10^4 \, Vm^{-1}$ is along the negative $z$-axis. If the charged particle continues moving along the $x$-axis,the magnitude of $\vec{B}$ is:

If a proton enters perpendicularly into a magnetic field with velocity $v$,the time period of revolution is $T$. If the proton enters with velocity $2v$,what will be the time period?

$A$ particle of mass $m$,charge $Q$,and kinetic energy $K$ enters a transverse uniform magnetic field of induction $B$. After $3$ $seconds$,the kinetic energy of the particle will be .......$K$.

$A$ charged particle moving with a velocity $\vec{v} = v_1 \hat{i} + v_2 \hat{j}$ in a magnetic field $\vec{B}$ experiences a force $\vec{F} = F_1 \hat{i} + F_2 \hat{j}$. Here $v_1, v_2, F_1, F_2$ are all constants. Then $\vec{B}$ can be