An electron,moving in a uniform magnetic field of induction of intensity $\vec{B}$,has its radius directly proportional to

$A$ charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle will be

$A$ particle of mass $m$ and charge $q$ is thrown from the origin at $t = 0$ with velocity $\vec{v} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ units in a region with a uniform magnetic field $\vec{B} = 2\hat{i}$ units. After time $t = \frac{\pi m}{qB}$,an electric field $\vec{E}$ is switched on such that the particle moves in a straight line with constant speed. $\vec{E}$ may be:

$A$ proton accelerated by a potential difference of $500 \; kV$ moves through a transverse magnetic field of $0.51 \; T$ as shown in the figure. The angle $\theta$ through which the proton deviates from the initial direction of its motion is......$^o$

$A$ 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

$A$ proton is projected with a speed of $2 \times 10^6 \ m/s$ at an angle of $60^{\circ}$ to the $x$-axis. If a uniform magnetic field of $0.104 \ T$ is applied along the $y$-axis,the path of the proton is: