A charged particle is moving with velocity $v$ in a magnetic field of induction $B$. The force on the particle will be maximum when
$v$ and $B$ are in the same direction
$v$ and $B$ are in opposite directions
$v$ and $B$ are perpendicular
$v$ and $B$ are at an angle of $45^\circ $
Given below are two statements: One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R).$
Assertion $(A)$ : In an uniform magnetic field, speed and energy remains the same for a moving charged particle.
Reason $(R)$ : Moving charged particle experiences magnetic force perpendicular to its direction of motion.
A charged particle is released from rest in a region of steady uniform electric and magnetic fields which are parallel to each other the particle will move in a
A proton and an alpha particle of the same enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the circular paths described by the alpha particle and proton is ....
An electron is projected with uniform velocity along the axis inside a current carrying long solenoid. Then :
A positive, singly ionized atom of mass number $A_M$ is accelerated from rest by the voltage $192 V$. Thereafter, it enters a rectangular region of width $w$ with magnetic field $B_0=0.1 \hat{k}$ Tesla, as shown in the figure. The ion finally hits a detector at the distance $x$ below its starting trajectory.
[Given: Mass of neutron/proton $=(5 / 3) \times 10^{-27} kg$, charge of the electron $=1.6 \times 10^{-19} C$.]
Which of the following option($s$) is(are) correct?
$(A)$ The value of $x$ for $H^{+}$ion is $4 cm$.
$(B)$ The value of $x$ for an ion with $A_M=144$ is $48 cm$.
$(C)$ For detecting ions with $1 \leq A_M \leq 196$, the minimum height $\left(x_1-x_0\right)$ of the detector is $55 cm$.
$(D)$ The minimum width $w$ of the region of the magnetic field for detecting ions with $A_M=196$ is $56 cm$.