A charged particle is moving with velocity $v$ in a magnetic field of induction $B$. The force on the particle will be maximum when

  • A

    $v$ and $B$ are in the same direction

  • B

    $v$ and $B$ are in opposite directions

  • C

    $v$ and $B$ are perpendicular

  • D

    $v$ and $B$ are at an angle of $45^\circ $

Similar Questions

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.

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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

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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 ....

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An electron is projected with uniform velocity along the axis inside a current carrying long solenoid. Then :

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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 $56 cm$.

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