When a charged particle moving with velocity $\vec{V}$ is subjected to a magnetic field of induction $\vec{B}$,the force on it is non-zero. This implies that the:

  • A
    Angle between $\vec{V}$ and $\vec{B}$ is necessarily $90^{\circ}$.
  • B
    Angle between $\vec{V}$ and $\vec{B}$ can have any value other than $90^{\circ}$.
  • C
    Angle between $\vec{V}$ and $\vec{B}$ can have any value other than $0^{\circ}$ and $180^{\circ}$.
  • D
    Angle between $\vec{V}$ and $\vec{B}$ is either $0^{\circ}$ or $180^{\circ}$.

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$A$ charged particle of specific charge $\alpha$ is released from the origin at time $t = 0$ with velocity $\vec{V} = V_o \hat{i} + V_o \hat{j}$ in a magnetic field $\vec{B} = B_o \hat{i}$. The coordinates of the particle at time $t = \frac{\pi}{B_o \alpha}$ are (specific charge $\alpha = q/m$):

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$A$ positive charge $q$ of mass $m$ is moving along the $+x$ axis with velocity $v$. We wish to apply a uniform magnetic field $B$ for time $\Delta t$ so that the charge reverses its direction,crossing the $y$ axis at a distance $d$. Then:

Derive the expression for the magnetic force acting on a moving charge $q$ with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$.

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