Question
  1. State the condition under which a charged particle moving with velocity v goes undeflected in a magnetic field B.
  2. An electron, after being accelerated through a potential difference of 104 V, enters a uniform magnetic field of 0.04 T, perpendicular to its direction of motion. Calculate the radius of curvature of its trajectory.

Answer

  1. The force experienced $\vec{F}=q(\vec{v}\times\vec{B})$
The charge will go undeflected when $\vec{v}$ is parallel or

antiparallel to$\vec{B}\because\vec{F}=0$
  1. The radius of electron
$eV=\frac{1}{2}mv^2$

$\frac{mv^2}{r}=qvB$

$\therefore\text{ }r=\frac{1}{B}\sqrt{\frac{2mV}{e}}$

$=\Bigg[\sqrt{\frac{2\times9.1\times10^{-31}\times10^4}{1.6\times10^{-19}}}\times\frac{1}{0.04}\Bigg]m$

$=8.4\times10^{-3}m$

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