Question: Q. 10. (i) State the condition under which a charged particle moving with velocity $v$ goes undeflected in a magnetic field $B$.
(ii) An electron, after being accelerated through a potential difference of $10^{4} \mathrm{~V}$, enters a uniform magnetic field of $0.04 \mathrm{~T}$, perpendicular to its direction of motion. Calculate the radius of curvature of its trajectory. U [Foreign I 2017]
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Solution:
Ans. (i) Condition for charge going undeflected 1
(ii) Formula for radius $1 / 2$
Calculation of radius $1 \frac{1}{2}$
(i) The force experienced, $\vec{F}=q(\vec{v} \times \vec{B}) \quad 1 / 2$
The charge will go undeflected when $\vec{v}$ is parallel or antiparallel to $\vec{B}$
$\because \quad \vec{F}=0$.
[Alternatively,
If $\vec{v}$ makes an angle of $0^{\circ}$ or $180^{\circ}$ with $\vec{B}$ ]
(ii) The $K E$ of electron
