Magnetics Ques 19
- A proton, an electron and a helium nucleus, have the same energy. They are in circular orbits in a plane due to magnetic field perpendicular to the plane.
Let $r _p, r _e$ and $r _{He}$ be their respective radii, then
(2019 Main, 10 April I)
(a) $r _e<r _p=r _{He}$
(c) $r _e<r _p<r _{He}$
(b) $r _e>r _p=r _{He}$
(d) $r _e>r _p>r _{He}$
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Answer:
Correct Answer: 19.(c)
Solution:
Formula:
Magnetic Force Acting On A Moving Point Charge:
- When a moving charged particle is placed in a magnetic field B.
Then, the net magnetic force acting on it is
Here,
$ \begin{aligned} \mathbf{F} _m & =q(\mathbf{v} \times \mathbf{B}) \\ \mathbf{F} _m & =q v B \sin \theta \\ \boldsymbol{\theta} & =90^{\circ} \\ \mathbf{F} _m & =q v B \end{aligned} $
Also, due to this net force, the particle transverses a circular path, whose necessary centripetal force is being provided by $F _m$, i.e.
$\frac{m v^{2}}{r} =q v B $
$\Rightarrow \quad r =\frac{m v^{2}}{q v B}=\frac{m v}{q B} $
$\Rightarrow \quad r \propto m$
So, for electron,
$ r _e=\frac{m _e v}{e B} $
$ r _e \propto m _e $
For proton,
$ \begin{aligned} & r _p=\frac{m _p v}{e B} \\ \text { or } \quad & r _p \propto m _p \end{aligned} $
For He-particle,
$ r _{He}=\frac{4 m _p v}{2 e B}=\frac{2 m _p v}{e B} $
Clearly, $\quad r _{He}>r _p$ $ \quad\left(\because r _{He}=2 r _p\right) $
and we know that, $m _p>m _e$
$ [\because m _p \approx 10^{-27} kg] $
$[ m _e \approx 10^{-31} kg]$
$ \begin{array}{lc} \therefore & r _p>r _e \\ \Rightarrow & r _{He}>r _p>r _e \end{array} $
BETA
