Force on a charged particle moving in a magnetic field

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A proton and an $α - particle,$ accelerated through the same potential difference, enter a region of uniform magnetic field normally. If the radius of the proton orbit is 10 cm, then that of $α -$ orbit is

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Solution

$\begin{array}{l} \frac{1}{2} {mu}^{2} = Q . V \Rightarrow u = \sqrt{\frac{2 QV}{m}}, r = \frac{u}{ω} = \sqrt{\frac{2 QV}{m}} \times \frac{m}{QB} = \frac{1}{B} . \sqrt{\frac{2 Vm}{Q}} \\ ∴ \frac{r_{\propto}}{r_{P}} = \sqrt{\frac{m_{α}}{m_{p}}} . \sqrt{\frac{Q_{P}}{Q_{α}}} = \sqrt{4} . \frac{1}{\sqrt{2}} = \sqrt{2} \Rightarrow r_{\propto} = 10 \sqrt{2} cm \end{array}$

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Force on a charged particle moving in a magnetic field

Remember concepts with our Masterclasses.

A proton and an α−particle, accelerated through the same potential difference, enter a region of uniform magnetic field normally. If the radius of the proton orbit is 10 cm, then that of α-orbit is

12 mu2=Q.V ⇒ u=2QVm, r=uω=2QVm × mQB=1B . 2VmQ∴ r∝rP=mαmp . QPQα=4 . 12=2 ⇒ r∝=102 cm

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A proton and an $α - particle,$ accelerated through the same potential difference, enter a region of uniform magnetic field normally. If the radius of the proton orbit is 10 cm, then that of $α -$ orbit is