Force on a charged particle moving in a magnetic field

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A proton and an α-particle after being accelerated through the same potential difference enter a region of uniform magnetic field and describe circular paths of radii $R_{p} and R_{α} Then R_{p} : R_{α}$ is equal to

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Solution

$\frac{1}{2} {mv}^{2} = QV \Rightarrow v = \sqrt{\frac{2 QV}{m}} Angular velocity, ω = \frac{QB}{m} \Rightarrow r = \frac{v}{ω} = \frac{m}{QB} . \sqrt{\frac{2 QV}{m}} = \frac{1}{B} \sqrt{\frac{2 mV}{Q}} ∴ \frac{r_{p}}{r_{α}} = \sqrt{\frac{m_{p}}{m_{α}}} . \sqrt{\frac{Q_{α}}{Q_{p}}} = \sqrt{\frac{1}{4}} . \sqrt{2} = \frac{1}{\sqrt{2}}$

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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 after being accelerated through the same potential difference enter a region of uniform magnetic field and describe circular paths of radii Rp and Rα Then Rp : Rα is equal to

12mv2=QV⇒v=2QVm Angular velocity, ω=QBm ⇒r=vω=mQB.2QVm=1B2mVQ ∴rprα=mpmα.QαQp=14.2=12

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A proton and an α-particle after being accelerated through the same potential difference enter a region of uniform magnetic field and describe circular paths of radii $R_{p} and R_{α} Then R_{p} : R_{α}$ is equal to