Magnetic Effects & Magnetism

JEE Main level, single correct, medium

160 of 276

Question 160 of 276

A proton and an α\alpha- particle (with their masses in the ratio of 1:41:4 and charges in the ratio of 1:21:2 ) are accelerated from rest through a potential difference V.V. If a uniform magnetic field (B)\left(B\right) is set up perpendicular to their velocities, the ratio of the radii rp:rα{r}_{p}:{r}_{\alpha } of the circular paths described by them will be:

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Answer: (B) 1:21:\sqrt{2}

The kinetic energy of a charge qq of mass mm in potential is given by,
12mv2=qV\dfrac{1}{2}m{v}^{2}=qV
mv=2mqVmv=\sqrt{2mqV}
Then, the radius of the charge will be,
r=2mqVqBr=\dfrac{\sqrt{2mqV}}{qB}
⇒r∝mq\Rightarrow r\propto \sqrt{\dfrac{m}{q}}
Hence, the ratio of radius of proton to α\alpha-particle,

rprα=\dfrac{{r}_{p}}{{r}_{\alpha }}=14×21=12\sqrt{\dfrac{1}{4}\times \dfrac{2}{1}}=\dfrac{1}{\sqrt{2}}.

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