Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

138 of 276

Question 138 of 276

A deuteron and a proton moving with equal kinetic energy enter into to a uniform magnetic field at right angle to the field. If rd{r}_{d} and rp{r}_{p} are the radii of their circular paths respectively, then the ratio rdrp\dfrac{{r}_{d}}{{r}_{p}} will be x:1\sqrt{x}:1 where xx is _____ .

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Answer: 2

As we know, radius of circular path followed by charged particle under perpendicular magnetic field is given by,
R=mvqBR=\dfrac{mv}{qB} and KE=12mv2KE=\dfrac{1}{2}m{v}^{2}
Therefore,
⇒R=2m(KE)qB\Rightarrow R=\dfrac{\sqrt{2m\left(KE\right)}}{qB} Then, R∝mqR\propto \dfrac{\sqrt{m}}{q}
⇒RdRp=mdmp×qpqd=21×11=2:1\Rightarrow \dfrac{{R}_{d}}{{R}_{p}}=\sqrt{\dfrac{{m}_{d}}{{m}_{p}}}\times \dfrac{{q}_{p}}{{q}_{d}}=\sqrt{\dfrac{2}{1}}\times \dfrac{1}{1}=\sqrt{2}:1
Thus, x=2x=2.

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