Magnetic Effects & Magnetism

JEE Main level, single correct, medium

124 of 276

Question 124 of 276

Two particles XX and YY having equal charges are being accelerated through the same potential difference. Thereafter, they enter normally in a region of uniform magnetic field and describes circular paths of radii R1{R}_{1} and R2{R}_{2} respectively. The mass ratio of XX and YY is :

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Answer: (B) (R1R2)2{\left(\dfrac{{R}_{1}}{{R}_{2}}\right)}^{2}

Energy gained due to potential difference by charged particle:
KE=qVKE=qV
Relation between momentum & kinetic energy:
p=2m(KE)p=\sqrt{2m\left(KE\right)}
Now, radius of the circular path under magnetic field will be given by,
R=mvqB=pqB=2m(KE)qB=2mqVqBR=\dfrac{mv}{qB}=\dfrac{p}{qB}=\dfrac{\sqrt{2m(KE)}}{qB}=\dfrac{\sqrt{2mqV}}{qB}
Therefore,
R∝m⇒m∝R2R\propto \sqrt{m} \Rightarrow m\propto {R}^{2}
⇒m1m2=(R1R2)2\Rightarrow \dfrac{{m}_{1}}{{m}_{2}}={\left(\dfrac{{R}_{1}}{{R}_{2}}\right)}^{2}

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