JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 27)
A proton and an $$\alpha $$-particle (with their masses in the ratio of 1 : 4 and charges in the ratio of 1 : 2) are accelerated from rest through a potential difference V. If a uniform magnetic field (B) is set up perpendicular to their velocities, the ratio of the radii rp : r$$\alpha $$ of the circular paths described by them will be ;
$$1:\sqrt 3 $$
1 : 3
$$1:\sqrt 2 $$
1 : 2
Explanation
KE = q$$\Delta $$V
r = $${{\sqrt {2mq\Delta V} } \over {qB}}$$
r $$ \propto $$ $$\sqrt {{m \over q}} $$
$${{{r_p}} \over {{r_ \propto }}}$$ = $${1 \over {\sqrt 2 }}$$
r = $${{\sqrt {2mq\Delta V} } \over {qB}}$$
r $$ \propto $$ $$\sqrt {{m \over q}} $$
$${{{r_p}} \over {{r_ \propto }}}$$ = $${1 \over {\sqrt 2 }}$$
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