JEE MAIN - Physics (2022 - 24th June Evening Shift - No. 8)
A proton, a deutron and an $$\alpha$$-particle with same kinetic energy enter into a uniform magnetic field at right angle to magnetic field. The ratio of the radii of their respective circular paths is :
1 : $$\sqrt 2 $$ : $$\sqrt 2 $$
1 : 1 : $$\sqrt 2 $$
$$\sqrt 2 $$ : 1 : 1
1 : $$\sqrt 2 $$ : 1
Explanation
$$\therefore$$ $$r = {{mv} \over {qB}} = {{\sqrt {2m(KE)} } \over {qB}}$$
$$ \Rightarrow {r_1}:{r_2}:{r_3} = {{\sqrt {{m_1}} } \over {{q_1}}}:{{\sqrt {{m_2}} } \over {{q_2}}}:{{\sqrt {{m_3}} } \over {{q_3}}}$$
$$ = {{\sqrt 1 } \over 1}:{{\sqrt 2 } \over 1}:{{\sqrt 4 } \over 2}$$
$$ = 1:\sqrt 2 :1$$
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