A proton and an $\alpha$-particle, having kinetic energies Kp and K$\alpha$ respectively, enter into a magnetic field at right angles.
The ratio of the radii of trajectory of proton to that of $\alpha$-particle is 2 : 1. The ratio of Kp : K$\alpha$ is :
Solution
Radius, $r = {{mv} \over {qB}} = {{\sqrt {2mK} } \over {qB}}$<br><br>$\Rightarrow K = {{{r^2}{q^2}{B^2}} \over {2m}}$<br><br>$\therefore$ $${{{K_p}} \over {{K_\alpha }}} = {\left( {{{{r_p}} \over {{r_\alpha }}}} \right)^2} \times {\left( {{{{q_p}} \over {{q_\alpha }}}} \right)^2} \times {{{m_\alpha }} \over {{m_p}}}$$<br><br>$$ = {\left( {{2 \over 1}} \right)^2} \times {\left( {{1 \over 2}} \right)^2} \times 4$$<br><br>$= 4$
About this question
Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law
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