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 :
Solution
<p>$\therefore$ $r = {{mv} \over {qB}} = {{\sqrt {2m(KE)} } \over {qB}}$</p>
<p>$$ \Rightarrow {r_1}:{r_2}:{r_3} = {{\sqrt {{m_1}} } \over {{q_1}}}:{{\sqrt {{m_2}} } \over {{q_2}}}:{{\sqrt {{m_3}} } \over {{q_3}}}$$</p>
<p>$= {{\sqrt 1 } \over 1}:{{\sqrt 2 } \over 1}:{{\sqrt 4 } \over 2}$</p>
<p>$= 1:\sqrt 2 :1$</p>
About this question
Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law
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