Question

A proton and a deuteron with the same initial kinetic energy enter a magnetic field in a direction perpendicular to the direction of the field. The ratio of the radii of the circular trajectories described by them is

• A. $1:2$
  
• B. $1:1$
  
• C. $1:\sqrt{2}$
• D. $1:4$
  

Answer 1

Answer: (C)

$1:\sqrt{2}$

Let the mass of proton be $m$, then the mass of deutron will be $2m$.

Also we know that force in magnetic field is $F=qvB$.

And in the case of circular motion, we have $\frac{m{v}^2}{r}=qvB$.

On further solving we have radius of circular path $r=\frac{mv}{qB}-\left(i\right)$.

Now let $v_1$ and $v_2$ be the respective velocities of the particles.

Also given that $\frac{m{v}_1^2}{2}=K$ for proton and for deutron we have $\frac{2m{v}_2^2}{2}=K$.

$⟹$ $v_1=\sqrt{\frac{2K}{m}}$ and $v_2=\sqrt{\frac{K}{m}}$

We have $r_1=\frac{m}{qB}\frac{\sqrt{2K}}{\sqrt{m}}$ and $r_2=\frac{2m}{qB}\frac{\sqrt{K}}{\sqrt{m}}$

Dividing both we have $\frac{r_1}{r_2}=\frac{1}{\sqrt{2}}$.