Question

A proton and an alpha particle follow the same circular path in a transverse magnetic field. Their kinetic energy is in the ratio __________.

Answer 1

Hint: Magnetic field is the space around the magnet where the effect of magnet can be felt by another magnet or iron piece. Magnetic fields can also be produced by a moving charge whose intensity can be determined by the velocity and magnitude of charge. The S.I unit of magnetic field is Tesla (T) whereas the C.G.S unit is Gauss (G).

Formula used:
$r=\dfrac{mv}{Bq},\ K.E = \dfrac12 mv^2$

Complete answer:
We know; mass of the alpha particle is four times the mass of the proton i.e. $m_{\alpha} = 4\times m_p$. As it is given that both particles are following the same circular path, so the radius of both will be equal. Also, charge of an alpha particle is ‘+2’ whereas that of a proton is ‘+1’, so $q_{\alpha} = 2 q_p$. Now, as radius of a charge ‘q’ having mass ‘m’ moving in a magnetic field of intensity ‘B’ with a velocity ‘v’ is $r=\dfrac{mv}{Bq}$, thus we have;
$v = \dfrac{Brq}{m}$
Thus $K.E. = \dfrac 12mv^2 = \dfrac 12 m\left( \dfrac{Brq}{m} \right)^2= \dfrac{B^2 r^2 q^2}{2m}$
Now, we can directly see that $K.E. \propto \dfrac{q^2}{m}$
Thus $\dfrac{K.E_{\alpha}}{K.E_{p}} = \left( \dfrac{q_{\alpha}^2}{q_p^2} \right) \left( \dfrac{m_p}{m_{\alpha}}\right)$
Putting: $\dfrac{q_{\alpha}}{q_p} = 2\ and\ \dfrac{m_p}{m_{\alpha}} = \dfrac{1}{4}$; we get,
$\dfrac{K.E_{\alpha}}{K.E_{p}} = \left( \dfrac{2}{1} \right)^2 \left( \dfrac{1}{4 }\right) = \dfrac{4}{4} = 1$
Which suggests the Kinetic energy of both the particles is the same.
Hence the ratio of kinetic energy of an alpha particle and a proton is 1.

Note:
Since the moving charge produces a magnetic field, hence if an external magnetic field is applied, the moving charge will interact with the field and hence it experiences a force or magnetic force. This force is given by $F = q(\vec v\times \vec B)$. This relation of force suggests that the direction of force is perpendicular to the velocity of charge as $\vec v\times \vec B$ is a cross product means the resultant will be perpendicular to both velocity and magnetic field. Thus force is perpendicular to velocity. And hence the path of the charge will be circular.