A charged particle enters a uniform magnetic | vedclass

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(B) The radius of the circular path of a charged particle in a uniform magnetic field is given by $r = \frac{mv}{qB}$.From the figure,the radius $r$ o

Vedclass · Accepted Answer

$1, 2$ only are correct

Vedclass · Answer

(B) The radius of the circular path of a charged particle in a uniform magnetic field is given by $r = \frac{mv}{qB}$.From the figure,the radius $r$ of the path is decreasing as the particle moves.Statement $[1]$: If the magnetic field strength $B$ increases,the radius $r$ decreases $(r \propto \frac{1}{B})$. Thus,this is a possible explanation.Statement $[2]$: As the particle moves through air,it loses kinetic energy $(K = \frac{1}{2}mv^2)$ due to collisions and ionization of air molecules. Since $r = \frac{\sqrt{2mK}}{qB}$,a decrease in kinetic energy $K$ leads to a decrease in radius $r$. Thus,this is a possible explanation.Statement $[3]$: If the particle loses charge $q$,the radius $r$ would increase $(r \propto \frac{1}{q})$,which contradicts the observed path. Therefore,this statement is incorrect.Thus,statements $[1]$ and $[2]$ are correct.

Similar Questions

In a Thomson mass spectrograph,the electric field and magnetic field are applied:

$A$ particle of charge $q$ and mass $m$ is moving along the $x-$ axis with a velocity $v,$ and enters a region of electric field $E$ and magnetic field $B$ as shown in the figures below. For which figure the net force on the charge may be zero?

Choose the correct statement.

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