An $α$ -particle describes a circular path of radius $r$ in a magnetic field $B$ . The radius of the circular path described by the proton of same energy in the same magnetic field is :

\frac{r}{2}

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r

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\sqrt{2} r

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2 r

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Solution

The correct option is B $r$
For $α -$ particle,
$M = 4 m$
$q = 2 e$
So, $r = \frac{\sqrt{2 K_{e} M}}{q B}$
= $\frac{\sqrt{2 K_{e} m}}{e B}$

For proton $M = m$ ,
$q = e$
$∴ R = \frac{M v}{q B}$
= $\frac{\sqrt{2 K_{e} M}}{q B}$
= $\frac{\sqrt{2 K_{e} m}}{e B}$
$= r$

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An α -particle describes a circular path of radius r in a magnetic field B. The radius of the circular path described by the proton of same energy in the same magnetic field is :

The correct option is B rFor α− particle,M=4mq=2eSo,r=√2KeMqB=√2KemeBFor proton M=m , q=e∴R=MvqB=√2KeMqB=√2KemeB=r

An $α$ -particle describes a circular path of radius $r$ in a magnetic field $B$ . The radius of the circular path described by the proton of same energy in the same magnetic field is :

The correct option is B $r$
For $α -$ particle,
$M = 4 m$
$q = 2 e$
So, $r = \frac{\sqrt{2 K_{e} M}}{q B}$
= $\frac{\sqrt{2 K_{e} m}}{e B}$

For proton $M = m$ ,
$q = e$
$∴ R = \frac{M v}{q B}$
= $\frac{\sqrt{2 K_{e} M}}{q B}$
= $\frac{\sqrt{2 K_{e} m}}{e B}$
$= r$