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(B) The radius of the circular path of the charged particle in the magnetic field is given by $R = \frac{mV}{qB}$.For the particle to enter Region $II

Vedclass · Accepted Answer

The particle enters Region $III$ only if its velocity $V < \frac{qlB}{m}$.

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(B) The radius of the circular path of the charged particle in the magnetic field is given by $R = \frac{mV}{qB}$.For the particle to enter Region $III$,the radius of the path must be greater than the width of Region $II$,i.e.,$R > l$.Substituting the expression for $R$,we get $\frac{mV}{qB} > l$,which implies $V > \frac{qlB}{m}$. Thus,option $A$ is correct and option $B$ is incorrect.The path length of the particle in Region $II$ is the arc length of the circular path. This length is maximum when the particle completes a semi-circle before exiting Region $II$,which occurs when $R = l$,or $V = \frac{qlB}{m}$. Thus,option $C$ is correct.The time spent in the magnetic field is $t = \frac{	heta}{\omega}$,where $\omega = \frac{qB}{m}$ is the angular frequency. For a particle returning to Region $I$,it traces a semi-circle,so $	heta = \pi$. Thus,$t = \frac{\pi}{\omega} = \frac{\pi m}{qB}$,which is independent of velocity $V$. Thus,option $D$ is correct.

$A$ particle of mass $m$ and charge $q$,moving with velocity $V$,enters region $II$ normal to the boundary as shown in the figure. Region $II$ has a uniform magnetic field $B$ perpendicular to the plane of the paper. The length of the region $II$ is $l$. Choose the incorrect option.

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