A charged particle sits at the centre of a circular region containing a magnetic field, $\vec{B}$ . The field $\vec{B} = \vec{B} (r)$ depends only on radial position r, and is perpendicular to the plane of circle. The total magnetic flux through the circle is zero. The particle is given a kick so that it leaves the circular region, then at the instant it leaves, its velocity makes an angle $n \times \frac{π}{4}$ with outward radial direction. Find n.

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Detailed Solution

$m (2 \frac{d r}{d t} \times \frac{d θ}{d t} + r \frac{d w}{d t}) = q \frac{d r}{d t} B \Rightarrow 2 m r \frac{d r}{d t} \times \frac{d θ}{d t} + r^{2} \frac{d w}{d t} = q r B \frac{d r}{d t}$

$\Rightarrow m \frac{d}{d t} (r^{2} w) = \frac{q B r d r}{d t}$

Change in angular momentum of particle
= $\int_{r = 0}^{r = R} q B r d r = 0$
$\Rightarrow \frac{d θ}{d t} = 0$ at time when particle leaves the region. So, velocity of particle is radially outward.