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Q. A charge q is accelerated through a potential difference V. It is then passed normally through a uniform magnetic field, where it moves in a circle of radius r. The potential difference required to move it in a circle of radius 2r is

KCETKCET 2018Moving Charges and Magnetism

Solution:

Radius of circular path: $r = \frac{\sqrt{2mqV}}{qB} $
$ r \propto\sqrt{V} $ where B is constant
or $ V \propto r^{2} $
$ \frac{V_{2}}{V_{1}} = \left(\frac{r_{2}}{r_{1}}\right)^{2} $
$\frac{V_{2}}{V}= \left(\frac{2r}{r}\right)^{2} 4$
$ \Rightarrow V_{2} = 4 V $