A ring of mass m and radius r and made of an insulating material carries uniformly distributed charge. Initially it rests on a frictionless horizontal tabletop with its plane vertical. The charge on the ring so that it starts rolling on entering completely into the region of the magnetic field is $\frac{\sqrt{N} m v_{0}}{r B}$ , then find the value of ‘N’.

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Detailed Solution

$m r^{2} β = \int_{0}^{0} \frac{2 q}{2 π} d α v B r \cos α$
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$β = \frac{d w}{d} = \frac{q U B sin θ}{π} m$ ……….. (1)

$- m a = \int_{0}^{0} 2 \frac{q}{2 π} d α r w B \cos α$

$a = \frac{d v}{d t} = \frac{q r w B \sin θ}{π m}$ ……….. (2)

$\frac{d w}{d v} = \frac{v}{r^{2} w} \Rightarrow r^{2} \int_{0}^{w} w d w = - \int_{- v_{0}}^{v} v d v$

$\Rightarrow v = \frac{v_{0}}{\sqrt{2}}$ $\frac{v d v}{d x} = \frac{q r w B \sin θ}{π m}$ $\int_{v_{0}}^{v_{0} / \sqrt{2}} \frac{v d v}{\sqrt{v_{0}^{2} - v^{2}}} = \frac{q r B}{m π} \int_{0}^{π} \sin^{2} θ d θ$ $\Rightarrow q = \frac{\sqrt{2} m v_{0}}{B r}$