A circular wire loop of radius $r$ can with stand a radial force $T$ before breaking. A particle of mass $m$ and charge $q (q > 0)$ is sliding over the wire. A magnetic field $c$ is applied normal to the plane of the wire. What maximum speed $v_{m a x}$ the particle can have before the loop breaks?

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$v_{m a x} = \frac{r}{m} [q B + \sqrt{q^{2} B^{2} + \frac{2 T m}{r}}]$

$v_{m a x} = \frac{r}{4 m} [q B + \sqrt{q^{2} B^{2} + \frac{T m}{r}}]$

$v_{m a x} = \frac{r}{2 m} [q B + \sqrt{q^{2} B^{2} + \frac{4 T m}{r}}]$

$v_{m a x} = \frac{r}{2 m} [q B + \sqrt{q^{2} B^{2} + \frac{8 T m}{r}}]$

answer is C.

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Detailed Solution

The centripetal force available is $= (T + q v B)$

From Newtons' second law

$(T + q v B) = \frac{m v^{2}}{r}$

Solving above equation for $v,$ we get

$v_{m a x} = \frac{r}{2 m} [q B + \sqrt{q^{2} B^{2} + \frac{4 T m}{r}}]$

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