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(C) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it follows a circular path of radius $R = \frac{mv}{qB} = \f

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$\frac{qB}{2} \left( \frac{y^2}{x} + x \right)$

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(C) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it follows a circular path of radius $R = \frac{mv}{qB} = \frac{p}{qB}$,where $p$ is the momentum.From the geometry of the circular arc,let $R$ be the radius of the path. The particle travels a horizontal distance $y$ and is deflected by a vertical distance $x$.Using the property of a circle,$R^2 = (R - x)^2 + y^2$.Expanding this,$R^2 = R^2 - 2Rx + x^2 + y^2$.This simplifies to $2Rx = x^2 + y^2$,so $R = \frac{x^2 + y^2}{2x} = \frac{x}{2} + \frac{y^2}{2x} = \frac{1}{2} \left( \frac{y^2}{x} + x \right)$.Since $p = qBR$,we substitute the value of $R$:$p = qB \cdot \frac{1}{2} \left( \frac{y^2}{x} + x \right) = \frac{qB}{2} \left( \frac{y^2}{x} + x \right)$.

$A$ particle having charge $q$ enters a region of uniform magnetic field $\vec{B}$ (directed inwards) and is deflected a distance $x$ after travelling a distance $y$. The magnitude of the momentum of the particle is:

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