Two charged particles of mass m and charge q | vedclass

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Question

(C) The radius of the circular path of a charged particle in a magnetic field is $R = \frac{mv}{qB}$.The time period of the circular motion is $T = \f

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$2(\frac{mv}{qB})^2$

Vedclass · Answer

(C) The radius of the circular path of a charged particle in a magnetic field is $R = \frac{mv}{qB}$.The time period of the circular motion is $T = \frac{2\pi m}{qB}$.The given time is $t = \frac{\pi m}{qB} = \frac{T}{2}$.This means each particle completes a semi-circle in time $t$.Let the initial velocity vectors be $\vec{v}_1$ and $\vec{v}_2$ at angles $0^{\circ}$ and $60^{\circ}$ with the $X$-axis respectively.After time $t = \frac{T}{2}$, the particles will be at a distance of $2R$ from the origin, diametrically opposite to their starting points.The position vector $\vec{r}_1$ will be at an angle of $90^{\circ}$ to the initial velocity $\vec{v}_1$ (along the $Y$-axis), and $\vec{r}_2$ will be at an angle of $90^{\circ}$ to the initial velocity $\vec{v}_2$.The angle between $\vec{r}_1$ and $\vec{r}_2$ is the same as the angle between $\vec{v}_1$ and $\vec{v}_2$, which is $60^{\circ}$.The magnitude of the position vector is $|\vec{r}_1| = |\vec{r}_2| = 2R = \frac{2mv}{qB}$.The dot product is $\vec{r}_1 \cdot \vec{r}_2 = |\vec{r}_1| |\vec{r}_2| \cos(60^{\circ}) = (2R)(2R) \cos(60^{\circ}) = 4R^2 	imes \frac{1}{2} = 2R^2$.Substituting $R = \frac{mv}{qB}$, we get $\vec{r}_1 \cdot \vec{r}_2 = 2(\frac{mv}{qB})^2$.

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