The region between y = 0 and y = d contains a magnetic field . A particle of mass m and charge q enters the region with a velocity. , the acceleration of the charged particle at the point of its emergence at the other side is:

Engineering

Physics

Magnet and Magnetic Dipole

Question

The region between y = 0 and y = d contains a magnetic field $\vec{B} = B \hat{z}$ . A particle of mass m and charge q enters the region with a velocity. $\vec{v} = v \hat{i} . I f d = \frac{mv}{2 qB}$ , the acceleration of the charged particle at the point of its emergence at the other side is:

$\frac{qνB}{m} (\frac{\hat{i} + \hat{j}}{\sqrt{2}})$

$\frac{qνB}{m} (\frac{- \hat{j} + \hat{i}}{\sqrt{2}})$

$\frac{qνB}{m} (- \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j})$

$\frac{qνB}{m} (\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j})$

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A charged particle in a magnetic field moves in a circular path. The radius of curvature is given by $r = \frac{m v}{q B}$ . Given $d = \frac{m v}{2 q B} = \frac{r}{2}$ , the particle emerges after traversing a semicircular path of radius r/2. This corresponds to a 60° arc, not 180°. At emergence, its velocity vector is rotated by 60° from the initial x-direction. The acceleration (centripetal) is always perpendicular to the velocity. Calculating the components for a 60° rotation from the vertical centripetal direction gives the correct vector.

Final Answer: $\frac{q v B}{m} (- \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j})$