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(C) The radius of the circular path of a charged particle in a magnetic field is given by $R = \frac{mv}{qB}$.For the region $y > 0$, the radius is $R

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$2$

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(C) The radius of the circular path of a charged particle in a magnetic field is given by $R = \frac{mv}{qB}$.For the region $y > 0$, the radius is $R_1 = \frac{mv_0}{qB_1}$.For the region $y Given $B_2 = 4B_1$, we have $R_2 = \frac{mv_0}{q(4B_1)} = \frac{R_1}{4}$.The particle moves in a semicircle in the region $y > 0$ and then in a semicircle in the region $y The total displacement along the $x$-axis is $\Delta x = 2R_1 + 2R_2 = 2R_1 + 2(\frac{R_1}{4}) = 2R_1 + \frac{R_1}{2} = \frac{5R_1}{2}$.The time taken to complete a semicircle in region $y > 0$ is $t_1 = \frac{\pi m}{qB_1}$.The time taken to complete a semicircle in region $y The total time taken is $T = t_1 + t_2 = t_1 + \frac{t_1}{4} = \frac{5t_1}{4}$.The average speed along the $x$-axis is $v_{avg} = \frac{\Delta x}{T} = \frac{5R_1/2}{5t_1/4} = \frac{5R_1}{2} 	imes \frac{4}{5t_1} = \frac{2R_1}{t_1}$.Substituting $R_1 = \frac{mv_0}{qB_1}$ and $t_1 = \frac{\pi m}{qB_1}$, we get $v_{avg} = \frac{2(mv_0/qB_1)}{\pi m/qB_1} = \frac{2v_0}{\pi}$.Given $v_0 = \pi 	ext{ m s}^{-1}$, the average speed is $v_{avg} = \frac{2(\pi)}{\pi} = 2 	ext{ m s}^{-1}$.

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