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# (a) What voltage will accelerate electrons to a speed of $6.00 \times 10 ^ { - 7 } \mathrm { m } / \mathrm { s }$? (b) Find the radius of curvature of the path of a proton accelerated through this potential in a 0.500-T field and compare this with the radius of curvature of an electron accelerated through the same potential.

Solution

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a) Since the electric force is a conservative force, the total energy is conserved:

$K_1 + U_1 = K_2+U_2$

$\because K_1 = U_2 = 0$

$\therefore K_2= U_1$

$\because U_1 =eV$ and $K_2 = \dfrac{1}{2}mv^2$

$\therefore eV = \dfrac{1}{2}mv^2$

Rearranging:

$V=\dfrac{mv^2}{2e}=\dfrac{(9.1 \times 10^{-31})(6 \times 10^{-7})^2}{2(1.6 \times 10^{-19})}= 1.024 \times 10^{-24} \, \, \mathrm{V}$

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