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An alpha-particle (m=6.64×1027kg,q=3.2×1019C)\left( m = 6.64 \times 10 ^ { - 27 } \mathrm { kg }, q = 3.2 \times 10 ^ { - 19 } \mathrm { C }\right ) travels in a circular path of radius 25 cm in a uniform magnetic field of magnitude 1.5 T. (a) What is the speed of the particle? (b) What is the kinetic energy in electron-volts? (c) Through what potential difference must the particle be accelerated in order to give it this kinetic energy?

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Answered 2 years ago
Answered 2 years ago
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a)The trajectory of a charged particle in a uniform magnetic field is circular with radius given by:

r=mvqBr=\dfrac{mv}{qB}

Rearranging:

v=rqBm=(0.25)(3.2×1019)(1.5)6.64×1027=1.807×107m/sv=\dfrac{rqB}{m} = \dfrac{(0.25)(3.2 \times 10^{-19})(1.5)}{6.64 \times 10^{-27}}=1.807 \times 10^{7} \, \mathrm{m/s}

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