The most common isotope of a single nucleus (stripped of its electrons) is accel

Question

The most common isotope of a single nucleus (stripped of its electrons) is accelerated from rest through a potential difference of 1000V and then fired into a uniform 58 mT B-field (directed out of the page) such that the nucleus is traveling perpendicular to the B-field. The nucleus is observed to travel in a circular path of radius 11.4412 cm. a.) The uniform magnetic field is generated by a Helmholtz coil, each coil having a radius of 30 cm and containing 585 turns. How much current is passing through the coils? b.) What element is the unknown nucleus?

Explanation / Answer

a) B = (4/5)^(3/2)*(mu*n*I/R) = 17533.93*I

solving we get I = 58*10^-3/17533.93 = 3.307 microAmpere

b) r = mv/qB

v = sqrt(2qV)

r = (m/B)*(sqrt(2V/q))

m = r*B/(sqrt(2V/q)) = 593.525*E-16 kg

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