A deuteron is accelerated from rest through a 10-kV potential difference and the

Question

A deuteron is accelerated from rest through a 10-kV potential difference and then moves perpendicularly to a uniform magnetic field with B = 1.6 T. What is the radius of the resulting circular path? (deuteron: m = 3.3 times 10^-27 kg, q = 1.6 times 10^-19 C). Give your answer in milimeters) Two long straight parallel wires separated by a distance of 20 cm carry currents of 30 A and 40 A in opposite directions. What is the magnitude of the resulting magnetic field at a point that is 15 cm from the wire carrying the 30-A current and 25 cm from the other wire? (Give your answer in micro Tesla: 1 mu T = 10^-6T

Explanation / Answer

1A)

KE= W

½*mv^2 = qV

½*(3.3*10^-27)*v^2 = (1.6*10^-19)*10*10^3 =>     v= 9.85*10^5 m/s

EB = EC

qvB = mv^2/r

qB= mv/r

r= mv/qB

r= (3.3*10^-27*9.85*10^5)/( 1.6*10^-19*1.6) =    0.013m

1B)

Bnet = B1+B2 = (0/2)*(I1/r1 + I2/r2) = [(4*10^-7)/(2)]*[(30)/(0.15) - (40)/(0.25)] = 8.0*10^-6 T

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