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) As R = mv/qB also v = sqrt[2qV/m]

R = (m/qB)*sqrt[2qV/m] = (1/B)*sqrt[2Vm/q]

= (1/1.6)*sqrt[2*10^4*3.3*10^-27/1.6*10^-19] = 1.3*10^-2 m

R = 13 mm

1B) B(I) = 2*(10)*(I/R)

With given values the point does not seem to be in the plane of the wire. So youneed to have vector sum of the two fields.

Angle, between them is given by cosine rule
cos = [{0.15² + 0.25² - 0.20²}/(2*0.15*0.25)] = 0.045
|B(30)| = 2*(10)*(30/0.15) = 4*10 T and
|B(40)| = 2*(10)*(40/0.25) = 3.2*10 T

Magnitude of required resultant |B| is given by
|B| = [4² + 3.2* + 2*4*3.2*cos ] = [4² + 3.2* + 2*4*3.2*0.045] = 5.233*10 T

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