You shoot a deuteron into a uniform magnetic field. The deuteron has an initial

Question

You shoot a deuteron into a uniform magnetic field. The deuteron has an initial speed of 1.7 times 107 m/s. The initial trajectory, moving horizontally from left to right, is set so that it will enter the magnetic field directly in front of you. If the direction of the magnetic field is horizontal, away from the observer (you). (The deuteron carries a charge of q = +e and a mass of m = 3.344 times 10^-27 kg) After entering the field, will the deuteron move in a clockwise (cw) or counterclockwise (ccw) orbit in a vertical plane? [enter cw or ccw, all lower case] If the deuteron moves in the orbit with a diameter of 29.4 cm, what is the magnitude of the magnetic field?

Explanation / Answer

direction of initial force is given by vXB
v---> right
B--> away
F--> up
It will give rise to CCW motion
Answer: CCW

R = 29.4 /2 cm = 14.7 cm = 0.147 m
R =m*V / (q*B)
0.147 = (3.344*10^-27)*(1.7*10^7) / (1.6*10^-19*B)
B = 2.42 T
Answer: 2.42 T

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