As shown in the figure below, a charged particle (mass = 1.34 x 10-25 kg) is acc

Question

As shown in the figure below, a charged particle (mass = 1.34 x 10-25 kg) is accelerated from rest from initial (105 V) to final (0 V) voltage as shown. The particle then enters a region of uniform magnetic field with magnitude 4.73 T. If the magnitude of the particle's charge is 1.6x10-19 C, what is the distance, d, shown in the figure? Assume that any losses of the particle's energy are negligible. Write your answer to three significant figures.

Please thoroughly explain the steps.

As shown in the figure below, a charged particle (mass = 1.34 times 10-25 kg) is accelerated from rest from initial (105 V) to final (0 V) voltage as shown. The particle then enters a region of uniform magnetic field with magnitude 4.73 T. If the magnitude of the particle's charge is 1.6 times 10-19 C, what is the distance, d, shown in the figure? Assume that any losses of the particle's energy are negligible. Write your answer to three significant figures.

Explanation / Answer

R=mv/qb=(2*K.E)^.5/q*b=2.36*10^11 m

D=2*r=4.72*10^11 m

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