A beam of protons traveling at 1.50km/s enters a uniform magnetic field, traveli

Question

A beam of protons traveling at 1.50km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction (the figure (Figure 1) ). The beam travels a distance of 1.10cm while in the field.

Part A

What is the magnitude of the magnetic field?

A beam of protons traveling at 1.50km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction (the figure (Figure 1) ). The beam travels a distance of 1.10cm while in the field.

Part A

What is the magnitude of the magnetic field?

B = T A beam of protons traveling at 1.50km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction (the figure (Figure 1) ). The beam travels a distance of 1.10cm while in the field. Part A What is the magnitude of the magnetic field? B = T

Explanation / Answer

it is basically based on a principle that the magnetic force exerted = centripetal force

so q*v*B=m*v^2/r

so it comes r=mv/qB

so here it is a quarter of circular perimeter

so 2*pi*r/4=1.10cm

r=0.700cm

so here B=mv/qr

so B=(1.67*10^-27*1.50*10^3)/(1.6*10^-19*0.00700m)=2.23*10^-3 T

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