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Chapter 4: Q24E (page 913)

A beam of protons traveling at 1.20 km/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 (Fig. E27.24). The beam travels a distance of 1.18 cm while in the field. What is the magnitude of the magnetic field?

Short Answer

Expert verified

The magnitude of the magnetic field isB=1.67×10-3T

Step by step solution

01

Radius of curvature for a particle moving in a magnetic field

The radius of curvature for a particle moving in a magnetic field is given by

R=mvqB

02

Determine the magnitude of the magnetic field

The beam travels a distance of 1.8 cm

The particle covered the angle quarter of the circle

Therefore,θ=π/2

Therefore, the radius of the path will be

R=sθ=1.18π/2=0.751

Therefore, the radius of curvature for a particle moving in a magnetic field is

R=mvqB

Or

B=mvqR

Substitute the given values we get

B=(1.67×10-27kg)(1200m/s)(1.60×10-19C)(0.00751m)=1.67×10-3T

Therefore, the magnitude of the magnitude of the magnetic field isB=1.67×10-3T

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