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Chapter 37: Q96P (page 1152)

Question:A 2.50 MeV electron moves perpendicularly to a magnetic field in a path with a 3.0 cm radius of curvature. What is the magnetic field B? (See Problem 53)

Short Answer

Expert verified

The magnetic field for the electron is 0.178T.

Step by step solution

01

Identification of given data

The energy of electron is E=2.50MeV

The radius of curvature is r=3cm

The magnetic field for the electron is found by equating the necessary centripetal force by magnetic force on the electron.

02

Determination of magnetic field for electron

The magnetic field for electron is given as:

B=2mEqr

Here, q is the charge of electron and its value is 1.6×10-19C , m is the mass of electron and its value is 9.1×10-31kg

Substitute all the values in the above equation.

B=29.1×10-31kg2.50MeV1.6×10-13J1MeV1.6×10-19C3cm10-2m1cmB=0.178t

Therefore, the magnetic field for the electron is 0.178T.

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