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Chapter 4: Q23E (page 913)

An electron in the beam of a cathode-ray tube is accelerated by a potential difference of 2.00Kv Then it passes through a region of transverse magnetic field, where it moves in a circular arc with radius What is the magnitude of the field?

Short Answer

Expert verified

The magnitude of the field is8.384×10-4T .

Step by step solution

01

Definition of magnetic field

The term magnetic field may be defined as the area around the magnet behave like a magnet.

02

Determine the magnitude of the field

The formula of radius for pathR=mvqBby the kinetic energy formulav=2eVmand q = e , the radius formula is

R=meB2eVmR=1B2eVmB=2mVeR2

Put all values

B=29.11×10-31kg2.00×103V1.60×10-19kg0.180m2B=8.384×10-4T

Hence, the magnitude of the field is8.384×10-4T .

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