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Chapter 4: Q3E (page 912)

In a 1.25-T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50μC and initially moving northward atis deflected toward the east. (a) What is the sign of the charge of this particle? Make a sketch to illustrate how you found your answer. (b) Find the magnetic force on
the particle.

Short Answer

Expert verified

The Magnetic field force is 0.0505N

Step by step solution

01

Important Concepts

The magnetic field force is given by

FB=q(v×B)F=qvBsinθ

Where q is the charge of the particle, V is the velocity and B is the magnetic field

02

Application

Using the given data we get

FB=qvBsinθFB=8.50×10-6C1.25T4.75×103m/ssin90FB=0.0505N

Using the right hand thumb rule, to v and B, thumb Points to east hence

The magnetic field force is 0.0505N and its direction is towards east.

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