   Chapter 19, Problem 6P

Chapter
Section
Textbook Problem

A proton moves perpendicular to a uniform magnetic field B → at a speed of 1.00 × 107 m/s and undergoes an acceleration of 2.00 × 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.

To determine
The magnitude and direction of the field.

Explanation

Given info: The speed of the proton is 1.00×107ms-1 . The proton is moving perpendicular to the magnetic field. The proton undergoes an acceleration of 2.00×1013ms-2 in the positive x direction.

Explanation:

The magnetic force on a charged particle is given by,

F=qvBsinθ       (1)

• q is the charge on the particle
• v is the velocity of the particle
• B is the magnetic field
• θ is the angle between the velocity and the magnetic field

Solving for B ,

B=Fqvsinθ       (2)

From Newton’s second law, the force on the proton will be,

F=ma       (3)

• m is the mass of the proton
• a is the acceleration of the proton

From (2) and (3),

B=maqvsinθ       (4)

Substitute 1.00×107ms-1 for v , 2.00×1013ms-2 for a , 1.60×1019C for q , 90° for θ and 1

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