Question

A proton moves perpendicular to a uniform magnetic field B at a speed of 2.30 107 m/s and experiences an acceleration of 1.70 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.

Answer 1

Answer:

Explanation:

Given the following :

Speed (V) = speed of 2.30×10^7 m/s

Acceleration (a) = 1.70×10^13 m/s^2

Using the right hand rule provided by  Lorentz law:

B = F / qvSinΘ

Where B = magnitude of the magnetic field

v = speed of the particle

Θ = 90° (perpendicular to the field)

q = charge of the particle

SinΘ = sin90°  = 1

Note F = ma

Therefore,

B = ma / qvSinΘ

Mass of proton = 1.67 × 10^-27

Charge = 1.6 × 10^-19 C

B = [(1.67 × 10^-27) × (1.70 × 10^13)] / (1.6 × 10^-19) × (2.30 × 10^7) × 1

B = 2.839 × 10^-14 / 3.68 × 10^-12

B = 0.7715 × 10^-2

B = 7.72 × 10^-3 T

2) Magnetic field will be in the negative y direction according to the right hand thumb rule.

Since Velocity is in the positive z- direction, acceleration in the positive x - direction, then magnetic field must be in the negative y-direction.