Question

A proton moves perpendicular to a uniform magnetic field at a speed of 1.75 107 m/s and experiences an acceleration of 2.25 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:

B = 0.013(-j) T

Explanation:

Given that,

The speed of a proton, $v=1.75\times 10^7\ m/s$

Acceleration experienced by the proton,$a=2.25\times 10^3\ m/s$

We need to find the magnitude and the direction of the magnetic field. At equilibrium,

$ma=qvB\\\\B=\dfrac{ma}{qv}\\\\B=\dfrac{1.67\times 10^{-27}\times 2.25\times 10^{13}}{1.6\times 10^{-19}\times 1.75\times 10^{7}}\\\\B=0.013\ T$

The velocity is in +z direction, force in +x direction, then the field must be in -y direction.