A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 1.80 x 10^7 m/s and experiences an acceleration of 2.20 x 10^13 m/s2 in the positive x-direction when its velocity is in the positive z-direction.

Determine the magnitude and direction of the field.

Explanation:

Given that,

A proton moves in a magnetic field

Charge of proton q = + 1.609 × 10^-19C..

Mass of proton m = 1.67 × 10^-27 kg.

Speed of proton v = 1.8 × 10^7 m/s

Acceleration experience a = 2.2 × 10¹³ m/s²

The acceleration is in positive x-direction

Then, a = 2.2 × 10¹³ •i m/s²

The speed moves along positive direction

Then, v = 1.8 × 10^7 •k m/s

Magnetic Field and it's direction?

The force that cause the proton to be in motion in the magnetic field is given as

F = qvBSinθ

Making B subject of formulas

B = F / qvSinθ

The velocity and the field are perpendicular, then θ=90°

Also, from newton second law, F=ma

Then,

B = ma / qvSinθ

Sin90 = 1

So,

B = 1.67 × 10^-27 × 2.2 × 10¹³ / 1.609 × 10^-19 × 1.8 × 10^7

B = 0.0127 T

B = 1.27 × 10^-2 T

Direction

Using the right-hand thumb rule, since the acceleration is in the positive x-direction and the velocity is in the positive z-direction, then, the magnetic field must be in the negative y-direction

Prove

F = q (V × B). Cross product of V and B

m(a) •i = q(V •k × B )

We know that, j × k = i and k × j = -i,

Since, k × j = -i, then k × -j = i

So, B must be in the negative y direction

Therefore,

B = —1.27 × 10^-2 •j T

0.863333

86%

hope this

i will answer your question if you answer mine