Question

(a) A cosmic ray proton moving toward the Earth at $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$experiences a magnetic force of $\mathbf{1}\mathbf{.}\mathbf{70}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{16}\mathbf{}\mathbf{N}}$. What is the strength of the magnetic field if there is a ${45}^{\circ }$ angle between it and the proton’s velocity? (b) Is the value obtained in part (a) consistent with the known strength of the Earth’s magnetic field on its surface? Discuss.

Short answer

(a) The earth's magnetic field is $\mathbf{3}\mathbf{.}\mathbf{01}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{5}\mathbf{}}\mathit{T}$.

(b) The value obtained is consistent with the known value of the earth's magnetic field.

Step-by-step solution

Given information

The speed of the cosmic ray of the proton is$\mathit{V}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}\mathbf{}}\mathit{m}\mathbf{/}\mathit{s}$

The charge on the proton is$\mathit{q}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{19}\mathbf{}}\mathit{C}$

The magnetic force is$\mathit{F}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{70}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{16}\mathbf{}}\mathit{N}$

The angle between the magnetic field and velocity is${\mathbf{45}}^{\mathbf{\circ }}$

Step 2:Defintion of the magnetic field.

We use the magnetic field as a tool to describe how the magnetic force is distributed in the space around and within something magnetic in nature.

Determining the strength of the magnetic field.    

(a)

The magnetic force can be determined from the strength of the magnetic field and velocity such that,

$\mathit{\theta }\mathbf{(}\mathbf{=}\mathit{q}\mathit{v}\mathit{B}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Equation (1) can be rearranged and given values are substituted to obtain,

$\mathit{B}\mathbf{=}\frac{\mathbf{F}}{\mathbf{\theta }\mathbf{y}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{(}}$

width="381">=1.70×10‐16N(1.6×10‐19C)×(5.00×107m/s)×sin(45∘)

$\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{01}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{5}}\mathbf{}\mathit{T}$

Therefore, the earth's magnetic field is$\mathbf{3}\mathbf{.}\mathbf{01}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{5}}\mathbf{}\mathbf{}\mathit{T}$.

(b)

The result is consistent with the known value of the earth's magnetic field which is $\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{5}}\mathbf{}\mathbf{}\mathit{T}$

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