Question

A proton moves in the magnetic field $\overrightarrow{B}=0.50\hat{ı}\mathrm{T}$with a speed of $1.0\times {10}^7\mathrm{m}/\mathrm{s}$in the directions shown in FIGURE. For each, what is magnetic force $\overrightarrow{F}$on the proton? Give your answers in component form.

Short answer

(a) The magnetic force on the proton$=0.56\times {10}^{-12}\hat{j}\mathrm{N}$$=0.56\times {10}^{-12}\hat{j}\mathrm{N}$

(b) The magnetic force on the proton $=8\times {10}^{-13}\hat{k}\mathrm{N}$$=8\times {10}^{-13}\widehat{k}\mathrm{N}$

Step-by-step solution

Introduction

Ampère's invention is essentially a long, protracted wire that transmits an electrical current. A little low compass needle prints out a sequence of concentric circular loops within the plane perpendicular to an electrical conductor wire, Ampère expertise in the areas.

Find the rate vector

The moving charge is polarized by the field, which is depending on the speed vector. The flux increases as the angle between the speed and the magnetic field increases. Equation gives the magnetic attraction.

$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$

Where $q$is the charge. The angle between the velocity and the magnetic field is $45°$in the $xz$plane

$\overrightarrow{v}=$$\left(1\times {10}^7\mathrm{m}/\mathrm{s}\right)\cos {45}^{\circ }$$\hat{i}$$+\left(1\times {10}^7\mathrm{m}/\mathrm{s}\right)\sin 45°$$\hat{k}$

The field of force has component $\overrightarrow{B}=0.50\hat{i}\mathrm{T}$, so we plug the values for $\overrightarrow{B},q$and $\overrightarrow{v}$into equation to induce $\overrightarrow{F}$where $(\hat{i}\times \hat{i})=0$

$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$

$=\left(1.6\times {10}^{-19}\mathrm{C}\right)\left[\left(1\times {10}^7\mathrm{m}/\mathrm{s}\right)\cos 45°\hat{i}+\right.$$\left.\left.\left(1\times {10}^7\mathrm{m}/\mathrm{s}\right)\sin 45°\hat{k}\right)\times (0.50\hat{i}\mathrm{T})\right]$

$=$$\left(1.6\times {10}^{-19}\mathrm{C}\right)\left(0.5\times {10}^7\sin 45°\right.$$(\hat{k}\times \hat{i}))$

$=0.56\times {10}^{-12}\hat{j}\mathrm{N}$

find a field of force that is travelling in the opposite direction

The angle between the rate and also the field is $90°$but its direction is within the negative $y$-direction so therefore the velocity vector is

$\overrightarrow{v}=\left(-1\times {10}^7\mathrm{m}/\mathrm{s}\right)\hat{j}$

The magnetic flux has component $\overrightarrow{B}=0.50\hat{i}\mathrm{T}$, so we plug the values for $\overrightarrow{B},q$and $\overrightarrow{v}$into equation to induce$\overrightarrow{F}$where $(\hat{j}\times \hat{i})=-\hat{k}$

$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$

$=\left(1.6\times {10}^{-19}\mathrm{C}\right)$$\left[\left(-1\times {10}^7\mathrm{m}/\mathrm{s}\right)\right.$$\hat{j}$$\times (0.50\hat{i}\mathrm{T})]$

$=\left(1.6\times {10}^{-19}\mathrm{C}\right)\left[-0.5\times {10}^7\right.$$(\hat{j}\times \hat{i})]$

$=8\times {10}^{-13}\hat{k}\mathrm{N}$