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Expert-verified Found in: Page 829 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Question: A proton travels through uniform magnetic and electric fields. The magnetic field is $\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{=}}{\mathbf{-}}{\mathbf{2}}{\mathbf{.}}{\mathbf{5}}\stackrel{\mathbf{^}}{\mathbf{i}\mathbf{}}{\mathbf{mT}}$.At one instant the velocity of the proton is $\stackrel{\mathbf{\to }}{\mathbf{v}}{\mathbf{=}}{\mathbf{2000}}{\mathbf{}}\stackrel{\mathbf{^}}{\mathbf{j}}{\mathbf{}}{\mathbf{m}}{\mathbf{/}}{\mathbf{s}}$ At that instant and in unit-vector notation, what is the net force acting on the proton if the electric field is (a) role="math" localid="1663233256112" ${\mathbf{4}}{\mathbf{.}}{\mathbf{00}}\stackrel{\mathbf{^}}{\mathbf{k}}{\mathbf{}}{\mathbf{V}}{\mathbf{/}}{\mathbf{m}}$, (b) ${\mathbf{-}}{\mathbf{4}}{\mathbf{.}}{\mathbf{00}}\stackrel{\mathbf{^}}{\mathbf{k}}{\mathbf{}}{\mathbf{V}}{\mathbf{/}}{\mathbf{m}}$ and (c) ${\mathbf{4}}{\mathbf{.}}{\mathbf{00}}\stackrel{\mathbf{^}}{\mathbf{i}}{\mathbf{}}{\mathbf{V}}{\mathbf{/}}{\mathbf{m}}$ ?

1. $\stackrel{\to }{F}=1.44×{10}^{-18}\stackrel{^}{\mathrm{k}}\mathrm{N}$
2. $\stackrel{\to }{F}=1.60×{10}^{-19}\stackrel{^}{\mathrm{k}}\mathrm{N}$
3. $\stackrel{\to }{F}=\left(6.41×{10}^{-19}\right)\stackrel{^}{\mathrm{i}}\mathrm{N}+\left(8.01×{10}^{-19}\right)\stackrel{^}{\mathrm{k}}\mathrm{N}$
See the step by step solution

## Step 1: Given

$\stackrel{\to }{\mathrm{B}}=-2.5\stackrel{^}{\mathrm{i}}\mathrm{mT}$

$\stackrel{\to }{\mathrm{v}}=2000\stackrel{^}{\mathrm{j}}\mathrm{m}/\mathrm{s}$

## Step 2: Determining the concept

The total force acting on the charged particle is the sum of the forces due to electric and magnetic fields.

Formulae are as follow:

Force acting on the charged particle due to electric field,

${F}_{e}=qE$

Force acting on the charged particle due to magnetic field,

${F}_{m}=qvB$

Where, ${F}_{m}$ is magnetic force, v is velocity, B is magnetic field, q is charge of particle.

## Step 3: (a) Determining the net force acting on the proton if the electric field is 4.00k^ V/m

The net force on the proton when $\stackrel{\to }{E}=4.00\stackrel{^}{\mathrm{k}}\mathrm{V}/\mathrm{m}$ :

$\stackrel{\to }{\mathrm{F}}=\mathrm{q}\left(\stackrel{\to }{\mathrm{E}}+\stackrel{\to }{\mathrm{V}}×\stackrel{\to }{\mathrm{B}}\right)\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{F}}=1.6×{10}^{-19}\left(4.00\stackrel{^}{\mathrm{k}}+2000\stackrel{^}{\mathrm{j}}×\left(-2.5×{10}^{-3}\right)\stackrel{^}{\mathrm{i}}\right)\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{F}}=1.6×{10}^{-19}\left(4.00\stackrel{^}{\mathrm{k}}+5.00\stackrel{^}{\mathrm{k}}\right)\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{F}}=1.44×{10}^{-18}\stackrel{^}{\mathrm{k}}\mathrm{N}$

Hence, the net force acting on the proton is $\stackrel{\to }{F}=1.44×{10}^{-18}\stackrel{^}{k}\mathrm{N}$

## Step 4: (b) Determining the net force acting on the proton if the electric field is -4.00k^ V/m

Then net force on the proton when $\stackrel{\to }{E}=-4.00\stackrel{^}{\mathrm{k}}\mathrm{V}/\mathrm{m}$:

$\stackrel{\to }{\mathrm{F}}=\mathrm{q}\left(\stackrel{\to }{\mathrm{E}}+\stackrel{\to }{\mathrm{V}}×\stackrel{\to }{\mathrm{B}}\right)\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{F}}=1.6×{10}^{-19}\left(-4.00\stackrel{^}{\mathrm{k}}+2000\stackrel{^}{\mathrm{j}}×\left(-2.5×{10}^{-3}\right)\stackrel{^}{\mathrm{i}}\right)\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{F}}=1.6×{10}^{-19}\left(-4.00\stackrel{^}{\mathrm{k}}+5.00\stackrel{^}{\mathrm{k}}\right)\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{F}}=1.60×{10}^{-19}\stackrel{^}{\mathrm{k}}\mathrm{N}$

Hence, the net force acting on the proton is $\stackrel{\to }{\mathrm{F}}=1.60×{10}^{-19}\stackrel{^}{\mathrm{k}}\mathrm{N}$

## Step 5: (c) Determining the  net force acting on the proton if the electric field is-4.00i^ V/m

The net force on the proton when $\stackrel{\to }{E}=4.00\stackrel{^}{\mathrm{i}}\mathrm{V}/\mathrm{m}$

$\stackrel{\to }{\mathrm{F}}=\mathrm{q}\left(\stackrel{\to }{\mathrm{E}}+\stackrel{\to }{\mathrm{V}}×\stackrel{\to }{\mathrm{B}}\right)\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{F}}=1.6×{10}^{-19}\left(4.00\stackrel{^}{\mathrm{i}}+2000\stackrel{^}{\mathrm{j}}×\left(-2.5×{10}^{-3}\right)\stackrel{^}{\mathrm{i}}\right)\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{F}}=1.6×{10}^{-19}\left(4.00\stackrel{^}{\mathrm{i}}+5.00\stackrel{^}{\mathrm{k}}\right)\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{F}}=\left(6.41×{10}^{-19}\right)\stackrel{^}{\mathrm{i}}\mathrm{N}+\left(8.01×{10}^{-19}\right)\stackrel{^}{\mathrm{k}}\mathrm{N}$

Hence, the net force acting on the proton is $\stackrel{\to }{\mathrm{F}}=\left(6.41×{10}^{-19}\right)\stackrel{^}{\mathrm{i}}\mathrm{N}+\left(8.01×{10}^{-19}\right)\stackrel{^}{\mathrm{k}}\mathrm{N}$.

Therefore, the values of net force due to different electric fields can be determined by taking the vector sum of forces due to the electric and magnetic field. ### Want to see more solutions like these? 