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Q 10P

Expert-verifiedFound in: Page 829

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** $\overrightarrow{\mathbf{B}}{\mathbf{=}}{\mathbf{-}}{\mathbf{2}}{\mathbf{.}}{\mathbf{5}}\hat{\mathbf{i}\mathbf{}}{\mathbf{mT}}$.**At one instant the velocity of the proton is $\overrightarrow{\mathbf{v}}{\mathbf{=}}{\mathbf{2000}}{\mathbf{}}\hat{\mathbf{j}}{\mathbf{}}{\mathbf{m}}{\mathbf{/}}{\mathbf{s}}$** **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" ${\mathbf{4}}{\mathbf{.}}{\mathbf{00}}\hat{\mathbf{k}}{\mathbf{}}{\mathbf{V}}{\mathbf{/}}{\mathbf{m}}$**, (b) ${\mathbf{-}}{\mathbf{4}}{\mathbf{.}}{\mathbf{00}}\hat{\mathbf{k}}{\mathbf{}}{\mathbf{V}}{\mathbf{/}}{\mathbf{m}}$ and (c) ${\mathbf{4}}{\mathbf{.}}{\mathbf{00}}\hat{\mathbf{i}}{\mathbf{}}{\mathbf{V}}{\mathbf{/}}{\mathbf{m}}$** **?**

- $\overrightarrow{F}=1.44\times {10}^{-18}\hat{\mathrm{k}}\mathrm{N}$
- $\overrightarrow{F}=1.60\times {10}^{-19}\hat{\mathrm{k}}\mathrm{N}$
- $\overrightarrow{F}=\left(6.41\times {10}^{-19}\right)\hat{\mathrm{i}}\mathrm{N}+\left(8.01\times {10}^{-19}\right)\hat{\mathrm{k}}\mathrm{N}$

$\overrightarrow{\mathrm{B}}=-2.5\hat{\mathrm{i}}\mathrm{mT}$

$\overrightarrow{\mathrm{v}}=2000\hat{\mathrm{j}}\mathrm{m}/\mathrm{s}$

**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,

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

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

Hence, the net force acting on the proton is $\overrightarrow{F}=1.44\times {10}^{-18}\hat{k}\mathrm{N}$

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

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

Hence, the net force acting on the proton is $\overrightarrow{\mathrm{F}}=1.60\times {10}^{-19}\hat{\mathrm{k}}\mathrm{N}$

The net force on the proton when $\overrightarrow{E}=4.00\hat{\mathrm{i}}\mathrm{V}/\mathrm{m}$

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

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

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.

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