Question

An electron that has velocity v=(2.0\times10^6\ m/s)i+(3.0\times10^6\ m/s)j moves through the uniform magneticfield B=(0.30T)i-(0.15T)j a) Find the force on the electron. b) Repeat your calculation for a proton having the same velocity.

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asked 2021-02-23
An electron that has velocity \(\displaystyle{v}={\left({2.0}\times{10}^{{6}}\ \frac{{m}}{{s}}\right)}{i}+{\left({3.0}\times{10}^{{6}}\ \frac{{m}}{{s}}\right)}{j}\) moves through the uniform magneticfield \(\displaystyle{B}={\left({0.30}{T}\right)}{i}-{\left({0.15}{T}\right)}{j}\)
a) Find the force on the electron.
b) Repeat your calculation for a proton having the same velocity.

Answers (1)

2021-02-25
\(\displaystyle{F}_{{B}}={q}{v}\times{B}\) is the equation to use: in words, v "crosses" B
\(\displaystyle\therefore\) in vector notation (instead of unit vector notation) \(\displaystyle{\frac{{{b}}}{{{c}}}}\) it's the same and its easier for me to type in:
a) \(\displaystyle{F}_{{B}}={q}{v}\times{B}\)
\(\displaystyle={q}{\left({<}{2000000}{i},{3}{000}{000}{j}{>}{x}{<}{0.03}{i},-{0.15}{j}{>}\right)}\)
\(\displaystyle={q}{\left(-{390000}{k}\right)}\)
\(\displaystyle={\left(-{1.602}\times{10}^{{{19}}}\right)}{\left(-{390000}{k}\right)}\)
\(\displaystyle={6.25}\times{10}^{{-{14}}}{N}\)
b) \(\displaystyle{F}_{{B}}={q}{v}\times{B}\)
\(\displaystyle={q}{\left({<}{2}{000000}{i},{3000000}{j}{>}{x}{<}{0.03}{i},-{0.15}{j}{>}\right)}\)
\(\displaystyle={\left(+{1.602}\times{10}^{{{19}}}\right)}{\left(-{390000}{k}\right)}\)
\(\displaystyle=-{6.25}\times{10}^{{-{14}}}{N}\)
pointed into the plane
As the texbook points out, answers are correct.
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