(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Earth's magnetic field near the equator is 5.0x10^{-5}T due north. Find the acceleration of an electron traveling northeast at 2.0x10^{5}m/s. The charge on an electron is -1.602x10^{-19}C and it's mass is 9.109x10^{-31}kg.

2. Relevant equations

Fm = |q|vB*sin(theta)

a = Fm/m

3. The attempt at a solution

Fm = |-1.602x10^{-19}C|*2.0x10^{5}m/s*5.0x10^{-5}T*sin(45)

Fm = 1.13x10^{-18}N

a = Fm/m = 1.13x10^{-18}N/9.109x10^{-31}kg = 1.24x10^{12}m/s^{2}

For some reason, I lost 1 point on this problem. I'm not sure where, how or why...

Should the acceleration also have a direction? If so, what would that be? Thank you for any input...

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# Homework Help: Acceleration of an Electron in Earth's Magnetic Field

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