Question

An electron has an instantaneous velocity of ~v=(106xˆ + 2 × 106 zˆ) m/s. It is in a region of the Earth’s magnetic field where the magnetic field is approximately constant. Assume that B~ =1×10−4zˆ T. Determine the number of orbits the electron completes while it travels1000 m in the z-direction.

Answer 1

Hint1 :

z component of the velocity contributes to the linear motion along the direction of magnetic field

Answer 2

so its the whole V x B thing?

Answer 3

i was thinking this but I didn't understand because of the different directions
\[R=\frac{mv}{qB}\]

Answer 4

\[q(v\times B) = \dfrac{mv^2}{R} \implies R = \dfrac{mv}{qB\sin\theta}\]

right ?

Answer 5

uh right

Answer 6

v=(106xˆ + 2 × 106 zˆ) m/s

Notice that x component is the perpendicular component,\(v\sin\theta\)
and z component is the parallel component, \(v\cos \theta\)

Answer 7

since B is along z

Answer 8

Here electron has travelled 1000 m along z axis.

Perhaps start by finding the time taken ?

Answer 9

I missed a few classes and legitimately know nothing

Answer 10

time = distance / speed

looks you have flipped the ratio, check once

Answer 11

oops
\[\frac{1000m*s}{2*10^{6}m}=5*10^{-4}s\]

Answer 12

next try finding how much time it takes for one revolution

Answer 13

isn't that kinda the same as angular frequency?
radians/s => \[\frac{2\pi }{5*10^{-4}s}\]

Answer 14

how ?
time taken for one revolution is just \(\dfrac{2\pi R}{v}\)

you can find \(R\) using the formula that you have posted few replies back..

Answer 15

\[\frac{2\pi mv}{qvBsin\theta}\]

Answer 16

\[\frac{2\pi mv\color{red}{\sin\theta}}{qvBsin\theta} = \dfrac{2\pi m}{qB}\]

Answer 17

Notice, the time period should be independent of the speed

Answer 18

No matter what the speed is, all the charged particles of same m/q ratio take "same time" to complete one revolution

Answer 19

ok I understand that

Answer 20

\[\frac{2\pi m}{qB}=\frac{2\pi(9.10*10^{-31}kg)}{1.6*10^{-19}C*1*10^{-4}T}=3.57*10^{-7}\]

Answer 21

So we have this so far :
1) It takes \(5 \times 10^{-4}\) s to travel 1000m in the z direction
2) It takes \(3.57\times 10^{-7}\)s to complete one revolution

Based on these, can we figure out the number of revolutions ?

Answer 22

\[\frac{5*10^{-4}}{3.57*10^{-7}}=1400.56\]

Answer 23

Perfect!

Answer 24

i was supposed to get 1397

Answer 25

and I don't understand why its ok to say that the magnetic field of earth is a constant

Answer 26

https://www.wolframalpha.com/input/?i=%281000%2F%282*10^6%29%29+%2F+%282 \pi*9.11*10^%28-31%29%2F%281.6*10^%28-19%29*10^%28-4%29%29%29

Answer 27

The question asks us to assume the magnetic field is constant. So why do we care ?

Answer 28

The electron only travelled one km. The Earth's radius is 6,400 km. The magnetic field is not going to change noticeably over such a small distance.

Answer 29

I was asked to explain it.. guess I didn't post that