Electrons in Earth's upper atmosphere have typical speeds near 5.87 ✕ 105 m/s. HINT (a) Calculate the magnitude of Earth's magnetic field if an electron's velocity is perpendicular to the magnetic field and its circular path has a radius of 6.81 ✕ 10−2 m. T (b) Calculate the number of times per second that an electron circles around a magnetic field line.

a) 4.8×10^-5T

b) 7.38×10^-7seconds

Explanation:

The centripetal force in a magnetic field is given by:

qvB= mv2/r

Where q= charge of the ion

m= mass of the ion

r= radius of the circular path

B= magnetic field

V= velocity

Rearranging to make B subject of formular gives:

B=mv/r

B= [(9.11×10^-31)×(5.87×10^5)]/ [(1.6×10^-19)×(6.81×10^-2)]

B = (5 34×10^-25)/(1.099×10^-20)

B= 4.86×10^-5T

b) T =( 2 pi × r)/v

T = 2 × 3.142 x (6.81×10^-2) / (5.87×10^5

T = 0.428/ (5.87×10^5)

T = 7.38 × 10^-7seconds

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

the answer to this is proletariats.

hope this : )