An electron moving with a speed of 4.0 * 105 m>s in the positive x direction experiences zero magnetic force. When it moves in the positive y direction, it experiences a force of 3.2 * 10-13 N that points in the positive z direction. What are the direction and magnitude of the magnetic field

Explanation:

Given that,

Charge on an electron is

q=1.609×10^-19C

Speed of electron V=4×10^5 m/s in positive x direction

V = 4×10^5 •i m/s

At this point F=0N

When the electron is moving in positive y direction

V =4×10^5 •j m/s

F = 3.2×10^-13 N in positive z direction.

F = 3.2×10 ^-13 •k N

What is the direction and magnitude of the magnetic field B?

The force in a magnetic field is given as

F = q (V ×B)

The only time V×B is zero is when both V and B are in same direction

This shows that B is in the positive x direction or negative x direction

i.e B= Bx •i

Now applying this to the second condition

F= q(V×B)

3.2×10^-13 •k =1.609×10^-19 × (4×10^5•j ×Bx •i)

Note that, j×i =-k

3.2×10^-13 •k / 1.609×10^-19 = (4×10^5Bx •(j×i))

1988812.927 •k =- 4×10^5Bx•k

1988812.927/4×10^5 •k =-Bx •k

4.972 •k =-Bx •k

Then, Bx = -4.972 •k Telsa

Then, the magnitude of B is

B=4.972 T

And its direction is negative z direction.

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