Aproton moves along the x-axis with vx=1.0×107 m/s. as it passes the origin, what are the strength and direction of the magnetic field at the (x, y, z) positions (a) (1 cm, 0 cm, 0 cm), (b) (0 cm, 1 cm, 0 cm), and (c) (0 cm, −2 cm, 0 cm)?

The magnetic field in x axis is 0.

The magnetic field in y axis is .

The magnetic field in y axis is .

Step-by-step explanation:

Given that,

Speed

(a). For position (1,0,0),

A proton moves along the x-axis and the position of the particle is in x axis so the angle is 0°

We need to calculate the magnetic field

Using formula of magnetic field

(b). For position (0,1,0),

A proton moves along the x-axis and the position of the particle is in y axis so the angle is 90°

Using formula of magnetic field

(c).  For position (0,-2,0),

A proton moves along the x-axis and the position of the particle is in -y axis so the angle is 270°

Using formula of magnetic field

Hence, The magnetic field in x axis is 0.

The magnetic field in y axis is .

The magnetic field in y axis is .

answer: 16

step-by-step explanation: first you would add 6+6 which equals 12 then add 2+2 which equals 4. then add 12 + 4 which is 16.

hope this ^_^