Question

A proton moving in the positive x direction with a speed of 8.9 105 m/s experiences zero magnetic force. When it moves in the positive y direction it experiences a force of 1.7 10-13 N that points in the positive z direction. Determine the magnitude and direction of the magnetic field.

Answer 1

The magnitude of magnetic field is 1.19 T and its direction magnetic is in negative x direction.

Step-by-step explanation:

Given that,

Speed of a proton,
`v=8.9* 10^5\ m/s` (due +x direction)

When it moves in the positive y direction it experiences a force of,
`F=1.7* 10^(-13)\ N` (due positive z direction)

We need to find the magnitude and direction of the magnetic field. The magnetic force is given by :

`F=q(v* B)\\\\B=(F)/(qv)\\\\B=(1.7* 10^(-13))/(1.6* 10^(-19)* 8.9* 10^5)\\\\B=1.19\ T`

For direction :

`F=q(v* B)\\\\k=(+q)(j* B)\\\\B=-i`

So, the magnitude of magnetic field is 1.19 T and its direction magnetic is in negative x direction.