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

Answer:

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

Explanation:

Given that,

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

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

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

$F=q(v\times B)\\\\B=\dfrac{F}{qv}\\\\B=\dfrac{1.7\times 10^{-13}}{1.6\times 10^{-19}\times 8.9\times 10^5}\\\\B=1.19\ T$

For direction :

$F=q(v\times B)\\\\k=(+q)(j\times B)\\\\B=-i$

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