Question

A positive charge of 0.087 C moves horizontally to the right at a speed of 638.121 m/s and enters a magnetic field directed vertically downward. If it experiences a force of 90.446 N, what is the magnetic field strength ?

Answer 1

The force of a moving particle in a magnetic field can be written as:

`F=qvBsin(\theta)`

One important aspect of the problem is when it tells us that the particle is moving horizontally, and the magnetic field is vertically downward. This gives us the information we need about the angle between the vectors, as it can be seen in the following drawing:

Thus, replacing the information in our formula, we get:

`90.446=0.087*638.121*B*sin(90°)`

Then, our magnetic field B is:

`B=(90.446)/(0.087*638.121*1)=1.629T`

Our final answer is B=1.629T