In a certain town, you will find that the magnetic field has a strength of about
ID: 1606293 • Letter: I
Question
In a certain town, you will find that the magnetic field has a strength of about 5X19^-3 T. The vector is directed northward, but it is not paralledl to the ground. It actually dips 70 degrees below the horizontal. (i.e. it is only 20 degrees away to the north from pointing straight into the ground.) In a labratory, a proton is moving horizontally, parallel to the floor, and due east at a speed of 5X10^6 m/s. What are the magnitude and direction of the magnetic force on the proton? (Hint; start by drawing a picture.)
Explanation / Answer
horizontal component of the magnetic field = 5 * 10^-3 * cos(20)
force = charge * velocity * magnetic field
force = 1.6 * 10^-19 * 5 * 10^6 * 5 * 10^-3 * cos(20)
force = 3.7587705 * 10^-15 N