Question

Dean is hunting in the Northwest Territories at a location where earths magnetic field is 7.0x10^-5T. He shoots by mistake at a duck decoy, and the rubber bullet he is using acquires a charge of 2.0x10^-12C as it leaves his gun at 300 m/s, perpendicular to earths magnetic field. What is the magnitude of the magnetic force acting on the bullet?

Answer 1

The magnetic force acting on an object of charge q is

`F=qvB \sin \theta`
where
q is the charge
v is the speed of the object
B is the magnetic field intensity

`\theta` is the angle between the directions of v and B.

In our problem, the direction of the bullet is perpendicular to the magnetic field, so
`\theta=90^(\circ)` and
`\sin \theta=1`, so we can ignore it in the formula.

Therefore, we can use the data of the problem to calculate the magnetic force on the bullet:

`F=qvB=(2.0 \cdot 10^(-12)C)(300 m/s)(7.0 \cdot 10^(-5) T)=4.2 \cdot 10^(-14) N`