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