Dean is hunting in the Northwest Territory at a location where the earth's magnetic field is 7.0x 10^-5 T. He shoots at a duck decoy by mistake, and the rubber bullet he is using acquires a charge of 2.0 x 10 ^ -12 C as it leaves his gun at 300m/s, perpendicular to the earth's magnetic field. What is the magnitude of the magnetic force acting on the bullet?
To find the magnitude of the magnetic force acting on the bullet, we can use the formula:
F = q * v * B * sin(theta)
Where:
- F is the magnetic force
- q is the charge of the object
- v is the velocity of the object
- B is the magnetic field
- theta is the angle between the velocity vector and the magnetic field vector
In this case, the bullet is moving perpendicular to the earth's magnetic field, so the angle theta is 90 degrees. Therefore, sin(theta) = 1.
Plugging the given values into the formula, we have:
q = 2.0 x 10^-12 C
v = 300 m/s
B = 7.0 x 10^-5 T
theta = 90 degrees
F = (2.0 x 10^-12 C) * (300 m/s) * (7.0 x 10^-5 T) * sin(90)
Using a scientific calculator, we find that sin(90) = 1.
F = (2.0 x 10^-12 C) * (300 m/s) * (7.0 x 10^-5 T) * 1
Simplifying the expression, we get:
F = (2.0 x 10^-12 C) * (300 m/s) * (7.0 x 10^-5 T)
Our final step is to calculate the product of these values:
F = (2.0 x 10^-12 C) * (300 m/s) * (7.0 x 10^-5 T)
Multiplying the numbers, we get:
F = 4.2 x 10^-9 N
Therefore, the magnitude of the magnetic force acting on the bullet is 4.2 x 10^-9 N.