Dean is hunting in the Northwest Territory at a location where the earth's magnetic field is 8.0x 10^-5 T. He shoots at a duck decoy by mistake, and the rubber bullet he is using acquires a charge of 3.0 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 calculate the magnitude of the magnetic force acting on the bullet, you can use the formula:
F = q * v * B * sin(θ)
Where:
F is the magnitude of the magnetic force
q is the charge of the bullet
v is the velocity of the bullet
B is the magnitude of the magnetic field
θ is the angle between the velocity vector and the magnetic field vector.
In this case, the bullet is leaving the gun perpendicular to the earth's magnetic field, so the angle (θ) between the velocity vector and the magnetic field vector is 90 degrees.
Plugging in the given values:
q = 3.0 x 10^-12 C
v = 300 m/s
B = 8.0 x 10^-5 T
θ = 90 degrees
F = (3.0 x 10^-12 C) * (300 m/s) * (8.0 x 10^-5 T) * sin(90 degrees)
To calculate sin(90 degrees), we know that sin(90 degrees) = 1.
F = (3.0 x 10^-12 C) * (300 m/s) * (8.0 x 10^-5 T) * 1
F = 7.2 x 10^-12 N
Therefore, the magnitude of the magnetic force acting on the bullet is 7.2 x 10^-12 Newtons.
To calculate the magnitude of the magnetic force acting on the bullet, you can use the formula for the magnetic force on a charged particle moving through a magnetic field:
F = q * v * B * sin(θ)
Where:
F is the magnetic force
q is the charge of the particle
v is the velocity of the particle
B is the magnetic field strength
θ 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 θ = 90°, and sin(θ) = 1.
Now, let's calculate the magnetic force:
F = (3.0 x 10^-12 C) * (300 m/s) * (8.0 x 10^-5 T) * (1)
F = 7.2 x 10^-11 N
Therefore, the magnitude of the magnetic force acting on the bullet is 7.2 x 10^-11 N.
F= qvB
= (3 x 10^-12 C)(300 m/s)(8.0 x 10^-5 T)
= ?