A proton, traveling with a velocity of 6.8 × 106 m/s due east, experiences a magnetic force that has a maximum magnitude of 8.6 × 10-14 N and direction of due south. What are the magnitude and direction of the magnetic field causing the force? If the field is up, then enter a number greater than zero. If the field is down, then enter a number less than zero.
To find the magnitude and direction of the magnetic field causing the force on the proton, we can use the formula for the magnetic force on a moving charged particle:
F = q * v * B * sin(θ)
Where:
F is the magnetic force acting on the particle,
q is the charge of the particle (in this case, the charge of a proton),
v is the velocity of the particle,
B is the magnitude of the magnetic field, and
θ is the angle between the velocity and the magnetic field.
From the given information, we know that:
F = 8.6 × 10^-14 N (magnitude of the magnetic force)
v = 6.8 × 10^6 m/s (velocity of the proton)
θ = 90 degrees (since the force is perpendicular to the velocity)
Since sin(90 degrees) = 1, the equation simplifies to:
F = q * v * B
Rearranging the equation, we can solve for the magnetic field (B):
B = F / (q * v)
Now let's substitute the known values:
B = (8.6 × 10^-14 N) / (1.602 × 10^-19 C * 6.8 × 10^6 m/s)
The elementary charge of a proton is approximately equal to 1.602 × 10^-19 C.
B = 7.07 × 10^-6 T
Therefore, the magnitude of the magnetic field causing the force on the proton is 7.07 × 10^-6 Tesla.
Since the force is directed due south, we can infer that the magnetic field must be directed opposite to the force, which is due north. This corresponds to a negative value in the Cartesian coordinate system. Therefore, the direction of the magnetic field causing the force is down, or less than zero.