At a certain location, the horizontal component of the earth's magnetic field is 2.3 x 10^-5 T, due north. A proton moves eastward with just the right speed, so that the magnetic force on it balances its weight. Find the speed of the proton.
To find the speed of the proton, we need to use the equation for the magnetic force on a moving charged particle in a magnetic field.
The magnetic force on a moving charged particle can be calculated using the equation:
F = qvBsinθ
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, and
θ is the angle between the velocity of the particle and the magnetic field.
In this case, the proton is moving eastward while the magnetic field is directed due north. This means the angle θ between the velocity and the magnetic field is 90 degrees.
Since the magnetic force balances the weight of the proton, we can equate the magnetic force with the weight of the proton:
F = mg
Where:
m is the mass of the proton, and
g is the acceleration due to gravity.
The charge of a proton (q) is equal to the elementary charge (e), which is 1.6 x 10^-19 C.
Given that the horizontal component of the earth's magnetic field (B) is 2.3 x 10^-5 T, and the mass of a proton (m) is approximately 1.67 x 10^-27 kg, and the acceleration due to gravity (g) is approximately 9.8 m/s^2, we can now solve for the velocity (v).
Let's substitute the values into the equation:
qvBsinθ = mg
(1.6 x 10^-19 C) * v * (2.3 x 10^-5 T) * sin(90°) = (1.67 x 10^-27 kg) * (9.8 m/s^2)
Since sin(90°) = 1, the equation simplifies to:
(1.6 x 10^-19 C) * v * (2.3 x 10^-5 T) = (1.67 x 10^-27 kg) * (9.8 m/s^2)
Now, solve for v:
v = [(1.67 x 10^-27 kg) * (9.8 m/s^2)] / [(1.6 x 10^-19 C) * (2.3 x 10^-5 T)]
Calculating this expression will give us the speed of the proton.