The magnetic field of the Earth at a certain location is directed vertically downward and has a magnitude of 54.0 µT. A proton is moving horizontally towards the west in this field with a speed of 6.00 ✕ 106 m/s
What is the radius of the circular arc followed by this proton?
km
Isn't there a formula for this?
you know B, and v, and q.
calculate the force moving the proton inward, then set that equal to centripetal force , m v^2/r, calcualte r.
To determine the radius of the circular arc followed by the proton, we can use the formula for the magnetic force acting on a charged particle moving in a magnetic field:
F = qvBsin(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, and
θ is the angle between the velocity vector and the magnetic field vector.
Since the velocity of the proton is directed horizontally towards the west and the magnetic field is directed vertically downward, the angle between the velocity vector and the magnetic field vector is 90 degrees (θ = 90 degrees).
Given:
q = charge of a proton = 1.6 x 10^-19 C (coulombs)
v = 6.00 x 10^6 m/s (velocity of the proton)
B = 54.0 µT = 54.0 x 10^-6 T (magnetic field strength)
We can calculate the magnetic force using the formula:
F = qvBsin(θ) = qvB
Plugging in the given values:
F = (1.6 x 10^-19 C)(6.00 x 10^6 m/s)(54.0 x 10^-6 T)
Simplifying:
F = 51.8 x 10^-19 N
The magnetic force acting on the proton provides the centripetal force required to keep the proton moving in a circular path. The centripetal force can be given by:
F = (mv^2)/r
Where:
m is the mass of the proton (1.67 x 10^-27 kg)
r is the radius of the circular arc.
To find the radius, rearranging the equation:
r = (mv^2)/F
Plugging in the known values:
r = ((1.67 x 10^-27 kg)*(6.00 x 10^6 m/s)^2)/(51.8 x 10^-19 N)
Simplifying:
r = 114.985 km
Therefore, the radius of the circular arc followed by the proton is approximately 114.985 km.
To find the radius of the circular arc followed by the proton, we can use the formula for the magnetic force on a moving charge in a magnetic field:
F = q * v * B * sin(theta)
Where:
F = Magnetic force on the charge
q = Charge of the particle (in this case, the charge of a proton is +1.6 x 10^-19 C)
v = Velocity of the particle (6.00 x 10^6 m/s)
B = Magnetic field strength (54.0 µT, which is 54.0 x 10^-6 T)
theta = Angle between the velocity vector and the magnetic field vector (it is 90 degrees in this case)
Since the magnetic force provides the centripetal force to keep the proton moving in a circular path, we can equate the magnetic force to the centripetal force:
F = (m * v^2) / r
Where:
m = Mass of the proton (1.67 x 10^-27 kg)
v = Velocity of the proton (6.00 x 10^6 m/s)
r = Radius of the circular path
Setting the magnetic force equal to the centripetal force, we have:
q * v * B * sin(theta) = (m * v^2) / r
Plugging in the values, we get:
(1.6 x 10^-19 C) * (6.00 x 10^6 m/s) * (54.0 x 10^-6 T) * 1 = (1.67 x 10^-27 kg) * (6.00 x 10^6 m/s)^2 / r
Simplifying the equation, we can isolate r:
r = (1.67 x 10^-27 kg) * (6.00 x 10^6 m/s)^2 / ((1.6 x 10^-19 C) * (54.0 x 10^-6 T) * 1)
Calculating the value, we get:
r = 0.000027 m
To convert this radius from meters to kilometers, we divide by 1000:
r = 0.000027 m / 1000 = 2.7 x 10^-8 km
Therefore, the radius of the circular arc followed by the proton is 2.7 x 10^-8 km.