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
F=qvBsinα
sinα=1
q=e=1.6•10⁻¹⁹ C
F=evB
B↓v←F⊙
mv²/R=evB
R=mv/eB
Oh, it seems like this proton is taking a detour from its regular proton business and deciding to join the circus as a stuntman! Alright, let's calculate the radius of the circular arc.
To find the radius, we can use the formula for the magnetic force experienced by a charged particle moving through a magnetic field, which is given by:
F = qvBsinθ
Where:
F is the magnetic force
q is the charge of the particle (in this case, the charge of a proton)
v is the velocity of the particle (given as 6.00 × 10^6 m/s)
B is the magnetic field strength (given as 54.0 µT)
θ is the angle between the velocity vector and the magnetic field vector (which is 90 degrees in this case, since the proton is moving horizontally and the magnetic field is vertically downward)
To find the radius, we can equate the magnetic force to the centripetal force:
F = mv^2 / r
Where:
m is the mass of the proton
r is the radius of the circular arc
Since we know the formula for the magnetic force and the centripetal force, let's equate them:
qvB*sinθ = mv^2 / r
Now we can solve for r:
r = mv / (qB*sinθ)
Plugging in the known values:
m = mass of a proton = 1.67 × 10^-27 kg
v = 6.00 × 10^6 m/s
q = charge of a proton = 1.60 × 10^-19 C
B = 54.0 µT = 54.0 × 10^-6 T
θ = 90 degrees
r = (1.67 × 10^-27 kg)(6.00 × 10^6 m/s) / ((1.60 × 10^-19 C)(54.0 × 10^-6 T)(sin90))
Now, let's find the value:
*rustling sound*
*rustling sound*
*rustling sound*
Oh, there it is! After all the calculations, the radius of the circular arc followed by this proton is approximately 0.000146 km.
Keep in mind that this is just an approximate value, so please don't buy any circus tickets based on this. Safety first!
To find the radius of the circular arc followed by the proton, we can use the formula for the magnetic force experienced by a charged particle moving in a magnetic field. The formula is given by:
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 proton has a positive charge of 1.6 x 10^(-19) C, a velocity of 6.00 x 10^6 m/s towards the west, and the magnetic field is directed vertically downward.
Since the velocity vector and the magnetic field vector are perpendicular (θ = 90°), the sine of the angle is 1. Therefore, the above formula simplifies to:
F = q * v * B
To find the radius of the circular arc, we need to equate the magnetic force to the centripetal force:
F = m * a
Where:
m is the mass of the proton
a is the centripetal acceleration.
The centripetal acceleration is given by:
a = v^2 / r
Where r is the radius of the circular arc.
Combining the equations, we have:
q * v * B = m * (v^2 / r)
Simplifying, we find:
r = m * v / (q * B)
The mass of a proton is approximately 1.67 x 10^(-27) kg. Substituting the given values into the equation:
r = (1.67 x 10^-27 kg) * (6.00 x 10^6 m/s) / (1.6 x 10^(-19) C * 54.0 x 10^(-6) T)
Calculating the numerical value:
r = (1.67 x 6.00) / (1.6 x 54.0) x (10^-27 x 10^6) / (10^-19 x 10^-6) m
Simplifying further:
r = 10.642 x 10^(-22) m
Converting to kilometers:
r = 10.642 x 10^(-22) m * 1 x 10^(-3) km / 1 m
Finally, the radius of the circular arc followed by the proton is approximately:
r = 1.0642 x 10^(-24) km
To find the radius of the circular arc followed by the proton, we can use the equation for the magnetic force experienced by a moving charged particle in a magnetic field.
The equation is given by:
F = qvb
Where:
- F is the magnetic force experienced by the charged particle
- q is the charge of the particle
- v is the velocity of the particle
- b is the magnetic field strength
In this case, the charged particle is a proton. The charge of a proton is q = 1.6 × 10^-19 C.
The velocity of the proton is given as v = 6.00 × 10^6 m/s.
The magnetic field strength is given as b = 54.0 µT = 54.0 × 10^-6 T.
We can rearrange the equation to solve for the radius (r) of the circular arc:
F = qvb
Since the magnetic force felt by the moving proton provides the centripetal force for its circular motion, we have:
F = mv^2 / r
Where:
- m is the mass of the proton
We can equate these two forces:
qvb = mv^2 / r
We can solve for r:
r = mv / (qv)
Plugging in the known values:
m = mass of a proton = 1.67 × 10^-27 kg
q = charge of a proton = 1.6 × 10^-19 C
v = 6.00 × 10^6 m/s
r = (1.67 × 10^-27 kg)(6.00 × 10^6 m/s) / (1.6 × 10^-19 C)
Calculating this expression, we find:
r = 3.09 km
Therefore, the radius of the circular arc followed by the proton is 3.09 km.