Moving fast proton of 4x106 m / s through a 1.7 T magnetic field experiences a force
8x10-13N magnetic. What is the angle between the velocity vector associated with the proton and the magnetic field?
To find the angle between the velocity vector and the magnetic field, we can use the formula for the magnetic force experienced by a charged particle moving through a magnetic field:
F = q(v x B)
Where:
F represents the magnitude of the magnetic force,
q is the charge of the particle,
v is the velocity vector of the particle, and
B is the magnetic field vector.
In this case, we are given:
F = 8x10^-13 N
q = charge of proton = 1.6x10^-19 C
v = velocity of proton = 4x10^6 m/s
B = 1.7 T
To find the angle θ, we can rearrange the formula:
F = qvBsin(θ)
Rearranging for θ, we have:
θ = sin^(-1)(F / qvB)
Plugging in the given values:
θ = sin^(-1)(8x10^-13 N / (1.6x10^-19 C * 4x10^6 m/s * 1.7 T))
Now, let's calculate the angle θ:
θ = sin^(-1)(8x10^-13 / (1.6x10^-19 * 4x10^6 * 1.7))
θ = sin^(-1)(0.25)
Using a calculator, we can find the inverse sine function:
θ ≈ 14.5 degrees
Therefore, the angle between the velocity vector associated with the proton and the magnetic field is approximately 14.5 degrees.
To determine the angle between the velocity vector associated with the proton and the magnetic field, we can use the equation for the magnetic force on a moving charged particle:
F = q * v * B * sinθ
Where:
- F is the magnitude of the magnetic force on the particle,
- q is the charge of the particle,
- v is the velocity vector of the particle,
- B is the magnetic field strength, and
- θ is the angle between the velocity vector and the magnetic field.
In this case, we are given:
- F = 8x10^-13 N,
- q = charge of a proton = 1.6x10^-19 C (Coulombs),
- v = 4x10^6 m/s (velocity of the proton), and
- B = 1.7 T.
We need to find θ.
Rearranging the equation, we get:
sinθ = F / (q * v * B)
Substituting the given values:
sinθ = 8x10^-13 N / (1.6x10^-19 C * 4x10^6 m/s * 1.7 T)
Now we can calculate sinθ:
sinθ ≈ 0.2955
To find θ, we can use the inverse sine (sin^-1) function. Taking sin^-1 of both sides:
θ ≈ sin^-1(0.2955)
Using a calculator or trigonometric tables, we find:
θ ≈ 17.2 degrees
Therefore, the angle between the velocity vector associated with the proton and the magnetic field is approximately 17.2 degrees.