A beta particle (fast moving electron) is released by a nuclear decay and starts travelling at 1.5x10^6 m/s [NΘE] in the Earth's magnetic field which has a value of 3.0x10^-5 T. If the force experienced by the beta particle is 4.34x10^-18 N, determine the angle between the direction of travel and the Earth's magnetic field. What is the direction of the force acting on the particle?
Using the formula f=qvBsin(theta):
the f value would be 4.34x10^-18 N
the B value would be 3.0x10^10^-5 T
the v value would be 1.5x10^6 m/s
the question asks for the angle (sin theta)
but I don't know what to put in place of q
q is the electron charge
-1.6*10^-19 Coulombs
You are supposed to know that, The charge on a proton is equal and opposite in sign.
In the equation f = qvBsin(theta), 'q' represents the charge of the beta particle. The charge of an electron is typically -1.6 x 10^-19 Coulombs.
Using this value for 'q', we can calculate the angle (theta) between the direction of travel and the Earth's magnetic field.
Let's substitute the values into the equation:
4.34 x 10^-18 N = (-1.6 x 10^-19 C) * (1.5 x 10^6 m/s) * (3.0 x 10^-5 T) * sin(theta)
Now, solve for sin(theta):
sin(theta) = (4.34 x 10^-18 N) / [(-1.6 x 10^-19 C) * (1.5 x 10^6 m/s) * (3.0 x 10^-5 T)]
sin(theta) ≈ -0.0904
To determine the angle, take the inverse sine (sin^-1) of -0.0904:
theta ≈ sin^-1(-0.0904) ≈ -5.2 degrees
Therefore, the angle between the direction of travel and the Earth's magnetic field is approximately -5.2 degrees.
Now let's determine the direction of the force acting on the particle:
Since the force experienced by the beta particle is given as 4.34 x 10^-18 N, in the given equation, the force is perpendicular to the velocity (v) and the magnetic field (B). The force will act in a direction perpendicular to both, following the right-hand rule. The direction of the force can be determined by using the right-hand rule:
Curl the fingers of your right hand in the direction of the velocity (v), then curl them toward the magnetic field (B). The direction in which your thumb points is the direction of the force acting on the particle.
Therefore, the direction of the force acting on the beta particle would be in the direction perpendicular to both the velocity and the magnetic field out of the screen or into the page.
To find the angle between the direction of travel and the Earth's magnetic field, we need to determine the value of q which represents the charge of the beta particle (fast-moving electron).
The formula f = qvBsin(theta) relates the force experienced by the particle (f), the charge of the particle (q), its velocity (v), the magnetic field strength (B), and the angle between the direction of travel and the magnetic field (theta).
Given that f = 4.34x10^-18 N, B = 3.0x10^-5 T, and v = 1.5x10^6 m/s, we can rearrange the formula as:
sin(theta) = f / (qvB)
We can now solve for q:
q = f / (vBsin(theta))
Since we're looking for the angle and not the charge, we need to re-arrange the formula to solve for sin(theta):
sin(theta) = f / (qvB)
Now, plug in the given values:
sin(theta) = (4.34x10^-18 N) / [(1.5x10^6 m/s)(3.0x10^-5 T)]
Calculate the value of sin(theta) using a calculator. Once you have the value of sin(theta), you can use the inverse sine function (sin^(-1)) to find the angle theta.
As for the direction of the force acting on the particle, it can be determined by the right-hand rule. If the direction of the magnetic field is known, we can use the right-hand rule to determine the direction of the force acting on a positive charge moving in that magnetic field. The direction of the force will be perpendicular to both the velocity of the particle and the magnetic field, according to the cross-product rule.