A positron (q = +e) moves at 5.00 × 10^7 m/s in a magnetic field of magnitude 0.850 T. The magnetic force on the positron has a magnitude of 3.20 × 10^−12 N.
What is the component of the positron’s velocity perpendicular to the magnetic field?
I found the answer! I will post it here for anyone else who may also have issues figuring it out.
V(perpendicular) = F/qB
V(perpendicular) = (3.20 x 10^-12)/ [(1.602 x 10^-19)(0.85)]
V(perpendicular) = 2.35 x 10^7 m/s
To find the component of the positron's velocity perpendicular to the magnetic field, we can use the formula for the magnitude of the magnetic force on a charged particle:
F = q v B sin(theta)
where F is the magnitude of the magnetic force, q is the charge of the particle, v is its velocity, B is the magnitude of the magnetic field, and theta is the angle between the velocity and the magnetic field.
We can rearrange this formula to solve for sin(theta):
sin(theta) = F / (q v B)
Plugging in the given values:
F = 3.20 × 10^−12 N
q = +e (the charge of the positron is the elementary charge, e = 1.60 × 10^−19 C)
v = 5.00 × 10^7 m/s
B = 0.850 T
sin(theta) = (3.20 × 10^−12 N) / ((1.60 × 10^−19 C) * (5.00 × 10^7 m/s) * (0.850 T))
Now we can calculate sin(theta):
sin(theta) = 0.3011764706
To find theta, we can take the inverse sine (sin^(-1)) of sin(theta):
theta = sin^(-1)(0.3011764706)
Using a calculator, we find:
theta = 17.50 degrees
Therefore, the component of the positron's velocity perpendicular to the magnetic field is 17.50 degrees.
To find the component of the positron's velocity perpendicular to the magnetic field, we need to use the formula for the magnetic force on a charged particle:
F = qvBsinθ
where F is the magnitude of the magnetic force, q is the charge of the particle, v is the velocity of the particle, B is the magnitude of the magnetic field, and θ is the angle between the velocity vector and the magnetic field vector.
From the given information, we can see that the magnetic force (F) is 3.20 × 10^−12 N and the charge (q) is +e, which is equivalent to 1.6 × 10^−19 C.
The magnitude of the magnetic field (B) is given as 0.850 T.
Now, we can rearrange the formula to solve for the angle (θ):
θ = arcsin(F / (qvB))
θ = arcsin((3.20 × 10^−12 N) / ((1.6 × 10^−19 C)(5.00 × 10^7 m/s)(0.850 T)))
Calculating this expression will give us the angle θ.
Finally, to find the component of the positron's velocity perpendicular to the magnetic field, we can use the formula:
v_perpendicular = v * sinθ
Substituting the given values, including the angle θ found above, will give us the answer.