A proton moves at right angles to a magnetic
field of 0.0609 T with a speed of 8.16×10
6 m/s.
Find the magnitude of the acceleration of
the proton.
Answer in units of m/s
2
To find the magnitude of the acceleration of the proton, we can use the equation:
a = qvB/m
Where:
a is the acceleration,
q is the charge of the proton (1.6 x 10^-19 C),
v is the speed of the proton (8.16 x 10^6 m/s),
B is the magnetic field (0.0609 T),
and m is the mass of the proton (1.67 x 10^-27 kg).
Plugging in the given values, we can calculate the acceleration:
a = (1.6 x 10^-19 C) * (8.16 x 10^6 m/s) * (0.0609 T) / (1.67 x 10^-27 kg)
The charge of the proton and the speed are both positive values, which means the proton will experience a centripetal acceleration towards the center of the circular motion.
Simplifying the equation:
a = 7.52 x 10^13 m/s^2
Therefore, the magnitude of the acceleration of the proton is approximately 7.52 x 10^13 m/s^2.
To find the magnitude of the acceleration of the proton, we can use the formula for the force on a charged particle moving in a magnetic field:
F = q * (v x B)
Where:
F is the force on the proton
q is the charge of the proton
v is the velocity of the proton
B is the magnetic field
Since the proton has a positive charge (q = +1.6 x 10^-19 C), the direction of the force on the proton will be perpendicular to both the velocity and the magnetic field.
Now, let's calculate the magnitude of the acceleration using the formula:
a = F / m
Where:
a is the acceleration of the proton
F is the force on the proton
m is the mass of the proton
Since the mass of a proton (m) is 1.67 x 10^-27 kg, we can substitute the values into the formulas:
F = q * (v x B)
F = (1.6 x 10^-19 C) * (8.16 x 10^6 m/s) * (0.0609 T)
To compute F, we need to calculate the cross product of the velocity (v) vectors and the magnetic field (B) vectors:
v x B = v * B * sin(theta)
Where:
v is the velocity of the proton
B is the magnetic field
theta is the angle between v and B
Since the proton moves at right angles to the magnetic field (theta = 90 degrees), sin(theta) will be equal to 1. Therefore:
F = q * v * B
Now, let's substitute the values and calculate F:
F = (1.6 x 10^-19 C) * (8.16 x 10^6 m/s) * (0.0609 T)
After calculating, we find that F is equal to 7.4496 x 10^-14 N.
Now, let's calculate the acceleration using the formula:
a = F / m
a = (7.4496 x 10^-14 N) / (1.67 x 10^-27 kg)
After calculating, we find that the magnitude of the acceleration of the proton is approximately 4.459 x 10^12 m/s^2.
Therefore, the magnitude of the acceleration of the proton is 4.459 x 10^12 m/s^2.