A proton moves at a speed of 5.9 × 107
m/s at right angles to a magnetic field with a
magnitude of 0.095 T.
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 for the magnetic force on a charged particle:
F = qvB
where:
F = magnetic force
q = charge of the particle
v = velocity of the particle
B = magnetic field strength
Since the proton carries a charge of +1.6 x 10^-19 C, and its velocity is perpendicular to the magnetic field, the angle between the velocity and the magnetic field is 90 degrees.
So, the magnetic force acting on the proton can be given as:
F = qvB sin(90°)
Since sin(90°) = 1, the equation simplifies to:
F = qvB
The force experienced by the proton is equal to the centripetal force:
F = (mv^2) / r
where:
m = mass of the proton
v = velocity of the proton
r = radius of the circular path
We can equate the two expressions for force:
qvB = (mv^2) / r
Simplifying, we get:
qvB = mv^2 / r
From this equation, we can find the magnitude of the acceleration (a) of the proton:
a = v^2 / r
To find the magnitude of the acceleration, we need to determine the radius of the circular path. Given that the velocity is perpendicular to the magnetic field, the proton will move in a circular path with a radius that is determined by the force balance between the magnetic force and the centripetal force:
qvB = (mv^2) / r
Simplifying further, we have:
r = (mv) / (qB)
Substituting the given values:
m = mass of a proton = 1.67 x 10^-27 kg
v = velocity of the proton = 5.9 x 10^7 m/s
q = charge of a proton = 1.6 x 10^-19 C
B = magnetic field strength = 0.095 T
Let's substitute these values into the equation to find the radius:
r = [(1.67 x 10^-27 kg) x (5.9 x 10^7 m/s)] / [(1.6 x 10^-19 C) x (0.095 T)]
r = 0.000610 m ≈ 6.1 x 10^-4 m
Now, we can substitute the value of the velocity and the radius into the equation for acceleration to find the magnitude of the acceleration:
a = (5.9 x 10^7 m/s)^2 / (6.1 x 10^-4 m)
a = 3.587 x 10^18 m/s^2
Therefore, the magnitude of the acceleration of the proton is approximately 3.587 x 10^18 m/s^2.
To find the magnitude of the acceleration of the proton, we can use the formula:
F = q * v * B
where:
F is the magnetic force on the proton,
q is the charge of the proton (which is the elementary charge, e, equal to 1.6 × 10^-19 C),
v is the velocity of the proton, and
B is the magnitude of the magnetic field.
The magnetic force acting on a charged particle moving perpendicular to a magnetic field can be written as:
F = m * a
where:
m is the mass of the proton, and
a is the acceleration of the proton.
Since we want to find the magnitude of the acceleration (|a|), we can set the equations for the magnetic force equal to each other:
q * v * B = m * |a|
Solving for |a|, we have:
|a| = (q * v * B) / m
Now let's substitute the given values into the equation and calculate the magnitude of the acceleration:
q = 1.6 × 10^-19 C (elementary charge)
v = 5.9 × 10^7 m/s (velocity of the proton)
B = 0.095 T (magnitude of the magnetic field)
m = mass of the proton = 1.67 × 10^-27 kg (approximately)
|a| = (1.6 × 10^-19 C * 5.9 × 10^7 m/s * 0.095 T) / 1.67 × 10^-27 kg
Calculating the expression on the right-hand side, we find:
|a| ≈ 5.2 × 10^15 m/s^2
Therefore, the magnitude of the acceleration of the proton is approximately 5.2 × 10^15 m/s^2.