A singly charged positive ion has a mass of
3.35 × 10−26 kg. After being accelerated
through a potential difference of 128 V, the
ion enters a magnetic field of 0.784 T, in a
direction perpendicular to the field.
Calculate the radius of the path of the ion
in the field.
Answer in units of cm
To calculate the radius of the path of the ion in the magnetic field, we can use the concept of the centripetal force experienced by a charged particle moving in a magnetic field.
The centripetal force acting on the ion is provided by the magnetic force:
F = qvB
where F is the centripetal force, q is the charge of the ion, v is its velocity, and B is the magnetic field strength.
The velocity of the ion after being accelerated through a potential difference can be calculated using the equation:
v = √(2qV/m)
where V is the potential difference, q is the charge of the ion, and m is its mass.
Now let's substitute the given values into the equations:
q = 1 (since it is a singly charged ion)
V = 128 V
m = 3.35 × 10^(-26) kg
B = 0.784 T
First, calculate the velocity of the ion:
v = √(2 * 1 * 128 / 3.35 × 10^(-26)) = 1.472 × 10^7 m/s
Next, we can calculate the centripetal force:
F = qvB = (1) * (1.472 × 10^7) * (0.784) = 1.152 × 10^7 N
Since the centripetal force is equal to the magnetic force, we can equate them:
mv^2 / r = qvB
Rearranging the equation, we get:
r = mv / (qB)
Plugging in the values we know:
r = (3.35 × 10^(-26)) * (1.472 × 10^7) / ((1) * (0.784))
Simplifying the equation:
r = (3.35 * 1.472) / (0.784) * 10^(-26+7-1)
r = (4.9272 / 0.784) * 10^(-20)
r = 6.28 * 10^(-20) m
Finally, to convert the result to centimeters, divide by 0.01 (since there are 100 centimeters in a meter):
r = (6.28 * 10^(-20)) / 0.01 = 6.28 * 10^(-18) cm
Therefore, the radius of the path of the ion in the field is 6.28 × 10^(-18) cm.