A singly charged positive ion has a mass of 2.74 10-26 kg. After being accelerated through a potential difference of 268 V, the ion enters a magnetic field of 0.450 T, in a direction perpendicular to the field. Calculate the radius of the ion's path in the field.
cm
mv²/2 =eU
v=sqrt(2eU/m)=…
mv²/R=evB
R= mv/eB=…
To calculate the radius of the ion's path in the magnetic field, we can use the formula for the radius of a particle moving in a magnetic field:
r = (m * v) / (q * B)
Where:
- r is the radius of the path
- m is the mass of the ion
- v is the velocity of the ion
- q is the charge of the ion
- B is the magnetic field strength
First, let's find the velocity of the ion. We can use the formula for kinetic energy:
K.E. = (1/2) * m * v^2
Since the ion has been accelerated through a potential difference of 268 V, the kinetic energy gained is equal to the potential energy lost:
K.E. = q * V
Thus, we can equate these two equations:
(1/2) * m * v^2 = q * V
Now, we can rearrange the equation to solve for v:
v = sqrt((2 * q * V) / m)
Next, we can substitute the given values into the equation. The charge q of a singly charged positive ion is +e, where e represents the elementary charge, approximately 1.60 x 10^-19 C.
q = +e = +1.60 x 10^-19 C
V = 268 V
m = 2.74 x 10^-26 kg
Plug these values into the equation to find v:
v = sqrt((2 * (1.60 x 10^-19 C) * (268 V)) / (2.74 x 10^-26 kg))
Now, let's solve for v:
v ≈ 2.40 x 10^6 m/s
Finally, substitute the calculated velocity and the given values for m, q, and B into the equation for the radius:
r = (m * v) / (q * B)
r = ((2.74 x 10^-26 kg) * (2.40 x 10^6 m/s)) / ((1.60 x 10^-19 C) * (0.450 T))
Solve the equation to find the radius:
r ≈ 3.24 x 10^-2 m
Finally, convert the radius to centimeters by multiplying by 100:
r ≈ 3.24 x 10^-2 m * 100 cm/m
r ≈ 3.24 cm
Therefore, the radius of the ion's path in the magnetic field is approximately 3.24 cm.