A singly charged positive ion has a mass of 5.2x10^-27 kg. After being accelerated from rest through a potential difference of 750V, the ion enters a magnetic field of 0.65T. Calculate the radius of the path of the ion in the field.
eU=mv²/2
v=sqrt{2eU/m} = sqrt{2•1.6•10⁻¹⁹ •750/5.2•10⁻²⁷) =6.7•10⁴ m/s
mv²/R=evB
R=mv/eB =5.2•10⁻²⁷•6.7•10⁴/1.6•10⁻¹⁹•0.65 = 3.4•10⁻³ m
To calculate the radius of the path of the ion in the magnetic field, we can use the formula for the radius of a charged particle in a magnetic field:
r = (m*v) / (Q*B)
Where:
r - radius of the path
m - mass of the ion
v - velocity of the ion
Q - charge of the ion
B - magnetic field strength
First, let's find the velocity of the ion. We know that the ion was accelerated through a potential difference of 750V, so we can use the equation for the kinetic energy of a charged particle:
K.E = q*V
Where:
K.E - kinetic energy
q - charge of the ion
V - potential difference
Since the ion is a singly charged positive ion, q = 1.6 x 10^-19 C (charge of an electron) and V = 750V. Therefore,
K.E = (1.6 x 10^-19 C)*(750V)
K.E = 1.2 x 10^-16 J
Next, we can use the kinetic energy to find the velocity of the ion. The kinetic energy is also equal to 1/2 times the mass times the velocity squared:
K.E = (1/2)*m*v^2
Rearranging the equation to solve for velocity:
v = sqrt((2*K.E)/m)
Plugging in the values:
v = sqrt((2*(1.2 x 10^-16 J))/(5.2 x 10^-27 kg))
v = 1.92 x 10^6 m/s
Now we have the velocity of the ion. We can use this value along with the charge of the ion (Q), the magnetic field strength (B), and the mass of the ion (m) to calculate the radius of the path.
Plugging in the values into the formula:
r = ((m*v) / (Q*B))
r = ((5.2 x 10^-27 kg)*(1.92 x 10^6 m/s)) / ((1.6 x 10^-19 C)*(0.65 T))
r = 1.96 x 10^-3 m
Therefore, the radius of the path of the ion in the magnetic field is approximately 1.96 x 10^-3 meters.