A singly charged positive ion has a mass of 3.20*10^_26kg . after being accelerated from rest through a p.d of 833v,the ion enter magnetic field of 0.920T a long a direction perpendicular to the direction of the field.cal.the radius of the path.
a. compute the KE of the first pd.
b. calculte the force of the magnetic field on the ion
c. that force has to equal the centripetal force, mv^2/r, so calculate r, knowing velocity and mass.
5.6m
To find the radius of the path of the ion, we can make use of the equations for the centripetal force and the Lorentz force.
1. Determine the charge of the ion
Since the ion is singly charged and positive, its charge is +1.6 x 10^-19 C. (Note: Coulomb)
2. Calculate the velocity of the ion
We can use the equation for the potential difference (V) and kinetic energy (K.E.) to find the velocity (v) of the ion.
K.E. = qV
Where q is the charge of the ion and V is the potential difference.
Since the ion starts from rest, its initial kinetic energy is zero.
Therefore, the equation becomes: (1/2)mv^2 = qV
Rearranging the equation for v, v = sqrt((2qV)/m)
Plugging in the values, we have v = sqrt((2 * 1.6 x 10^-19 C * 833 V) / 3.20 x 10^-26 kg)
3. Calculate the radius of the path
The Lorentz force (F) acting on the ion in a magnetic field (B) is given by F = qvB.
This force provides the centripetal force required to keep the ion moving in a curved path.
Therefore, we can equate the Lorentz force to the centripetal force.
qvB = (mv^2) / r
Rearranging the equation, r = mv / (qB)
Plugging in the known values, we have r = (3.20 x 10^-26 kg)(sqrt((2 * 1.6 x 10^-19 C * 833 V) / 3.20 x 10^-26 kg)) / (1.6 x 10^-19 C)(0.920 T)
4. Solve for r
By evaluating the above expression, we can find the value of r, the radius of the ion's path.