A singly charged positive ion has a mass of 2.30x10^-26kg. After being accelerat

Question

A singly charged positive ion has a mass of 2.30x10^-26kg. After being accelerated through a potential difference of 234v, the ion enters a magnetic field of 0.465T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field. A singly charged positive ion has a mass of 2.30x10^-26kg. After being accelerated through a potential difference of 234v, the ion enters a magnetic field of 0.465T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field.

Explanation / Answer

First we need to know the speed of the ion after it has been accelerated.

The ion is singly charged, meaning it has a charge of e (the elementary charge) = 1.60 * 10^-19 V

Using W = QV, we can find the energy given to the ion after it has passed through the potential difference.

W = QV
= 1.60 * 10^-19 C * 234 V
= 3.744 * 10^-17J

This energy is in the form of kinetic energy, so we can work out the speed of the ion.

E_k = 0.5 m v^2, so rearranging the equation we get:
v = sqrt( 2 E_k / m)

so using the value for energy we worked out before, and putting in the value for the mass, we get:
v = sqrt ( 2 * 3.744 * 10^-17 / (2.30 * 10^-26))
= sqrt3255652174
= 24961.34...
= 57058.32 ms^-1

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In a magnetic field a charged ion will experience a force, as the ion moves in a direction perpendicular to the field the force will always be perpendicular to the ions direction of motion. The magnitude of the force will not change, and so the ion will be made to move in a circular path. This means we need to bring in the equations for motion in a circle.

The equation for circular motion is:
F = mv^2 / r

The force experienced by the ion in the field is:
F = BQv

Equating the two we end up witih:
BQv = mv^2 / r

Upon simplifying and solving for r:
r = mv / (BQ)

We now know all these values, so substitute them in and evaluate to find out r.

r = 2.30 * 10^-26 kg * 57058.32 ms^-1 / (0.465 T * 1.60 * 10^-19 C)
= 0.01763
= 0.0176 m
or
1.76 cm

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