An electron has approximate mass 9 * 10^-31 kg and charge of magnitude 1.6 * 10^-19 C.
An electron moves in the horizontal direction at velocity 1.9 * 10^6 m/s within a uniform magnetic field of .004 Tesla , with the field directed vertically upward. Describe its path and give its radius of curvature.
A charge q moving at velocity v in a magnetic field B will experience a force of magnitude F = q v B sin(`theta), where `theta is the angle between the velocity and magnetic field vectors. The direction of the force is perpendicular to both the velocity and the field, and is given by the right-hand rule, where the fingers of the right hand point in the direction of the velocity and the palm is oriented perpendicular to the magnetic field, opening in the direction of the field.
In the present case the velocity and the field are perpendicular so that sin(`theta) = 1, and F = q v B. Here the charge q is the charge -1.6 * 10^-19 C of an electron.
We thus have
The force on the moving charge will be perpendicular to its velocity, which is in a horizontal direction, and to the vertical magnetic field. The force must therefore be in the horizontal direction (since it is perpendicular to the vertical magnetic field) but perpendicular to the direction of the velocity. This force has no component in the direction of the velocity and therefore has no effect on the speed of the particle. The only effect of the force is therefore to change the direction of motion. This results in circular motion at the original speed, with a constant centripetal force of .01216 * 10^-13 N. The since initial velocity and force are both entirely in a horizontal plane the circle will also be in this plane.
The centripetal acceleration is therefore
Since centripetal acceleration = v^2 / r, we see that the radius of the resulting circle is
If the particle was a proton instead of an electron its mass would be nearly 2000 times as great. The magnitude of its acceleration would be the same as for an electron, since it carries the same charge and is assumed to have the same velocity, so its acceleration would be less by a factor of nearly 2000. The radius v^2 / a of the circle would therefore be about 2000 times that of the electron, or about 5.412 meters.
The directions of the forces will be opposite, due to the opposite charges on the particles. Thus one particle will circle in a clockwise direction while the other circles in a counterclockwise direction. The precise directions are found using the right-hand rule.
In general a mass m carrying charge q moving at velocity v in a magnetic field B, with velocity perpendicular to field, will move in a circle in response to the force F = q v B, which will be the centripetal force for the motion. The circle will be in a plane perpendicular to the magnetic field.
The centripetal acceleration will be F / m = q v B / m, which will be equal to v^2 / r, where r is the radius of the circle. Thus we have
q v B / m = v^2 / r,
which we easily solve for r to obtain
radius of circle: r = v * m / q.
Thus if we can observe the circular track left by a particle with known charge and velocity we can determine the mass of the particle.
We see that the radius is proportional to the mass of the particle (as for the proton vs. the electron in the present example), and also to its velocity; and is inversely proportional to the charge. Thus, for example, doubly charged ions of the same element, which have identical velocities, will have exhibit just about half the radius of curvature of singly charged ions moving at the same velocity.
It is worth noting that a charged particle passing through a combined electric and magnetic field, with particle velocity, electric field and magnetic field mutually perpendicular, will experience equal and opposite forces from the two fields and will therefore pass through the fields undeflected. Thus we can create a beam of particles of known velocity without knowing anything about their charges or masses.
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