Video: Motion of a Charged Particle in a Magnetic Field

A cosmic-ray electron moves at 7.5 × 10⁶ m/s perpendicular to Earth’s magnetic field at an altitude where the field strength is 1.0 × 10⁻⁵ T. What is the radius of the circular path the electron follows? Electron mass is 9.1 × 10⁻³¹ kg and the electron charge is −1.6 × 10⁻¹⁹ C.

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Video Transcript

A cosmic-ray electron moves at 7.5 times 10 to the sixth meters per second perpendicular to Earth’s magnetic field at an altitude where the field strength is 1.0 times 10 to the negative fifth tesla. What is the radius of the circular path the electron follows? Electron mass is 9.1 times 10 to the negative 31st kilograms and the electron charge is negative 1.6 times 10 to the negative 19th coulombs.

In this scenario, we have a charged particle and electron, moving along with a speed we can call 𝑣, in a direction perpendicular to a magnetic field surrounding the particle. Because this electron is a charged particle in a magnetic field, that means that, as it moves along, it will experience a magnetic force. And if we use our right-hand rule to figure out the direction of that force, then because the motion of positive charge — remember this is an electron — is in this direction that our green arrow shows and the magnetic field is into the page, that means the overall magnetic force direction on our negatively charged electron will be to the right.

As this force pushes the electron in that direction, something interesting starts to happen. The electron starts to move in a circular arc. And it will keep moving in that circular path if the magnitude of this attractive force towards the center of the circle remains constant.

Speaking of this force, which we’ve called 𝐹, we can tell a little bit more about it actually, because it’s due to the magnetic field acting on a charged particle. It’s a magnetic force. And based on that, we can recall a mathematical relationship between magnetic force and some of the parameters of this example. The magnitude of magnetic force acting on a charged particle with charge 𝑞 is equal to that charge multiplied by the velocity of the particle times the strength of the magnetic field that it’s in. And not only is this force a magnetic force, but, as we’ve seen, it’s a center-seeking or centripetal force as well. It causes the electron to move in a circular path.

So not only is 𝐹 sub 𝐵 equal to 𝑞 times 𝑣 times 𝐵, but since it’s also a centripetal or center-seeking force, it’s also equal to the mass of the particle in circular motion multiplied by its velocity squared divided by the radius of the circle. So we can do something really great here. Because our magnetic force is a centripetal force, we can say that 𝑞𝑣𝐵 is equal to 𝑚𝑣 squared over 𝑟, where 𝑞 is the charge of our electron, 𝑣 is its velocity, 𝐵 is the strength of the magnetic field the electron is in, 𝑚 is its mass, and 𝑟 is the radius of the circle that it moves in. Our goal is to solve for the radius of that circle. And we can do it by rearranging this equation to isolate 𝑟. If we multiply both sides of our equation by 𝑟 and divide both sides by 𝑞 times 𝑣 times 𝐵, then we get this expression for the circle radius. And we see that a factor of 𝑣 cancels from top and bottom.

So if we know the mass of the electron, its speed, its charge, and the strength of the magnetic field that it moves in, then we can solve for the radius 𝑟. As part of the problem statement, we’re given each of these four values. 𝑚, the electron’s mass, is 9.1 times 10 to the negative 31st kilograms. 𝑣, its speed, is 7.5 times 10 to the sixth meters per second. 𝑞, its charge, is negative 1.6 times 10 to the negative 19th coulombs. And 𝐵, the magnetic field, is 1.0 times 10 to the negative fifth tesla. If we plug all these values into our expression, we’re just about ready to calculate 𝑟, except for one thing. Notice the charge of our electron is a negative value.

So if we enter this expression on our calculator, we’ll come up with a negative radius. But that doesn’t make sense. The radius is clearly positive. What we’ll do is we’ll keep our charge magnitude but will take its absolute value, so it returns a positive number. This does no injustice to the accuracy of our result and actually delivers us from a physically unrealistic answer. So, without one caveat, we calculate 𝑟 and find that, to two significant figures, it’s 4.3 meters. That’s the radius of the circular arc in which this electron moves.

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