An electron follows a helical path in a uniform magnetic...

An electron follows a helical path in a uniform magnetic field of magnitude $0.300 \mathrm{~T}$. The pitch of the path is $6.00 \mu \mathrm{m}$, and the
magnitude of the magnetic force on the electron is $2.00 \times 10^{-15} \mathrm{~N}$. What is the electron's speed?

Key Concepts

Magnetic Force as Centripetal Force

For the component of the particle's motion that is perpendicular to the magnetic field, the magnetic force acts as the centripetal force required to keep the particle in its circular path. This establishes the relation q v_perp B = m v_perp^2 / r, which is used to relate the particle’s speed and the radius of its circular motion.

Pitch of a Helix

The pitch of a helical path is the linear distance that the particle travels along the magnetic field during one complete revolution around the circular path. It is directly related to the parallel component of the particle’s velocity and the period of the circular motion, as pitch = v_parallel × period.

Lorentz Force

The Lorentz force is the fundamental force acting on a charged particle moving in a magnetic field. It is given by F = q(v × B), meaning that the force is perpendicular to both the velocity of the particle and the magnetic field. This perpendicularity causes the particle to follow a curved, often circular or helical, path rather than accelerating along the field direction.

Helical Motion

Helical motion occurs when a charged particle has a velocity component both parallel and perpendicular to a uniform magnetic field. The perpendicular component causes the particle to undergo circular motion due to the Lorentz force, while the parallel component allows the particle to drift along the direction of the magnetic field, creating a helical trajectory.

Transcript

00:01 In this problem, it is given that an electron follows a helical path like this.

00:08 And we have given some data like the magnetic field in magnitude b is given, which is 0 .30 tesla.

00:21 And we have given the pitch which is 6 micrometer.

00:29 That means from this to this distance is equal to pitch.

00:34 Which is given 6 micrometer and we have given magnetic force which is equal to 2 into 10 to the power minus 15 newton now we have to find the electron speed since electron is moving in a helical path that means it has two component of velocity one is parallel and one is perpendicular this side parallel component we know that time period for a electron in a helical path 2 pi m divided by q into b this is the time period and we have given the pitch which will be equal to we know that distance is equal to velocity into time now velocity into time time is equal to 2 pi m by q b and pitch is given to us so from here we will get the the parallel component of velocity which is p into q into b divided by 2 pi m now which is given to us which is 6 micrometer and the charge of electron is 1 .6 into 10 to the power minus 19 and the magnetic field is given 0 .30 tesla divided by 2 pi m the mass of electron is 9 .11 into 10 to the power minus 31 kg so from this we get the parallel component of velocity which is equal to around after solving this 50 .3 into 10 to the power 3 meter per second this is the parallel component now for the vertical component we have given q force magnetic force will be equal to qv cross v so here v perpendicular into b is equal to force from here we get the perpendicular component of velocity which is equal to f upon e into b f is given which is 2 into 10 to the power minus 15 divided by e 1 .6 into 10...

Please give Ace some feedback