A charged particle isinjected into a uniform magnetic field such that its velocityvector is perpendicular to themagnetic field vector. Ignoring the particle's weight, the particlewill

A) follow a spiralpath.
B) move in a straight line.
C) move along a parabolic path.
D)follow a circular path.

A charged particle is injected into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vector. Ignoring the particle's weight, the particle will follow a circular path.

Option D

Explanation:

Magnetic force causes charged particles to move in spiral paths. The Particle accelerators keep the protons to follow circular paths when it is in the magnetic field. Velocity has a change in direction but magnitude remains the same when this condition exists.

The magnetic force exerted on the charged particle is given by the formula:

where

q is the charge

v is the velocity of the particle

B is the magnetic field

is the angle

In this problem, the velocity is perpendicular to the magnetic field vector, hence

= and sin =sin 90 degree = 1.

So applying the formula,

the force is simply

Also, the force is perpendicular to both B and v and so according to the right-hand rule, we have:

This means that the force acts as a centripetal force, so it will keep the charged particle in a uniform circular motion.

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).