Charge in Magnetic Fields - Question... (1 Viewer)

[Nuendo]

A single positively charged particle, mass 4.6x10^-27kg, enters a 4.0T magnetic field into the page at 2.5x10^6m/s. The field covers an area of 0.1mx0.1m. It enters at 90 degrees, 0.05m up from the bottom left corner.

a) Calculate the force on the particle while it is in the field.
b) Calculate the radius of the path the particle takes in the field.
c) Calculate the speed with which the particle exits the field.
d) Where did the particle exit the field?

 

[k02033]

a) F=qvb

b) First force due to B fields are always directed perpendicular to the velocity. And so this force will cause the particle to travel in circular motion. so we can say that force due to B field provides centripetal acceleration for the particle. F=mv^2/r And the another key is since the force is always perpendicular it will only change the direction of the velocity and not its magnitude. much like how gravity does not vary the horizontal speed but changes the direction of overall velocity. and so the particle will travel in the circle with the speed that it entered the B field with. plug in m and v solve for r

c) exits with what it entered with

d)0.05 +2r from bottom left

 

[k02033]

oh and u need to say that gravity on the particle is negligible due to its small mass.

 

[Nuendo]

I'm still a little unsure about two things...

For a), how do you solve for the force when you're not given the charge?

And for d) I keep getting 0.086m (using 0.018m as my r from part b)

 

[k02033]

when you're not given the charge?

you are not? then i am confused too

 

[k02033]

and how did you get r without knowing the charge? you will need q,v,b, and m in order to solve for r

 

[Nuendo]

and how did you get r without knowing the charge? you will need q,v,b, and m in order to solve for r

I just used the answers in the back of the book (surfing physics - shadwick *cringe*). I used his answer for r to solve the rest of the question.
I understand all that stuff about equating the magnetic force with that of centripetal, so I think my only problem is solving part a :mad1: - the rest is straight forward.

 

[Aerath]

Um, Shadwick left it out - but I assumed it was a proton? So 1.6 x 10^-19 C? =\
So:
a) 1.6 x 10^-12 N
b)1.8cm (or 0.018 m)
c) Speed is constant
d) 0.086m (Nuendo - you are correct).

 

[Tully B.]

I don't understand (d), though maybe that's because I haven't done any questions for "Ideas to Implementation" yet. I wouldn't have been able to guess how to do (b), but I do understand the process...

 

[Aerath]

I don't understand (d), though maybe that's because I haven't done any questions for "Ideas to Implementation" yet. I wouldn't have been able to guess how to do (b), but I do understand the process...

Tully: Question (d)

 

[Tully B.]

Tully: Question (d)

What's with the 0.5 in 0.5 + 2r? I can sort of get how it would be 2r...

 

[Aerath]

A single positively charged particle, mass 4.6x10^-27kg, enters a 4.0T magnetic field into the page at 2.5x10^6m/s. The field covers an area of 0.1mx0.1m. It enters at 90 degrees, 0.05m up from the bottom left corner.

Questions states that it starts 0.05m up from bottom left corner.

 

[Tully B.]

Ohhhhhhhh, I get it. Enters 0.05m from the bottom, use right hand rule to find direction, already know radius. Thanks for the help.