Stuck on a few magnetic field questions

Click For Summary
The discussion centers on two problems involving magnetic fields and particle motion. In Problem 1, the participant initially attempted to find acceleration using an electric field, but was corrected that no electric field exists, and they should apply the Lorentz force directly. For Problem 2, the participant calculated linear momentum but encountered issues with the results, prompting a request for clarification on the equations used. They eventually reported a calculated linear momentum of 4.94×10^-18 kg·m/s for part a. The conversation emphasizes the importance of understanding the Lorentz force and classical mechanics in solving these problems.
NutriGrainKiller
Messages
60
Reaction score
0
Problem 1:
A particle with mass 1.81*10^-3 kg and a charge of 1.22*10^-8 C has, at a given instant, a velocity v = 3.00*10^4 m/s jhat.

What is the magnitude of the particle's acceleration produced by a uniform magnetic field B=1.63T ihat + 0.980T jhat?
here is what i did:

knew to find the acceleration I needed to use a=QE/M, but since I didn't have E I had to find it. Used F(mag field)=Q*V*B*Sin(theta). I found theta from drawing the ijz axis and using trig, I got 59 degrees.

After I solved this I plugged it into F=QE to get E. Then plugged that into a=QE/M, and got an incorrect answer..not sure where I went wrong.


Problem 2:
A particle with charge 6.40 *10^-19 C travels in a circular orbit with radius 4.68 mm due to the force exerted on it by a magnetic field with magnitude 1.65 T and perpendicular to the orbit.

a) What is the magnitude of the linear momentum P of the particle?

b) What is the magnitude of the angular momentum L of the particle?

for a:
equations I used:

P=MV
R=MV/QB -> V= RQB/M

plugged 2nd into 1st, M cancelled, answer wasn't correct. eh?


for b:

not sure i know the equation for this problem...same as in classical physics?

thanks guys!
 
Physics news on Phys.org
NutriGrainKiller said:
Problem 1:

here is what i did:

knew to find the acceleration I needed to use a=QE/M, but since I didn't have E I had to find it. Used F(mag field)=Q*V*B*Sin(theta). I found theta from drawing the ijz axis and using trig, I got 59 degrees.

After I solved this I plugged it into F=QE to get E. Then plugged that into a=QE/M, and got an incorrect answer..not sure where I went wrong. Problem 2: for a:
equations I used:

P=MV
R=MV/QB -> V= RQB/M

plugged 2nd into 1st, M cancelled, answer wasn't correct. eh?for b:

not sure i know the equation for this problem...same as in classical physics?

thanks guys!

1) There's nothing wrong with your first problem except that you seem to be insisting there's an electric field. There isn't. The Lorentz force says:
\mathbf{F}=q\mathbf{E}+q\mathbf{v}X\mathbf{B}. Since there is no E, the first term drops out. Then you calculated F, as you did. Now simply apply F=ma.

2) Part a) looks right, I'm not sure why things aren't working. What answer did you get and what does your book say? For part b) I would assume that you aren't being relativistic or quantum mechanical, so yes use the Classical formula.

-Dan
 
Last edited:
topsquark said:
1) There's nothing wrong with your first problem except that you seem to be insisting there's an electric field. There isn't. The Lorentz force says:
\mathbf{F}=q\mathbf{E}+q\mathbf{v}X\mathbf{B}. Since there is no E, the first term drops out. Then you calculated F, as you did. Now simply apply F=ma.

2) Part a) looks right, I'm not sure why things aren't working. What answer did you get and what does your book say? For part b) I would assume that you aren't being relativistic or quantum mechanical, so yes use the Classical formula.

-Dan

thanks, figured the first one out!

for 2a i got 4.94×10−18
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
10
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 10 ·
Replies
10
Views
8K