Magnitude and direction of uniform magnetic field

AI Thread Summary
To ensure the proton passes through undeflected in the presence of a uniform electric field, the magnetic force must equal the electric force acting on it. The electric force can be calculated using F = qE, where q is the charge and E is the electric field strength. The magnetic force is given by F = qvB, where v is the velocity and B is the magnetic field strength. By setting these forces equal, the required magnetic field can be determined using the relationship B = E/v. The discussion emphasizes the importance of understanding the interplay between electric and magnetic fields in particle motion.
mwadhwa
Messages
2
Reaction score
0

Homework Statement


A proton, with mass 1.67*10^-27 kg and charge +1.6*10^-19C, is sent with velocity 5.0*10^4 m/s in the x-direction into a region where there is uniform electric field of magnitude 390 V/m in the y direction. What is the magnitude and direction of the uniform magnetic field in the region, if the proton is to pass through undeflected? Assume that the magnetic field has no x-component. Neglect gravitational effects.


Homework Equations





The Attempt at a Solution


I have no idea where to even begin with this one! Any helps is greatly appreciated!
 
Physics news on Phys.org
Think about controling velocity for a mass spectrometer. You know the E field and Velocity, how do they relate to the B field? (charge cancels in the balanced equation.)
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Find the magnitude of the electromagnetic force on electron?
Replies
3
Views
2K
How Does a Proton Behave in a Magnetic Field?
Replies
16
Views
6K
Solving Two Charged Plates Homework: Magnetism, Frequency & Charge Density
Replies
3
Views
2K
Magnetic Field Strength Calculation for Electron in Uniform Field
Replies
14
Views
6K
Understanding work in translation and rotation of a magnetic dipole
Replies
1
Views
2K
Does a Bent Wire Affect Induction Calculations?
Replies
7
Views
2K
The right hand rule in a magnetic field
Replies
2
Views
2K
Magnetic flux- is this the correct way to solve it?
Replies
3
Views
1K
Rate of Change of Magnetic Field in Square wire loop
Replies
5
Views
3K
Magnetic flux. Dont know where to start
Replies
1
Views
1K

Hot Threads

Recent Insights

Back
Top