Charge Acted By Magnetic Force AND Friction

[madmax101]

Homework Statement

Part 1: A charge particle having charge q enters a region in which there is a friction force proportional to the speed of the particle. The constant of proportionality of friction is "b". The particle stops traveling a distance of 10m, in a straight line.

Now, the experiment is repeated, but with magnetic field of unknown strength, in a direction perpendicular to entry velocity of the particle. The particle is found to stop at a point 6m from the entry point, then find the strength of magnetic field.

Part 2:The magnetic field is now doubled in magnitude. How far from the point of entry the particle come to rest?

Homework Equations

F=ma
F=qVB
S=ut+.5at^2
v^2=u^2 +2as
Perhaps something else as well??

The Attempt at a Solution

For the first case, the net force on the particle in the first case is simply the algebraic sum of the force due to the magnetic field and the friction i.e. F=quB-bu where B is magnetic field, u is the velocity, q= charge of the particle, now considering F=ma, we get a=(quB-bu)/m (where m is the mass of the particle), then we can use v^2=u^2+2as (a is retardation so V^2=u^2-2as)

so 0=u -2(qB-b)10/m (as we are given tht s=10m)
so mu=20 (qB-b) ------------->1

When the field is made perpendicular to the particle, the magnetic field will, at every point acts tangentially to the circular path of the particle, and the frictional force acts in a dirction exactly opposite to this ...atleast thts what I think

Tht's all I've been able to get upto ...haven't even started off on the second part!

Respected sirs, I'm a 17 yr old student, so please go easy on me if you feel that this problem is too simple or if my method is completely wrong

cheers!
Thanks for your help

 

[Jhero]

with this problem! I am a scientist who is responding to your forum post.

Part 1:

Your approach to finding the strength of the magnetic field is correct. We can use the equation F=quB-bu to find the acceleration of the particle, and then use the equation v^2=u^2+2as to find the velocity at the point where the particle comes to rest. Since we know the distance traveled (10m) and the initial velocity (u), we can solve for the acceleration:

a = (quB-bu)/m

Now, we can use this acceleration to find the final velocity at the point where the particle comes to rest:

v^2 = u^2 + 2as

Solving for v, we get:

v = √(u^2 + 2as)

Plugging in the values we know, we get:

v = √(u^2 + 2a(10)) = √(u^2 + 20(quB-bu)/m)

Since we know the final velocity is 0, we can set this equation equal to 0 and solve for B:

0 = √(u^2 + 20(quB-bu)/m)

Squaring both sides and rearranging, we get:

(u^2 + 20(quB-bu)/m)^2 = 0

Simplifying, we get:

u^2 + 20(quB-bu)/m = 0

Solving for B, we get:

B = (bu^2)/(20qu)

Plugging in the values we know, we get:

B = (bu^2)/(20qu) = (qbu)/(20mu)

Now, we can solve for B using the values given in the problem:

B = (qbu)/(20mu) = (q(6)(u))/(20m(10))

B = (3qu)/(200m)

So, the strength of the magnetic field is (3qu)/(200m).

Part 2:

To find how far the particle will travel when the magnetic field is doubled in magnitude, we can use the same approach as before. The only difference is that now the magnetic field is twice as strong, so we can substitute 2B for B in our equations. This will give us the new acceleration and final velocity of the particle. We can then plug these values into the equation v^2=u