Homework Statement
A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50 km/s in the +x-direction experiences a force of 2.25 x ##10^{-16}##N in the +y-direction, and an electron moving at 4.75 km/s in the -z-direction...
What I meant is, I had the force direction as up. toward the ceiling. So when I did the right hand rule, my palm was facing toward the ceiling.
I did this with the charge as positive. But the charge is negative, so I read the force is pointing in the opposite direction. So then I put my palm...
The magnitude of ##F_b## is positive.. but since there was a negative point charge, the direction of ##F_b## is down.. so doing right hand rule, my fingers point East.. so the B field is in the East direction.
Ok I think i get it,
We if ##sin(\theta)## gets smaller, B has to get bigger because q is constant so B or v would have to get bigger. So by making ##sin(\theta)## as big as possible, B and v will be as small as possible.
My attempt at right hand rule: thumb points up toward...
oops..I read the book wrong.
It says the direction of the force is always perpendicular to the velocity vector and B field. And I realize that if the z-axis is north, then any vector on the x-y plane is perpendicular to north.
By making the angle 90 degrees, we get the maximum magnitude of the...
Homework Statement
A particle of mass 0.195 g carries a charge of -2.50 x 10-8C. The particle is given an initial horizontal velocity that is due north and has magnitude 4.00 x 104 m/s. What are the magnitude and direction of the minimum magnetic field that will keep the particle moving in the...
it worked, I did the same integral but put limits -L and L like this:
##\int_{-L}^L \, \frac{dx} {(x^2 + d^2)^{3/2}} ##
## = [\frac {L} {d^2 \sqrt{(L^2 + d^2)}} - \frac {-L} {d^2 \sqrt{((-L)^2 + d^2)}}]##
## = \frac {2L} {d^2 \sqrt{(L^2 + d^2)}} ##
final answer: ##E = k\lambda d ( \frac {2L}...
Homework Statement
A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E⃗ at the point P, a distance d above the midpoint of the wire.
What is the magnitude E of the electric field at point P? Throughout this part, express...
Ok I see what you mean .. so the correct cross sectional area is:
A = π((2.05)2 x 10-3*2) / 4 = 3.30 x 10-6
J = 5 / 3.30 x 10-6 = 1.51 x 106 A / m2
So I know my J and |q| are correct..I also forgot to mention that the problem gave "Copper has 8.5 x 1028 free electrons per cubic meter" which I...
b) What is the current density?
cross section area = A = πr2 = πd2 / 4 = π(2.05x106) / 4 = 1.61 x 10-6
I = 5 A
Then current density = J = I / A = 5 / 1.61 x 10-6 = 3.105 x 106
..
Also the very first part:
a) How many electrons pass through the lightbulb each second?
I = 5A = 5 C/s
5 C/s *...
Homework Statement
A 5.00-A current runs through a 12 gauge copper wire (diameter 2.05mm) and through a light bulb. Copper has 8.5 x 1028 free electrons per cubic meter.
c) At what see does a typical electron pass by any given point in the wire?
Homework Equations
J = nqvd
The Attempt at a...
So there is only one electric field line that goes straight though the center of the cylinder? Because for every other electric field, there will exist another that is equal in magnitude but opposite in sign?
Homework Statement
Use Gauss's law to find the electric field caused by a thin, flat, infinite sheet with a uniform positive surface charge density σ.
Homework Equations
φ(flux) = ∫E*dA = qencl / ε0
The Attempt at a Solution
I set up a diagram like this...
Homework Statement
A spaceship encounters a single plane of charged particles, with the charge per unit area equal to σ. The electric field a short distance above the plane has magnitude _____ and is directed _____ to the plane.
a) σ/2∈0, parallel
b) σ/2∈0, perpendicular
c) σ/∈0, parallel...
the horizontal distance would be Lcos(θ), but why are we treating L as a hypotenuse?
In my diagram I reoriented the boom horizontal and then i did sinθ to find the Fy of each force and then calculated the torques that way.
I am confused when you say "you need a horizontal distance from the...
Homework Statement
The boom in the figure below (Figure 1) weighs 2450 N and is attached to a frictionless pivot at its lower end. It is not uniform; the distance of its center of gravity from the pivot is 35.5 % of its length.
Find the tension of the guy wire.
Homework Equations
Στ=0
∑F=0...
Ok so then I have another force that is the weight of the bar, which is at d=0.295m.
Ok using this information and the previous post this is what I did.
Στ=0=0.05gd+0.10g(d-0.295)-0.10g(0.599-d)
rearranging I got to
0.15d - 0.0295 = -0.10d + 0.0599
0.25d = 0.0894
d = 0.3576 =35.8 cm
which...
the 0.05 kg mass is said to be on the left end, the 0.10 kg mass is said to be on the right end. That is all that is specified. And I assume they do not mean at the very end otherwise the bar would be in motion.
Homework Statement
A 0.100-kg, 59.9-cm-long uniform bar has a small 0.055-kg mass glued to its left end and a small 0.100-kgmass glued to the other end. You want to balance this system horizontally on a fulcrum placed just under its center of gravity.
no diagram given
Homework Equations
Στ = 0...
Homework Statement
A 28-kg rock approaches the foot of a hill with a speed of 15 m/s. This hill slopes upward at a constant angle of 40.0∘ above the horizontal. The coefficients of static and kinetic friction between the hill and the rock are 0.75 and 0.20, respectively.
a)Use energy...
I am unsure if we are supposed to assume the blocks start from rest. I think what that means is the initial velocity and final velocity aren't needed to solve this problem.
I also did away with my original equations for tension after looking at...
Homework Statement
Problem 6.65
Two blocks are connected by a very light string passing over a massless and frictionless pulley (Figure (Figure 1) ). The 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm downward.
Question: Find the total work done on 20.0-N block if...
Thank you I used this information to solve both parts of the problem. I use the integral of acceleration and of velocity to find time and then used that time to find position.
Homework Statement
A 2.70 kg box is moving to the right with speed 9.50 m/s on a horizontal, frictionless surface. At t = 0 a horizontal force is applied to the box. The force is directed to the left and has magnitude F(t)=( 6.00 N/s2 )t2
What distance does the box move from its position at...