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    Group of particles in a magnetic field

    Ok.. I followed your right hand rule steps to find the direction, thank you
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    Group of particles in a magnetic field

    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...
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    Direction of magnetic field and moving particle

    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...
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    Direction of magnetic field and moving particle

    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.
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    Direction of magnetic field and moving particle

    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...
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    Direction of magnetic field and moving particle

    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...
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    Direction of magnetic field and moving particle

    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...
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    Electric Field Produced by a Finite Charged Wire

    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}...
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    Electric Field Produced by a Finite Charged Wire

    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...
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    Find the drift velocity

    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...
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    Find the drift velocity

    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 *...
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    Find the drift velocity

    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...
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    Field of an infinite plane sheet of charge

    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?
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    Field of an infinite plane sheet of charge

    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...
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    Finding magnitude and direction of an electric field

    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...
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    Statics:Find the Tension in a rope holding a boom

    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...
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    Statics:Find the Tension in a rope holding a boom

    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...
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    Where to place a weight on a bar so that the net torque is 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...
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    Where to place a weight on a bar so that the net torque is 0

    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.
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    Where to place a weight on a bar so that the net torque is 0

    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...
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    Conservation of energy/Finding maximum height

    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...
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    Find work given a pulley and two weights

    ok. Thank you for your help! I will do some more two weights/one pulley problems to make sure i got it
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    Find work given a pulley and two weights

    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...
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    Find work given a pulley and two weights

    Yes that is the figure, sorry i forgot to add it. I'll add it now. Never mind i waited too long to edit. But yes that is the figure.
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    Find work given a pulley and two weights

    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...
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    I don't understand force as a function of time

    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.
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    I don't understand force as a function of time

    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...
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