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Taking into account that my value for r is incorrect, is this the correct approach? If I plugged in the correct value for r, this would work? Thanks for all the assistance so far!

So, generically speaking, r = \sqrt{s^2 + <x,y>^2} where <x,y> will change depending on which segment I'm considering. I could integrate, B(<x,y>) = \frac{\mu_0 I}{4 \pi} \int^{a}_{0} \frac{ds}{|r|^2} using the generic value for r which will produce an equation that gives B as a function of...

Does r = \sqrt{s^2 + \frac{a^2}{16}}? --edit-- Correction: r = \sqrt{s^2 + \frac{x^2}{16}}

If I'm understanding correctly, B(\hat r) = \int \frac{\mu_0 I}{4 \pi} \frac{dl \times \hat r}{|r|^2} dl = ds \hat r = <x,y> Thus, B(<x,y>) = \frac{\mu_0 I}{4 \pi} \int \frac{ds \times \hat r}{|<x,y>|^2} ds \times \hat r = ds Integrate from 0 to a, B(x,y) = \frac{\mu_0 I}{4 \pi}...

I not sure what value to use for r in Biot-savart. Could I use a double integral and integrate over r as well considering that it is a changing value?

Homework Statement What is the magnitude and direction of magnetic field at point P shown in the figure attached if i = 10 A and a = 8.0 cm? Express the answer in vector notation. Homework Equations B = \frac{\mu_0I}{2\pi d} The Attempt at a Solution I'm really not sure how to...

I'm not sure I understand, Are you saying that q_1 E_{net} = v_1 and that the same is true for q2 and v2... so, v_1+v_2 =\Delta V Therefore w = Q \Delta V

Homework Statement Referring to the figure attached, how much work must be done to bring a particle, of charge Q = +16e and initially at rest, along the dashed line from infinity to the indicated point near two fixed particles of charges q1 = +4e and q2 = -2e? Distance d = 1.40 cm, theta...

That was the problem.. I was using the wrong values for the constants... I'll work it out and let you know if I have any more problems. Thanks for the help!

I got it at this point. Thanks for all your help. And just for the record.. you were picturing it right.

z_2 = 3.00 meters

http://www.wolframalpha.com/input/?i=int%28int%28-3b+dy%2C+1%2C+0%29dx%2C+1%2C+4+%29 http://www.wolframalpha.com/input/?i=int%28int%283b+dy%2C+1%2C+0%29dx%2C+1%2C+4+%29

Something isn't working out. Each side cancels the other side... including the sides which involve b. Looking at the left face, d \vec A = -i dy dz and \vec E \cdot d \vec A = -(10.00 + 2.00x) dy dz Integrating the left and right face each over z_1 to z_2 and 0 to y_2 gives 48 and...

So... I need to determine d \vec A and dot it with \vec E for each face of the shape. Integrate to remove the differentials, then sum the result of each integration to solve for b? Am I finally on the right track? Thank you so much for your help so far.

So... square the i and j components, sum them, and take the square root?

If I'm understanding correctly, \vec E \cdot d \vec A = 10.00 + 2.00x dy dz \hat i Correct?

Would it be, \int^{z_1}_{z_2} \int^{y_2}_{0} -2.00 \hat j + bz \hat k dy dz Sorry if I seem like a fish out of water.. but I actually haven't had multivariable calc... it isn't supposed to be required for the class ...

So I have: E_{NET} = \frac{k}{R^2} [(Q_1 - Q_2 cos \vartheta) \hat i - Q_2 sin \vartheta \hat j] I'm not sure how the pythagorean theorem comes into play though... sin^2 \vartheta + cos^2 \vartheta = 1

So you're saying that I won't use i, j, k for all components? Only those applicable. For example, sticking with the right face, I would use all of them (as it's in i, j and and k) but for the bottom I'd only use i and k? Thanks.

Homework Statement The box-like Gaussian surface of the attached figure encloses a net charge of +24.0 \epsilon_0 \mu C and lies in an electric field given by \vec E = [(10.0 + 2.00x) \hat i - 2.00 \hat j + bz\hat k]N/C with x and z in meters and b a constant. The bottom face is in...

Thanks for your help! Wouldn't E_1 be in the -\hat i direction? Concerning E_2, |E_2| (cos\vartheta -\hat i + sin\vartheta \hat j) I would take these and add them (vector wise) for an equation with theta to solve for?

Homework Statement The Figure below shows a plastic ring of radisu R = 50.0cm. Two small charged beads are on the ring. Bead 1 of charge +2.00 micro coulombs is fixed in the place at the left side of the ring. Bead 2 of charge +6.00 micro coulombs can be moved along the rignt. The two...

I think I see it now. mv = \frac{2nh}{r_n} Subbing mv into equation from above r_n = \frac{mv}{qB} I get r_n = \frac{2nh}{r_nqB} A bit more simplification yields r_n = \sqrt{\frac{2nh}{qB}} Is that correct?

\frac{mv^2}{r_n} = qVB sin \vartheta sin \vartheta = 0 as the angle of the particle with the B field is 90 degrees. So, simplifying I get, r_n = \frac{mv}{qB} How do I tie in this equation with the above?

Homework Statement A particle of charge q and a mass m, moving with a constant speed v, perpendicular to a constant magnetic field B, follows a circular path. If in this case the angular momentum about the center of this circle is quantized so that mvr_n = 2nh, determine the allowed radii for...

Great! I'll give this a shot and hope for the best. Thank you for your assistance.

Thanks... I believe this give me just what I needed! The assignment is due shortly so I will work out what I think and hopefully I won't hit any more roadblocks. Thank you for all of your help.

So it is as simple as: B = \frac{mg}{N I l} ?

The only relationship I can find where coils come into play is with the flux and emf. \epsilon = -N \frac{d \Phi_B}{d t} Even working on that equation a little doesn't give me any good relationship defining how N affects the B field. Is there an equation I'm missing?

It all makes since now. When I take d/dt of the equation, y becomes dy/dt which is velocity and adding a minus sign to both sides of the equation gives me the emf. I can get the current using I = \frac{\epsilon}{R} Thank you SO much! However, I do still have a question. There were...

So when I integrate (using limits from a to L), I get \Phi = \frac{\mu_0 I y}{2 \pi} \left[ ln(L) - ln(a) \right] I still do not know what y is really. Can I substitute y = \vec{v} = \frac{dy}{dt}. If I do that... how do I take the dt back out? It kind of seems to throw off what I'm...

Homework Statement The figure attached illustrates a setup that can be used to measure the magnetic field. A rectangular coil of wire contains N turns and has a width w. The coil is attached to the one arm of a balance and is suspended between the poles of a magnet. the field is uniform...

I appreciate your help so far... If I'm understanding correctly, The B field at distance r from the wire is equal to \frac{\mu_0 I}{2\pi r} The flux for a horizontal strip in the loop is d \Phi = B y dr = \frac{\mu_0 I}{2\pi r} y dr. If I integrate over the height of the loop, then I am...

The only direct equation in the book is #1 from my first post involving both B and dA which I only have one of... Trying to use some imagination here.. but what about this: \int \epsilon dt = \Phi_B \int IR dt = \Phi_B using I from the long wire... and R from the loop (?) but it really...

Homework Statement The figure attached shows a rod of length L = 10 cm that is forced to move a a constant speed of v = 5.00 m/sec along horizontal rails. The rod, rails, and connecting strip at the right form a conducting loop. The rod has a resistance of 0.4 \Omega and the rest of the loop...

Ok... so Forces for cylinder: Linear m_Agsin\vartheta - T = m_Aa => T = m_Agsin\vartheta -m_Aa Radial \tau = R \times F = R m_Aasin\vartheta = I_c\alpha => \alpha = \frac{R m_Aasing\vartheta}{I_c} Forces for block: T- m_Bg = m_Ba =>...

I'm not sure I follow what you mean by set up equations for each mass . My best guess leads me back to the same world I've been trapped in for nearly 3 hours... \tau = R \times F = Rm_Agsin\vartheta + T F_B = ma = m_Bg -T .:. Rm_Agsin\vartheta = -m_Bg There's no unknown in this equation...

So wouldn't that mean that the torque of the tape around the cylinder affect the tension in the tape? Would the torque generated by the cylinder apply a force to the tape?

Well mathematically \tau = r \times F ... with r being radius and F being force I suppose that means that torque is force applied radially. So wouldn't that mean that the torque of the tape around the cylinder affect the tension in the tape? Also, does the Force in the equation for torque...

Does that change your suggestions?

Well... I have had trouble with Torque vs Force... So I'm guessing \alpha = \frac{\tau}{I_c} isn't valid? If it is.. how to we calculate tau? Concerning Tension... Would T_{Net} = m_Agsin\vartheta - m_Bg ?

I've attached a picture of the set up... just for reference. Thanks for the assistance.

The force of the tension in the tape should be subtracted from the force of gravity compelling the cylinder to roll down the incline. So \tau = m_Ag sin\vartheta - T where T is tension. Correct? I'm calculating T = m_Bg as the m_B is free hanging and pulling on the tape. Wouldn't the...

Ok... scratch what I said earlier. \alpha = \frac{\tau}{I_c} \alpha = \frac{a_t}{R} So.. \frac{\tau}{I_c} = \frac{a_t}{R} which means \tau R = a_t I_c (*) A force diagram tells me \tau = m_Ag sin\vartheta - m_Bg The cylinder rolls down the incline \vartheta under the influence of...

Homework Statement A solid cylinder of weight 50 lb and radius 3.0 inches has a light thin tape wound around it. The tap passes over a light smooth fixed pulley to a 10 lb body hanging vertically in the air. If the plane on which the cylinder moves is inclined 30 degrees to the horizontal...

You're value for acceleration is fine in the y direction. How did you derive your equation for time?

Thanks for all the help and advice! I think we can mark this one off as solved. =D

I did state at the end of the reasoning that I'm assuming LA is negative, by virtue of the same reason... If LA was hitting in the same direction of LE then it would be positive. Perhaps the way I went about stating it was incorrect, but I did realize that LA was to be subtracted in the actual...

Ah... I forgot to factor in R_E... the missing length factor. It is part of the position vector. Good catch. It made it to the final solution. But it needs to be woven throughout the reasoning I suppose. I'll make edits.

I think I have finally got it! Using the picture it is easy to see the triangle formed between where the asteroid starts, where it strikes and CME. This gives us L_A = r_{AE} \times p_A = M_A V_A R_E sin \theta which is the angular momentum of the Asteroid w.r.t. CME. The angular...