The velocity field in a potential flow is governed by the two kinematic equations grad*u=0 and grad×u=0. The same two equations, grad*B=0 and grad×B=0, govern the distribution of static magnetic fields, and indeed many of the potential flows discussed in mathematics texts were originally derived as magnetic field distributions by nineteenth-century physicists. Where does Newton's second law enter into such velocity distributions?
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The velocity field in a potential flow is governed by the two
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- A 25 cm rod moves at 5.6 m/s in a plane perpendicular to a magnetic field of strength 0.34 T. The rod, the velocity vector, and the magnetic field vector are mutually perpendicular, as indicated in the accompanying figure. Calculate: (a) The magnetic force on an electron in the rod. (b) The potential difference between the ends of the rod.A proton, that is accelerated from rest through a potential of 14.0 kV enters the velocity filter, consisting of a parallel-plate capacitor and a magnetic field, shown below. P E ☑ B ☑ ☑ The E-field between the parallel capacitor plates is 1.1.105 N/C. What B-field is required so that the protons are not deflected? (Ignore relativistic effects for high velocities.) Submit Answer Tries 0/12 Send FeedbackA physics lab instructor is working on a new demonstration. She attaches two identical conducting balls with mass m = 0.190 g to threads of length L as shown in the figure. There are two strings in the figure. The top of each string is connected to the ceiling, and both strings are connected at the same point. The bottom of each string is connected to a spherical mass labeled m. Both strings have length Land hang at an angle of θto the vertical, with the two strings on opposite sides of the vertical. Both balls have the same charge of 7.60 nC, and are in static equilibrium when θ = 4.55°. What is L (in m)? Assume the threads are massless. Draw a free-body diagram, apply Newton's second law for a particle in equilibrium to one of the balls. Find an equation for the distance between the two balls in terms of L and θ, and use this expression in your Coulomb force equation. m (b) What If? The charge on both balls is increased until each thread makes an angle of θ = 9.10° with the…
- A particle with unit charge (q = 1) enters a constant magnetic field B = i + j with velocity v = 21k. Find the magnitude and direction of the force on the particle. Make a sketch of the magnetic field, the velocity, and the force. What is the magnitude of the force on the particle? 21√√2 (Type an exact answer, using radicals as needed.) What is the direction of the force on the particle? Select the correct choice below and fill in the answer box to complete your choice. A. The force is applied at a(n) B. The force is applied at a(n) O c. The force is applied at a(n) angle with the positive x-axis in the xz-plane. angle with the positive x-axis in the xy-plane. angle with the positive y-axis in the yz-plane.A proton is accelerated in a particle accelerator from rest through a potential difference of 0.130 kV. If the proton then passes through a uniform magnetic field of magnitude 0.120 T that is perpendicular to the proton's path, what is the radius of the path that the proton will take through the magnetic field? Give your answer in cm. (The mass of a proton is 1.672*10^-27 kg.)Consider a charged particle moving through a region in which the electric field is perpendicular to the magnetic field, with both fields perpendicular to the initial velocity of the particle (see figure below). Such a device is called a velocity selector because for one particular value of the particle speed select the particle will travel straight through the device without being deflected, whereas particles with speeds that are either less than or greater than Vselect will be deflected. Consider a case with E = 505 V/m and B = 0.11 T. B Perspective view AE = 'select v>F'select (a) When protons are used, what speed is selected? m/s (b) What is the speed selected for Ca²+ ions? m/s (c) If you were to use your selector with Fions, what change would be necessary?
- After being accelerated to a speed of 1.92×105 m/s , the particle enters a uniform magnetic field of strength 0.800 T and travels in a circle of radius 34.0 cm (determined by observing where it hits the screen as shown in the figure). The results of this experiment allow one to find m/q. Find the ratio m/q for this particle. Express your answer numerically in kilograms per coulomb.Let's take the Earth's magnetic field BE = 1.70 x 10-4 T pointing north. A metal cable attached to a space station stretches by 1.4 km. What would be the potential difference ( ΔΔ V) that develops between the ends of the cable if it’s traveling eastbound around Earth with a speed 4.4 km/s?Diamagnets have the property that they "dampen" the effects of an external magnetic field by creating an opposing magnetic field. The diamagnet thus has an induced dipole moment that is anti- aligned, such that the induced north pole is closer to the north pole creating the external field. An application of this is that diamagnets can be levitated 2. Now, the mathematics of generally describing a force by a non-uniform field on a dipole is a little beyond the scope of this course, but we can still work through an approximation based on energy. Essentially, whenever the theoretical loss of gravitational potential energy from "falling" no longer can "pay the cost" of increasing the magnetic potential energy, the object no longer wants to fall. Suppose a diamagnetic object floats above the levitator where the magnitude of the magnetic field is 12 T, which is inducing" a magnetic dipole moment of 4.6 LA - m² in the object. The magnetic field 1.9 mm below the object is stronger with a…
- In vacuum, magnetic force on a moving charged particle provides an acceleration to the particle. Thus, it also changes the kinetic energy of the particle. O True O False The direction of magnetic force is perpendicular both to the velocity of the particle and to the magnetic field. True O False The net magnetic force acting on any closed current-carrying loop in a uniform magnetic field is zero. True O False All electric field lines are continuous and always form closed loops. OTrue FalseThe diagram above shows segments of two long straight wires, carrying currents. This time I1 = 6.46, I2 = 2.03 A, and the two wires are separated by 1.10 cm. Now consider the charge q = 6.74 x 10^-6 C, located a distance of 6.31 cm to the right of wire I2, moving to the right at speed v = 43.2 m/s. What is the magnitude of the total magnetic force on this charge? 3.34E-09 N 2.00E-09 N 6.95E-09 N 4.17E-09 NA charged particle travelling at right angles to a magnetic field undergoes helical motion, involving circular movement around the magnetic field lines. The radius of the circular motion is given by: Rc = where symbols have their conventional meaning (you may want to look at Section 5.1.3 of Unit 11 if you are unsure of these). From the list of possible units, associate the correct unit with each symbol in the equation by dragging and dropping to complete the table below. Symbol Unit Rc m m v |q| B V Ō |q| B kg ns-1 m S ms-2 kg s-2 A-1 с JC-1 N no units W m VA-1 C-1 V-1