A ribbon-shaped conductor is oriented so that it carries a current which flows from top to bottom. The conductor is placed in a magnetic field pointing out of the paper. Make two sketches showing how the Hall Effect can be used to tell whether positive or negative charges are carrying the current. Make it clear what tells you the sign of charge.
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A ribbon-shaped conductor is oriented so that it carries a current which flows from top to bottom. The conductor is placed in a magnetic field pointing out of the paper. Make two sketches showing how the Hall Effect can be used to tell whether positive or negative charges are carrying the current. Make it clear what tells you the sign of charge.
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- Q. Consider the configuration shown on the right, the inner cylindrical conductor of radius 'a' surrounded by a cylindrical shell of inner radius 'b' and outer radius 'c'. The inner conductor and the outer shell each carry equal and opposite currents I. Where I is uniformly distributed through the conductors. a I a) Find the magnetic field intensity for r< a and aMagnetic and electric field lines are in many ways similar, with few crucial differences. Which of the properties listed below does NOT apply to both fields? O All of these properties apply to both E and B field lines. The density of lines is proportional to the field strength in that area. Field lines never intersect. Tangent to the line at a point gives you the direction of the field as that point. O Field lines form closed loops.A docs.google.com/forms/d/e. In the Hall Effect, the magnetic field is in z direction and the velocity is in x direction. ?What is the direction of the electric field z O Y O What does p/T represent? Where T is life time holes per second lost time holes Measurement of Hall coefficient in a semiconductor provides information on :the Sign and concentration of charge carriers Sign of charge carriers alone Mass and concentration of charge carriersConsider a straight conductor of length 9.7 cm. The conductor moves at right angles to a magnetic field of uniform strength B = 10-3 T generating e.m.f. of 2.5 × 10-³ V. Calculate the velocity of the straight conductor. Give your answer in SI units. Answer: Choose... +A circular closed, conducting loop of radius r is in the presence of a uniform magnetic field that points into the page, shown in the figure below. The strength of the magnetic field changes as a function of time, which is described by the following expression: B(t) = B1t? + Bo. You may assume that B1 and Bo are both positive numbers. The direction of the magnetic field stays constant. The total resistance of the conducting loop is R. Use this information to solve parts (a) - (d). Write your answers in terms of known quantities such as: r, R, B1, Bo, and t. B(t) = B,t? + Bo %3D R r (a) Write an expression for the magnetic flux through the loop, assuming that the area vector of the loop points out of the page. Is the flux increasing or decreasing over time? (b) Determine the magnitude of the induced electromotive force driven through the loop. (c) Determine the magnitude of the induced current driven through the loop. (d) In which direction does the induced current flow (clockwise or…An electrically conductive rod is moving through a uniform magnetic flux at a constant velocity at right angles, as shown by its cross-section in the diagram. The velocity is in the x-direction, the rod is of length 150 mm along z, and the magnetic flux density is 0.2 T in the positive y-direction. Speed If the potential difference (voltage) across the ends of the rod is 5.8 Volts, what is the magnitude of the velocity? m/s Not perpendicular If the rod is passing through the flux with its axis not at right angles what can you be sure of? The speed to generate 5.8 V would have to be smaller. • The speed to generate 5.8 V would have to be bigger. The speed to generate 5.8 V would be the same.A square loop of wire with sides of 20.0cm is half in and half out of a uniform magnetic field which points out of the page as shown. The loop is moving with a speed of 3.00m/s and the strength of the magnetic field is 0.100Tesla. The total resistance of the loop is 4.00Ω. Find the magnitude and directionof the current if the loop is moving in the following directions. a.in the +x direction b.in the -x direction c.in the +y direction d.in the -y directionConsider a long, horizontal Large Wire with current of 10 A running through it. We want to levitate a horizontal, thin, 0.50 m length of wire above it. If the thin wire has a mass of 10 grams, and a current of 300 mA, how far above the Large Wire will it hover (net force of zero) due to magnetic and gravitational forces? A. If the thin wire hovers above the Large Wire due to their magnetic fields, are their currents going the same direction, or opposite directions. Explain. B. Draw a diagram and label the directions of currents, and all other relevant quantities and vectors. C. Find the distance above the Large Wire the small thin wire will hover (net force of zero). D. Would your answers to parts A and C change if we wanted to find a distance below (rather than above) the Large Wire that the smaller thin wire could hover, due to their magnetic fields. Explain. Don't calculate any values but draw a new diagram and explain how this situation compares to the problem above.only parts d and eA U-shaped conductor is locked in place while a vertically aligned, 3.0m-long cylindrical conductor slides across it to the left with a speed of 4.0m/s. The conductors maintain contact at all times forming a closed loop. An external magnetic field of 3.0T passes through the conductors as shown (in the image) A) At the moment the two conductors form a perfect square, what is the magnetic Flux passing between them? Is this Flux increasing or decreasing? B) What are the directions of the induced current and the induced magnetic field the conductors generate? Explain how you know your answers.Consider a straight conductor of length 15 cm. The conductor moves at right angles to a magnetic field of uniform strength B = 10-³ T generating e.m.f. of 2,5 x 10-5 V. Calculate the velocity of the straight conductor. Give your answer in Sl units. Answer: m/s ♦