29A. Electromagnetic Induction

29a. Electromagnetic Induction

Conceptual Induction

Part A What is the direction of the induced current in the loop when the loop is above the solenoid, moving downward? clockwise Correct

Part B What is the direction of the induced current when the loop is at the midpoint of the solenoid and still moving downward? no currentCorrect

Part C What is the direction of the induced current when the loop is below the solenoid and moving downward? counterclockwise Correct

Introduction to Faraday's Law

Part A Consider the direction of the electric field in the figure. Assume that the magnetic field points upward, as shown. Under what circumstances is the direction of the electric field shown in the figure correct? if decreases with time Correct Part B Now consider the magnetic flux through a surface bounded by the loop. Which of the following statements about this surface must be true if you want to use Faraday's law to relate the magnetic flux to the line integral of the electric field around the loop? The surface can be any surface whose edge is the loop. Correct

You are free to take any surface bounded by the loop as the surface over which to evaluate the integral. The result will always be the same, owing to the continuity of magnetic field lines (they never start or end anywhere, since there are no magnetic charges). It is important to understand the vast differences between electric fields produced by changing magnetic fields via Faraday's law and the more familiar electric fields produced by charges via Coulomb's law. Here are some short questions that illustrate these differences. Part C When can an electric field be measured at any point from the force on a stationary test charge at that point? no matter how the field is generatedCorrect

In fact, this operation defines an electric field. Similarly, if the test charge is moving, it will measure magnetic fields. Part D When can an electric field that does not vary in time arise? in either of the above two cases Correct

Part E

When will the integral around any closed loop of the projection of the electric field along that loop be zero? only if the field is generated by the coulomb field of charges that are static or moving in a straight line Correct

The electric field generated by a uniformly moving charge is just the moving field of a static charge, which always has zero loop integral. Here is a simple quantitative problem that uses Faraday's law. Part F A cylindrical iron rod of infinite length with cross-sectional area is oriented with its axis of symmetry coincident with the z axis of a cylindrical coordinate system as shown in the figure. It has a magnetic field inside that varies according to . Find the theta component of the electric field at distance from the z axis, where is larger than the radius of the rod. Express your answer in terms of , , , , and any needed constants such as , , and . =(-A*B_1)/(2*pi*R)Correct

29b. Electromagnetic Induction

Rail Gun

Part A A conducting rod is free to slide on two parallel rails with negligible friction. At the right end of the rails, a voltage source of strength in series with a resistor of resistance makes a closed circuit together with the rails and the rod. The rails and the rod are taken to be perfect conductors. The rails extend to infinity on the left. The arrangement is shown in the figure.

There is a uniform magnetic field of magnitude , pervading all space, perpendicular to the plane of rod and rails. The rod is released from rest, and it is observed that it accelerates to the left. In what direction does the magnetic field point? out of the plane of the figureCorrect

Part B Assuming that the rails have no resistance, what is the most accurate qualitative description of the motion of the rod? The rod will accelerate but the magnitude of the acceleration will decrease with time; the velocity of the rod will approach but never exceed a certain terminal velocity. Correct

Part C What is the acceleration of the rod? Take to be the mass of the rod. Express your answer as a function of , , the velocity of the rod , , , and the mass of the rod . =(((V-v_r(t)*B*L)/R)*B*L)/mCorrect

Making the substitution , you obtain the dfferential equation

, which you can solve to find the velocity of the rod as a function of time:

To achieve a high acceleration, which is necessary for a useful gun, a magnetic field of large magnitude and a high voltage are advantageous. Part D What is the terminal velocity reached by the rod? =V/(B*L)Correct

A larger magnetic field increases the acceleration of the rod, but lowers the terminal velocity: a trade-off for rail gun engineers!

The Ampère-Maxwell Law

Part A First find , the line integral of around a loop of radius located just outside the left capacitor plate. This can be found from the usual current due to moving charge in Ampère's law, that is, without the displacement current. Find an expression for this integral involving the current and any needed physical constants such as .

=mu_0*I(t)Correct

Part B

Now find an expression for , the same line integral of around the same loop of radius located just outside the left capacitor plate as before. Use the surface that passes between the plates of the capacitor, where there is no conduction current. This should be found by evaluating the amount of displacement current in the Ampère-Maxwell law above.

Express your answer in terms of the electric field between the plates , , the plate area , and various physical constants.

=mu_0*epsilon_0*dE(t)/dt*ACorrect

A necessary consistency check Part C We now have two quite different expressions for the line integral of the magnetic field around the same loop. The point here is to see that they both are intimately related to the charge on the left capacitor plate. First find the displacement current in terms of . Express your answer in terms of , , and other physical constants. =dq(t)/dtCorrect

Part D Now express the normal current in terms of the charge on the capacitor plate . Express your answer in terms of , , and any needed physical constants. =dq(t)/dtCorrect

Using Gauss's law, you have shown that the displacement current from the changing electric field between the plates equals the current from the flow of charge through the wire onto that plate. This means that the Ampère-Maxwell law can consistently treat cases in which the normal current due to the flow of charge is not continuous. This realization was a great boost to Maxwell's confidence in the physical validity of his new displacement-current term.