Conceptual Test CH 33: Electromagnetic Induction

- Magnetic field is going out of the page --> Current is ccw - Magnetic flux is decreasing (because as the bar moves upward, the area is decreasing), therefore the induced current will combat it by moving ccw and magnetic field will be going out of the page aka increase the decreasing field. Book solution: The induced current will be counterclockwise (ccw). As the bar moves upward through the constant-magnetic- field region, the area of the loop decreases, so the flux through the loop decreases. By Lenz's Law, any induced current will tend to oppose this decrease. A ccw current will create a stronger magnetic field inside the loop, tending to increase the flux.

You need to push against a repulsive force. In its initial position, the loop encircles essentially no magnetic flux, whereas once the loop is positioned between the poles of the magnet, magnetic flux will pass through the loop. Lenz's Law tells us that the current flows in the direction required to oppose a change in flux, so the current will flow in the direction required to create a repulsive force on the copper wire. The current induced in the loop is therefore in the direction shown (into the page on the left side of the copper loop). The magnetic poles of the induced current loop are also shown in the diagram. The resulting magnetic force on the loop is to the right. Note that if the magnetic poles were reversed, the induced current would reverse as well so that the resulting force would still be to the right.

Yes, there is a net upward force. Lenz's Law tells us that a current will be induced in the loop to oppose the change in magnetic flux. The net force will be upward because the upper loop segment traverses the field B whereas the lower loop segment does not (the two side segments produce forces of equal magnitude but in opposite directions). We know the force must be upward because, by Lenz's Law, it must oppose the force due to gravity that is causing the change in flux.

a. The loop of wire will have a cw current to combat the increasing magnetic field out of the page. b. There will be no current because a changing magnetic flux is necessary to induce a current. c. The loop of wire will have a ccw direction to help the decreasing magnetic field.

a. There is no current. No magnetic flux is changing, so no magnetic field is induced. b. Cw. An increasing counterclockwise current in the lower loop creates an increasing upward- pointing magnetic field in the upper loop. Thus, a downward magnetic field is induced, resulting from a cw current in the upper loop. c. There will be no current because the current through the lower loop is constant. No changing in magnetic flux in the upper loop. d. Ccw. Immediately after the switch is closed the magnetic field will be decreasing in the lower loop. Thus, a upward magnetic field is induced, resulting from a ccw current in the upper loop.

a. Clockwise from above. The increasing flux through the loop is due to the upward-pointing field of the north pole of the magnet, so the induced magnetic field is downward. b. Yes, upward. The induced downward field is like a magnet with its north pole on the bottom, which is repulsed by the bar magnet. c. Yes, downward. By Newton's third law, the bar magnet is pushed down.

There is no current induced in the loop. As the bar magnet is pushed upward, the net magnetic flux through the loop is zero. In other words, all flux lines that enter the loop will, by symmetry, also exit the loop. Note that the magnet must be under the center of the loop for this symmetry argument to hold.

a. Left to right. The bar magnet's field inside the coil points from right to left. The flux through the coil is increasing. B induced will point left to right, and is caused by a current flowing left to right through the meter. b. No current. The flux through the coil is not changing. c. Right to left. Flux through the loop decreases as the magnet is withdrawn. Since B original is right to left, B induced is left to right. It is caused by a current flowing right to left through the meter by the RHR.

The magnetic field is out of the page and therefore has a current moving ccw. The given induced current is cw, suggesting that the magnetic field strength is increasing.

The energy stored in an inductor is U = (1/2)LI^2. To increase the energy to 2U, the current must increase by a factor of sqrt(2). Thus the required current is sqrt(2)*(2.0 A) = 2.8 A.

a. No. Since ?V = ?LdI /dt and ?V and L are both known, we can only find dI/dt, not I. b. Yes, through the right-hand inductor, since dI/dt = ?V/L. c. Yes, it is possible to tell: The current is decreasing. If dI/dt < 0 then ?VL > 0 so the input side is more negative and the potential from the induced current increases in the direction of the original current (see discussion accompanying Figures 33.40 and 33.41).

?a = L/R ?c = ?b > ?a For circuit b, note that the two inductors are in series. Recall that the inductance of an inductor is given by its turn density N/l. Because two identical inductor in series will have the same turn density but double the number of turns, the total inductance will be doubled. Another way to look at this is by considering the expression for inductance, which is proportional to N2/l. For two inductors in series we double N and l, so the total inductance increases by a factor of two.

a. Zero. The potential across the resistor is zero, and there is a zero initial current because the current in the inductor cannot change instantaneously. b. After the switch has been closed for a long time, the current in the inductor reaches the steady state (dI/dt = 0) so there is no voltage drop across the inductor. Kirchhoff's loop law then gives: ? - IR = 0 --> I = ?/R = 10V/5? = 2A