9478 / 2027
Electromagnetic Induction overview

Topic 5 of 6

Magnetic braking and energy

A conductor moving into a field can develop a current whose magnetic force opposes the motion. Trace the changing linkage and the complete current path before predicting a braking force.

A rigid rectangular conducting loop moves right into a bounded uniform field directed out of the page. Initially only its right vertical edge is inside the field. The area of the loop exposed to B increases, so outward flux increases.

Lenz's law requires an induced inward field. Viewed from the front, current is therefore clockwise. On the right vertical edge it flows down; in the outward external field, the force on that current-carrying conductor points left. It opposes entry.

Entering the field increases the linked outward flux

Entering the field increases the linked outward fluxThe uniform field is out of the screen, represented by teal dots, only to the right of the fixed vertical boundary at x150. A complete rectangular conducting loop spans x65 to x215 and y130 to y270, so only its right vertical edge is in the field. The loop moves right at three metres per second. Induced conventional current is clockwise: right along the top, down the active right edge, left along the bottom and up the left return. The magnetic force on the right edge points left. A separate applied pull points right on the same horizontal line, keeping speed steady; these two force arrows have equal lengths. The horizontal-edge magnetic forces cancel vertically in the stated symmetric model.Field boundaryDots: B outv = 3.00 m/sFmagAppliedll = 0.200 m; I = 0.400 A

Brown arrows follow the complete induced-current path. The right-edge current is down, so its magnetic force is left. At steady speed the applied pull balances that force. Horizontal-edge forces cancel vertically; Fv = 0.0960 W is transferred to I squared R heating.

The complete loop is now inside the uniform field

The complete loop is now inside the uniform fieldThe same rectangular loop, unchanged in size, is translated right to span x165 to x315, entirely to the right of the same x150 field boundary. It still moves right at three metres per second. The outward field is uniform and unchanging, so the total flux through the rigid loop is constant. No induced current arrows or magnetic braking-force arrows are drawn. After entry transients, this model has zero total induced e.m.f. and zero sustained current. No opening has been made in the conducting rectangle.Field boundaryDots: B outv = 3.00 m/sOutsideB = 0Unchanged linkage; no loop current

The linked area and B are now constant even though the loop moves. This closed rigid-loop control has no sustained induced current or braking force in the stated model. It does not describe every eddy-current path in a solid conductor near a field boundary.

The entering loop has a complete clockwise current path and a leftward force on its right edge. The horizontal-edge forces cancel vertically in the symmetric model. The second view keeps the whole loop inside the unchanged uniform field, where its linkage no longer changes. Motion and force arrows represent different quantities.

Worked entry and energy account

Derive the e.m.f. from changing area

Use B = 0.400 T, vertical edge length l = 0.200 m, speed v = 3.00 m/s and total loop resistance R = 0.600 Ω. Treat the field as uniform within its boundary, resistance as constant and self-inductance as negligible.

During an interval Δt while the loop is entering, its edge moves vΔt and the exposed area increases by lvΔt. The one-turn flux change therefore gives:

|Eind| = BΔA/Δt = Blv
= (0.400)(0.200)(3.00) = 0.240 V
I = |Eind|/R = 0.240/0.600
= 0.400 A, clockwise

The right edge is perpendicular to B, so its opposing force has magnitude:

F = BIl = (0.400)(0.400)(0.200)
= 0.0320 N

For steady speed, an applied pull of the same magnitude acts right. Its mechanical input power is:

Pin = Fv = (0.0320)(3.00) = 0.0960 W
Pheating = I2R = (0.400)2(0.600)
= 0.0960 W

The matching powers check the energy account. External work maintains the speed while the loop warms. Without that pull, and with no other energy input, the heating comes from a decrease in the loop's kinetic energy.

Identify when the braking stops

Once the entire rigid loop is inside the unchanged uniform field, its area, orientation and B are constant. Its total flux is unchanged even while it translates. In the stated negligible-self-inductance model there is no sustained induced e.m.f. around the loop, no current and no associated braking force.

The cancellation requires the whole-loop condition. During entry, only one vertical edge is in the field; after full entry, both are. Motion by itself is not sufficient evidence of induction. A stationary conductor does not acquire a persistent braking current merely because a static B is nonzero.

Extend the reasoning to a solid conductor

A moving solid conductor can contain closed induced-current paths within its material. These eddy currents can create opposing forces and heating where the field and motion change their linkage. Slots interrupt or restrict some current paths and can reduce the braking effect.

Apply the changing-linkage test to the actual geometry. The entering-loop result does not imply that every solid conductor translating anywhere within a uniform field must experience sustained magnetic braking.

Optional check A rigid closed loop has 0.400 A induced while entering a bounded uniform field. It then moves wholly inside the same unchanging uniform field, with its area and orientation fixed. Neglect self-inductance and other effects. What is its sustained model current there?
A rigid closed loop has 0.400 A induced while entering a bounded uniform field. It then moves wholly inside the same unchanging uniform field, with its area and orientation fixed. Neglect self-inductance and other effects. What is its sustained model current there?