Question

Find an expression for the power expended in pulling a conducting loop out of a magnetic field.

Answer 1

Figure (a): A loop is moving out of the magnetic field with velocity v 

1. Consider a loop ABCD moving with constant velocity `vec"v"` in a constant magnetic field B.
2. A current (i) is induced in the loop in a clockwise direction and the loop segments, being still in a magnetic field, experience forces, F₁, F_2, and F₃.
3. To pull the loop at a constant velocity `vec"v"` towards the right, it is required to apply an external force `vec"F"` on the loop so as to overcome the magnetic force of equal magnitude but acting in opposite direction.
   
   Figure (b): Induced emf (e), Induced current (i), and collective resistance (R) of the loop
4. The rate of doing work on the loop is, P = `("Work"("W"))/("time" ("t")) = ("Force" ("F") xx "displacement" ("d"))/("time" ("t"))`
   P = Force (F) × velocity (v)
   ∴ `vec"P" = vec"F" . vec"v"` ….(1)
5. Magnitude of magnetic flux through the loop is,
   Φ_B = B.A = B.Lx ….(2)
6. As the loop is moved to the right, the area lying within the magnetic field decreases, thus causing a decrease in the magnetic flux linked with the moving loop. As per Lenz's law, the decreasing magnetic flux induces a current in the loop. 
7. According to Faraday’s law, the magnitude of induced emf,
   |e| = `|("d"phi)/"dt"| = "d"/"dt" ("BLx")`
   = BL.`"dx"/"dt" = "BLv"` ….(3)
8. The magnitude of induced current i can be written using equation (3) as
   i = `|"e"|/"R" = "BLv"/"R"` ….(4)
9. The three segments of the current-carrying loop experience the deflecting forces `vec"F"_1`, `vec"F"_2`, and `vec"F"_3` in the magnetic field `vec"B"` in accordance with equation, `vec"F" = "i"vec"L" xx vec"B"`
10. From the symmetry, the forces `vec"F"_2` and `vec"F"_3` being equal and opposite, cancel each other. The remaining force `vec"F"_1` is directed opposite to the external force `vec"F"` on the loop, `vec"F"` = `-vec"F"_1`.
11. The magnitude of `|vec"F"_1|` can be written as `|vec"F"_1|` = iLB sin90 = iLB = `|vec"F"|` ….(5)
12. From equation (4), and equation (5)
    `|vec"F"|` = `|vec"F"_1|` = iLB
    = `"BLv"/"R"."LB" = ("B"^2"L"^2"v")/"R"` ….(6)
13. From equation (1) and (6), the rate of doing mechanical work (power), is given as,
    P = `vec"F".vec"v" = ("B"^2"L"^2"v")/"R"."v" = ("B"^2"L"^2"v"^2)/"R"` ….(7)
14. If current i is flowing in the closed circuit with collective resistance R, the rate of production of heat energy in the loop as we pull it along at constant speed v, can be written as,
    P = i² R ….(8)
15. From equation (4) and equation (8)
    P = `("BLv"/"R")^2."R"`
    P = `("B"^2"L"^2"v"^2)/"R"` ….(9)
16. Comparing equation (7) and equation (9), it can be found that the rate of doing mechanical work is exactly the same as the rate of production of heat energy in the circuit/loop.
17. Thus the work done in pulling the loop through the magnetic field appears as heat energy in the loop.