Lessons 4–5 · 2 lessons · N. Sh. Turdiyev, K. A. Tursunmetov, A. G. Ganiyev, K. T. Suyarov, J. E. Usarov, A. K. Avliyoqulov. Physics Grade 11, 1st edition. Niso Poligraf Publishing House, Tashkent, 2018
4
Work done in moving a current-carrying conductor in a magnetic field
Textbook: pp. 13–14
GoalCalculate the work of the Ampère force in moving a current-carrying conductor in a magnetic field with A = I·ΔΦ, and explain where the energy comes from.
New words
displacement · ko‘chishchange of magnetic flux · magnit oqimining o‘zgarishicurrent source · tok manbayileft-hand rule · chap qo‘l qoidasi
Explanation
In a uniform field a conductor of length l carrying current I and perpendicular to the field lines feels the force F = I·B·l, whose direction follows the left-hand rule. If the conductor moves a distance d along the force, the work is A = F·d = I·B·l·d. The product l·d is the area ΔS swept by the conductor and B·ΔS is the magnetic flux ΔΦ it has crossed, hence A = I·ΔΦ. If the conductor makes angle α with the field lines, F = I·B·l·sinα and the work is multiplied by sinα as well. This work is done not at the expense of the field's own energy but at the expense of the source that maintains the current in the circuit: the magnetic force itself (like the Lorentz force) creates no energy. The same principle is used in electric motors, moving-coil meters and electromagnetic locks.
Worked examples
A conductor 20 cm long carries 3 A; it is perpendicular to the field lines of B = 0.5 T and moves 10 cm along the Ampère force. ΔS = 0.2·0.1 = 0.02 m², ΔΦ = 0.5·0.02 = 0.01 Wb, A = I·ΔΦ = 3·0.01 = 0.03 J = 30 mJ.
While a circuit with current 8 A is moved in a field, the flux through the region it crosses changes by 5 mWb. A = I·ΔΦ = 8·5·10⁻³ = 0.04 J = 40 mJ. With half the current the work would also be halved.
Class activity
In your notebook draw two parallel rails with a rod across them (connected to a battery), mark the field as «into the page» and find the force direction with the left-hand rule. Only the teacher demonstrates the experiment, with a low-voltage battery.
Practice
1
What is the formula for the work in moving a current-carrying conductor in a field?
A = I·ΔΦ
2
I = 5 A and ΔΦ = 12 mWb. What is the work in mJ?
60
3
l = 50 cm, I = 4 A, B = 0.2 T (perpendicular), displacement 15 cm. What is the work in mJ?
60
4
At whose expense is the work done, and why does the field itself not supply the energy?
At the expense of the current source: the magnetic force only sets the direction; the energy comes from the source that maintains the current.