60
Problem solving
Textbook: p. 165
GoalSolve problems from the magnetic field chapter: the Ampère force, the Lorentz force and motor efficiency.
New words
Ampère force · Amper kuchiLorentz force · Lorens kuchiefficiency · FIKunits · birliklar
Explanation
First decide which phenomenon the problem is about: for a current-carrying conductor use F = BIl, for a moving charge F = qvB, for a motor P = IU and η = P_useful / P_total · 100 %. Remember the formulas are for the perpendicular case. Rearrange to find the needed quantity (B = F/(Il), I = F/(Bl), v = F/(qB)) and convert to SI: cm → m, mA → A. Check the force direction with the left-hand rule (for a negative charge the fingers point against the velocity). Judge whether the result makes sense; for example efficiency must not exceed 100 %.
Worked examples
A conductor l = 0.5 m, I = 10 A, B = 0.4 T: F = B·I·l = 0.4 · 10 · 0.5 = 2 N.
An electron with v = 5·10⁶ m/s in B = 2.5 T: F = 1.6·10⁻¹⁹ · 5·10⁶ · 2.5 = 2·10⁻¹² N = 2 pN. A motor takes 440 W and delivers 330 W: η = 330 : 440 · 100 % = 75 %.
Class activity
Make three problems of your own: one on the Ampère force, one on the Lorentz force and one on motor efficiency. Write the solutions in your notebook and swap with a classmate.
Practice
1
B = 0.6 T, I = 4 A, l = 0.5 m. The Ampère force in N?
1.2
2
F = 4 N, I = 10 A, l = 0.2 m. B in T?
2
3
A motor takes 300 W and delivers 240 W. Efficiency in %?
80
4
Why can a motor’s efficiency not be 100 %?
Part of the energy goes to heating of the windings and to friction.