41 
A toroid consists of a wooden core with a winding having two layers with 500 uniformly distributed turns each. The inside and outside radii of the core are 5 and 6 in. respectively, and the axial depth of the core is 4 in. Assume all the magnetic flux to be confined to the wooden core, and the mean length of the flux path to be the average of the inner and outer circumferences of the core.
(a) If the two layers of the windings are connected in series with their polarities such that the mmfs add and the current is 1 amp, determine (i) The magnetic flux in the core, (ii) The flux linkage, (iii) The selfinductance of the winding, (iv) The energy stored in the magnetic field.
(b) Repeat part (a) above if a current of 1 amp is passed through only one layer of the winding.
(c) Repeat part (b) above if the two layers are connected in parallel with the polarities such that the mmfs add and if the total current is 2 amp.
(d) Repeat part (a) for a core having a uniform relative permeability μ_{r} = 400.

42 
Determine the magnetic reluctance and the magnetic permeance of the toroid for the conditions specified in Problem 41 (a), (b), (c), and (d).

43 
A long, straight, round copper conductor of radius R m and length l m carries a current of I amp distributed uniformly over its cross section. Determine in terms of I and other suitable constants on the basis that the relative permeability of copper μ_{r} = 1
(a) The total magnetic flux within the conductor.
(b) The magnetic energy density within the conductor as function of the distance from the center of the conductor.

44 
The number of turns in the windings designated as circuits 1 and 2 in Fig. 44(b) is N_{1} = 100 and N_{2} = 200. A current i_{1} = 5 amp in circuit 1 produces an equivalent flux μ_{11} = 0,0032 weber, which links all N_{1} turns. Assume the magnetic circuit to be linear, i.e., the reluctance of the iron represented by the shaded area to be negligible. Calculate
(a) The selfinductance L_{11} of circuit 1.
(b) The mutual flux Φ21 if the reluctances are in the ratios R_{2}/R_{1} = 2 and R_{1}/R_{1} = 2/3
(c) The mutual inductance L_{21} between the two circuits.
(d) The coefficients k_{1}, k_{2} and the coefficient of coupling k.
(e) The selfinductance L_{22} of circuit 2.
(f) The energy stored in the fields associated with each of the three reluctances.
(g) The stored energy based on 1/2L_{11}i_{1}^{2} [How does this compare with the sum of the three values of part (f)?]

45 
Figure 49 shows an electromagnet with equal variable air gaps g. When the current in the 1000turn winding on the stationary member is 5 amp the flux through the movable member is Φ_{a} = 0.002 weber and the leakage flux Φ_{1} = 0.001 weber. The reluctance of the iron is negligible. The length of the variable air gaps g is increased 10 percent; the flux Φ_{a} through these air gaps is held constant at 0.002 weber. Neglect the effect of fringing.

Figure 49. Electromagnet for Problem 45 
(a) Calculate for the increased length of air gaps (i) The current. (ii) The leakage flux Φ_{1}.
(iii) The total energy stored in the field of the two air gaps g. (iv) The stored energy associated with the leakage flux Φ_{1}. (v) The selfinductance of the 200turn winding, (vi) The energy stored in the selfinductance (1/2Li^{2}).
(b) Calculate
(i) The mechanical energy input or output. (Which is it?) (ii) The electromagnetic energy input or output (Which is it?) during the increase in the length of the air gaps.
(c) If the original length of each of the gaps^ is 0.125 in., calculate the force.
(d) Neglect fringing and calculate the flux density in the air gaps g.
(e) Calculate the force on the basis of area and flux density and compare with that of part (c).
HINT : Use Eq. 371.
(f) Calculate the average voltage induced in the winding if the time during which the air gap is increased is 0.01 sec.

46 
When each of the variable air gaps^ in Fig. 49 has a length of 0.125 in. and the current in the 1,000turn winding is 5 amp, the fluxes Φ_{a} and Φ_{1} are 0.002 and 0.001 weber, respectively. The length of air gap g' is assumed to be fixed. Neglect fringing and the reluctance of the iron and calculate
(a) The length and the crosssectional area of the fixed air gap g' if the crosssectional areas of all three air gaps are equal.
(b) The permeance of the fixed air gap g'.
(c) The permeance of the two variable air gaps g in series as a function of the length of a single gap.
(d) The total permeance of the magnetic circuit as a function of the length of a single air gap g.
(e) The selfinductance of the 1,000turn winding as a function of the length of a single air gap g.
(f) The total force on the movable member or armature based on the relationship
for a current of 5 amp when the length of each air gap g is 0.125 in. (g) The force on the faces of the fixed air gap g' for the conditions of part (f).

47 
The selfinductance of two identical parallel straight nonmagnetic cylinders is given by
where d is the distance between centers (small in comparison with the length of the conductors), and where r is the radius of each conductor. During a short circuit on a power system, the current in two such conductors reaches a peak value of 100,000 amp. Calculate the force per meter length on each conductor if the radius of each conductor is 1 in. and the spacing is 6 in. between centers.

48 
Figure 410 shows a rotary electromagnet.
(a) Neglect the reluctance of the iron, leakage, and fringing and determine for the rotor position in Fig. 410(a) when the current in the stator

Figure 410. (a) Rotary electromagnet winding on stator only, for Problem 48, (b) rotor of rotary electromagnet with winding for Problems 49 and 410.

winding is 1.5 amp
(i) The flux in the air gap.
(ii) The flux in the iron.
(iii) The selfinductance of the stator.
(iv) The permeance of the magnetic circuit.
(v) The reluctance of the magnetic circuit.
(vi) The energy stored in the air gap.
(vii) The energy stored in the iron.
(b) Repeat part (a) above, but with the rotor displaced 15° (i) In the clockwise direction, (ii) In the counterclockwise direction.
(c) What is the torque in both parts of (b) above?
(d) Repeat part (a) above when the rotor is in the position shown in Fig. 410(a), taking into account the reluctance of the iron (USS Electrical Annealed Sheet Steel) if the mean length of the path in the iron is 8.5 in. Use Fig. 312 for determining the effect of the iron.

49 
Suppose that a winding of 67 turns were placed on the rotor [see Fig. 410(b)] of the electromagnet in Problem 48 and a current of 7.5 amp passed through this winding while the current in the stator winding was zero. Neglect leakage and fringing, but take into account the reluctance of the iron, and determine
(a) The flux in the air gap.
(b) The flux linkages of the rotor winding.
(c) The selfinductance of the rotor on the basis of Eq. 46.
(d) The permeance of the magnetic circuit.
(e) The flux linking the stator winding.
(f) The flux linkage of the stator winding due to the current in the rotor winding.
(g) The mutual inductance between stator and rotor on the basis of flux linkage per ampere.

410 
In the electromagnet of Fig. 410(a), assume a winding of 335 turns on the stator and a winding of 67 turns on the rotor. Neglect leakage, fringing, and the reluctance of the iron. Determine
(a) The flux that links the rotor winding when the stator winding carries 1.5 amp and when there is no current in the rotor winding.
(b) On the basis of part (a) above, mutual inductance between the stator and rotor windings.
(c) The flux that links the stator winding when the rotor winding carries 1.5 amp and when there is no current in the stator winding.
(d) On the basis of part (c) above, the mutual inductance between the two windings.
(e) The selfinductance of the electromagnet when the two windings are connected series aiding and when the current is 1.25 amp. Neglect fringing, leakage, and the reluctance of the iron.

411 
Figure 411 shows a cylindrical ironclad electromagnet with a cylindrical plunger. When actuated by the current in the winding, the plunger moves in

Figure 411. Cylindrical plunger magnet for Problem 411 
a brass guide tube of 0.03in. wall thickness. The length of the air gap g_{1} varies with the position of the plunger, whereas the gap g_{2} is fixed by the wall thickness of the brass tube and is assumed constant of length 0.03 in. The relative permeability of brass is 1. Neglect magnetic leakage, fringing and the reluctance of the iron.
(a) Calculate for length of gl = 0.125 and 0.500 in. (i) The magnetic reluctance, (ii) The magnetic permeance of the two gaps in series.
(b) Calculate the flux in the magnet for each airgap length in part (a) when the winding carries a constant current of 2.2 amp.
(c) Calculate the force on the plunger for each airgap length in part (a) when the current is 2.2 amp, based on Eq. 467.
(d) Express the selfinductance of the magnet as a function of g_{1} when g_{1} is in inches and when it is in meters.
(e) Calculate the mechanical energy associated with the movement of the plunger when the air gap gl is decreased from a length of 0.500 in. to 0.125 in. while the current remains constant at a value of 2.2 amp. Is motor action or generator action involved? Explain.
(f) Calculate the electrical energy required to hold the current constant at 2.2 amp while g_{1} is decreased from a length of 0.500 in. to 0.125 in.
(g) Calculate the change in the energy stored in the field for parts (e) and (f)
above,
(h) If the value of the current is 2.2 amp when gl = 0.500 in. and the flux
remains constant, what is
(i) The value of the current when g_{1} = 0,125 in.?
(ii) The electrical energy associated with the motion of the plunger?
(iii) The change in the stored energy from g_{1} = 0.500 in. to g_{1} =0.125 in.?

412 
Figure 412 shows a rotary electromagnet with a uniform air gap between the stator and the rotor. The selfinductance of the stator coils 11' and

Figure 412. Rotary electromagnet for Problem 412 
22' is L_{11} = L_{22} = 0.20 h and the selfinductance of the rotor coil 33' is 0.10 h. The mutual inductances between the stator coils and the rotor coil are L_{13} =0.125[1 (2θ/π)]h and L_{23} = 0.250θ/πh when the rotor is in the position shown.
(a) Calculate the torque when the currents are in the directions indicated and have the following values
i_{1} = +10 amp, i_{2} = +5 amp, i_{3} = +15 amp for θ = 30°.
(b) What is the value of Θ at which the torque is zero for the values of current in part (a) ?

413 
Figure 413 shows the magnetization curve of an ironcore reactor. The exciting winding has 400 turns and carries a direct current of 4 amp. The

Figure 413. Saturation curve of magnetic circuit with air gap for Problem 413.

leakage flux is negligible. Determine
(a) The magnetic flux.
(b) The magnetic flux linkage λ.
(c) The apparent inductance L_{a}.
(d) The effective inductance L_{e}.
(e) The differential inductance L_{d}.
(f) The range of currents for which the three inductances L_{a}, L_{e}, and L_{d} have the same value.

414 
The hypothetical magnetization curve shown in Fig. 414 approximates that of an electromagnet. Express the energy W_{Φ} stored in the magnetic field and the coenergy W_{Φ}' in terms of i_{1}, i_{2}, λ_{1} and λ_{2} for a current of i_{1}. amp and for a current of i_{2} amp.

415 
Find the change in the energy stored in the field, the change in the coenergy, and the electromagnetic energy input when the air gap in the electromagnet having the hypothetical magnetization curves shown in Fig. 415 changes from g_{1} to g_{2}
(a) While the mmf is held constant at 6.0 amp.
(b) While the flux linkage is held constant at 0.12 weber turn.
(c) If the locus of λ vs i during the change in gap is along the line ab.

Figure 414. Hypothetical magnetization curve for Problem 414 

416 
Calculate the average forces for parts (a), (b), and (c) in Problem 415 if the change in airgap length from g_{1} to g_{2} is 0.10 in. Does the length of the air gap increase or decrease in going from g_{1} to g_{2}? Explain.

417 
The noload current in a transformer winding is 9.1 amp when the voltage across the same winding is 6,600 v at 60 cycles. The real power is 9,300 w. Determine
(a) The reactive power.
(b) The Q of this transformer winding at no load on the basis of the real and reactive power.

418 
A choke coil has a selfinductance of 9,0 h. The effective resistance of the winding is 250 ohms at a frequency of 1000 cps. Determine the Q of the

Figure 415. Hypothetical magnetization curves for Problems 415 and 416 
choke for a frequency of
(a) 1000 cps.
(b) 60 cps on the basis that the selfinductance and the effective resistance have the same values as at 1000 cps.

419 
Prove that Eqs. 4109, 4110, and 4111 are equivalent.
