2. A wire coil of 25 turns has a cross-sectional area of 10 cm^2. The coil is placed in a uniform field of large magnet with B=0.20 T. The coil is suddenly rotated 90 degrees from an orientation parallel to the field to one perpendicular to the field. (a) If the time to flip the coil 0.50s, what is the average emf produced? (b) What will be the average current produced if the circuit of the coil is closed and the resistance is 75 O?
3. What emf is developed between the tips of the wings of a Boeing 767 jet liner in level flight at 850 km/h in a location where the vertical component of the earth's magnetic field is 1.75 x 10^-5 T? The distance between the wing tips of the 767 is 47.65 m.
4. A 50-turn loop of wire has a cross-sectional area of 0.010 m^2. It is rotated about an axis perpendicular to the earth's field of 5.0 x 10^-5 T. What is the peak value of the emf generated if the loop rotates at 3600 rotations per minute?
5. A transformer is made by winding a primary coil of 250 turns around an iron core. A secondary winding of 50 turns is made around the same core. If the primary voltage is 120 V, what is output voltage on the secondary?
6. A 0.85-H inductor carries a current that decreases at a rate of 0.12A/s. What is the induced emf?
8. A circular laser beam has a diameter of 1.5 mm and a power of 3.0 mW. (a) Assuming the light is uniformly spread over the circle, what is the energy density in the beam? (b) What are the values of the electric and magnetic field amplitudes in the beam?
Area of the coil A = 37 cm^2 = 37*10^(-4) m^2
change in flux density delta(B) = (9.3 * 10^(-3) - 6.5 * 10^(-3)) T
=> delta(B) = 0.0028 T
time t = 0.50s.
emf induced in the coil e = delta(B) *A/t
=> e = 0.0028*37*10^(-4)/0.5 = 2.072*10^(-5) V --Answer
N = 25 turns
cross-sectional area A = 10 cm^2 = 10*10^(-4) m^2
uniform field B=0.20 T
coil rotatatin =90 degrees from parallel to the field to perpendicular
(a) time t = 0.50s
initially the angle of normal to the coi with field = 90 degree,
angle of the normal to the coil with the magnetic field = 0 ...
The solution provides step-by-step calculations for various electromagnetics problems, including calculations for emf, magnetic fields, output voltage, and current.