Explanation:
The compass needle moved when the wire was connected to the battery. The important point here is that the needle is affected by the wire only when both ends of the wire are connected to the battery because only at this time is current flowing through the circuit.
Produce power through the circuit.
Answer:
η = 2.57%
Explanation:
Given:
Output power of the steam power plant, P(out) = 200 MW
Consumption of coal, m = 700 tons/h = (700 × 10³)/3600 kg/s= 194.44 kg/s
Heating capacity of the coal, Cv = 40000 kJ/kg
now,
Input power, P(in) = mCv
or
P(in) = 194.44 kg/s × 40000 = 7777.77 MW
thus,
efficiency, η = P(out)/P(in) = (200 MW) / (7777.77 MW) = 0.02571
or
η = 2.57%
Answer:
ΔΦ = -3.39*10^-6
Explanation:
Given:-
- The given magnetic field strength B = 0.50 gauss
- The angle between earth magnetic field and garage floor ∅ = 20 °
- The loop is rotated by 90 degree.
- The radius of the coil r = 19 cm
Find:
calculate the change in the magnetic flux δφb, in wb, through one of the loops of the coil during the rotation.
Solution:
- The change on flux ΔΦ occurs due to change in angle θ of earth's magnetic field B and the normal to circular coil.
- The strength of magnetic field B and the are of the loop A remains constant. So we have:
Φ = B*A*cos(θ)
ΔΦ = B*A*( cos(θ_1) - cos(θ_2) )
- The initial angle θ_1 between the normal to the coil and B was:
θ_1 = 90° - ∅
θ_1 = 90° - 20° = 70°
The angle θ_2 after rotation between the normal to the coil and B was:
θ_2 = ∅
θ_2 = 20°
- Hence, the change in flux can be calculated:
ΔΦ = 0.5*10^-4*π*0.19*( cos(70) - cos(20) )
ΔΦ = -3.39*10^-6
