Answer:
Φ = 5.589×10⁻⁵ Wb
Explanation:
The inductance of a coil is given as
L = e/(di/dt) ..................... Equation 1
Where L = inductance of the coil, e = induced e.m.f, di/dt = rate of change of current in the coil.
Also,
The inductance of each turn of the coil when a magnetic field is step up in the coil is
L = NΦ/i ................. Equation 2
Where N = number of turns, Φ = magnetic field, i = current.
equating equation 1 and equation 2
e/(di/dt) = NΦ/i
making Φ the subject of the equation,
Φ = (e×i)/N.(di/dt) .................. Equation 3
Given: e = 28.0 mV = 0.028 V, N = 501 turns, di/dt = 12.0 A/s, i = 4.00 a
Substitute into equation 3,
Φ = (0.028×4)/(12×501)
Φ = 0.112/2004
Φ = 5.589×10⁻⁵ Weber
Φ = 5.589×10⁻⁵ Wb
