Answer:
(a) VL = = 1.64758 V
(b) VL = 32.9517 V
(c) VL = 98.855 V
Explanation:
Given Data;
L = 1/6m = 0.0625m
N = 500
d = 3/4 inch = 1.905 cm = 0.01905m
Calculating the cross sectional area, we have
A =πd²/4
= π*0.01905²/4
= 2.8502*10⁻⁴ m³
Calculating the reductance, we have;
R₁ = 1/u₀urA
= 1/4.0*10⁻⁷ *1150* 2.8502*10⁻⁴
= 151736.987 a-b/wb
Calculating the inductance (L), we have
L = N²/R₁
= 500²/151736.987
L = 1.64758 H
Calculating the voltage induced across the coil, we have;
(a)
VL = L di/dt
but di/dt = 1 A/s
Therefore,
VL = 1.64758 * 1
= 1.64758 V
(b)
di/dt = 20mA/mS = 20 A/s
VL = 1.64758 * 20
= 32.9517 V
(c)
di/dt = 6mm/100us = 6*10⁻³/100*10⁻⁶ = 60 A/s
VL = 1.64758 * 60
= 98.855 V
