Problem 1 A coil measures ⁄ meter in length with 500 turns of ⁄ inch diameter, and has a metallic core with a relative permeability of 1150. For the first part of this problem determine the inductance of this coil. Then, find the voltage induced across the coil if the rate of change of current is:

a) 1 A/s
b) 20 mA/ms
c) 6 mA/100 μs

(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

answer:

the answer is d

explanation: