Consider a uniformly wound solenoid having N=210 turns, length l=0.18 m, and cross-sectional area A = 4.00 cm2. Assume l is much longer than the radius of the windings and the core of the solenoid is air. (A) Calculate the inductance of the solenoid. (B) Calculate the self-induced emf in the solenoid if the current it carries decreases at the rate of 120 A/s. Inductor

Explanation:

Number of turns

N = 210turns

Length of solenoid

l = 0.18m

Cross sectional area

A = 4cm² = 4 × 10^-4m²

A. Inductance L?

Inductance can be determined using

L = N²μA/l

Where

μ is a constant of permeability of the core

μ = 4π × 10^-7 Tm/A

A is cross sectional area

l is length of coil

L is inductance

Therefore

L = N²μA / l

L=210² × 4π × 10^-7 × 4 × 10^-4 / 0.18

L = 1.23 × 10^-4 H

L = 0.123 mH

B. Self induce EMF ε?

EMF is given as

ε = -Ldi/dt

Since rate of decrease of current is 120 A/s

Then, di/dt = —120A/s, since the current is decreasing

Then,

ε = -Ldi/dt

ε = - 1.23 × 10^-4 × -120

ε = 0.01478 V

ε ≈ 0.015 V

d. platinum would have the greatest temperature increase because it has the lowest specific heat.