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3 years ago

5

The heating coils in a hair dryer are 0.800 cm in diameter, have a combined length of 1.00 m, and a total of 400 turns. (a) What

is their total self-inductance assuming they act like a single solenoid?

2 answers:

VashaNatasha [74]3 years ago

7 0

Answer: 13.1 μH

Explanation:

Given

length of heating coil, l = 1 m

Diameter of heating coil, d = 0.8 cm = 8*10^-3 m

No of loops, N = 400

L = μN²A / l

where

μ = 4π*10^-7 = 1.26*10^-6 T

A = πd²/4 = (π * .008 * .008) / 4 = 6.4*10^-5 m²

L = μN²A / l

L = [1.26*10^-6 * 400 * 400* 6.5*10^-5] / 1

L = 1.26*10^-6 * 1.6*10^5 * 6.5*10^-5

L = 1.31*10^-5

L = 13.1 μH

Thus, from the calculations above, we can say that the total self inductance of the solenoid is 13.1 μH

lisov135 [29]3 years ago

4 0

Answer:

Their total self-inductance assuming they act like a single solenoid is 10.11 μH

Explanation:

Given;

diameter of the heating coil, d = 0.800 cm

combined length of heating coil and hair dryer, $l$ = 1.0 m

number of turns, N = 400 turns

Formula for self-inductance is given as;

$L = \frac{\mu_oN^2A}{l}$

where

μ₀ is constant = 4π x 10⁻⁷ T.m/A

A is the area of the coil:

A = πd²/4

A = π (0.8 x 10⁻²)²/4

A = 5.027 x 10⁻⁵ m²

$L = \frac{\mu_oN^2 A}{l } = \frac{4\pi *10^{-7}(400)^2 *5.027*10^{-5}}{1 } \\\\L =1.011 *10^{-5} \ H\\\\L = 10.11 \mu H$

Therefore, their total self-inductance assuming they act like a single solenoid is 10.11 μH

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