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Novosadov [1.4K]

3 years ago

12

A coiled telephone cord forms a spiral with 90.0 turns, a diameter of 1.30 cm, and an unstretched length of 57.0 cm. Determine t

he inductance of one conductor in the unstretched cord.

1 answer:

Vinvika [58]3 years ago

8 0

Answer:

2.36 μ H

Explanation:

Given,

Number of turns= 90

diameter = 1.3 cm = 0.013 m

unscratched length = 57 cm = 0.57 m

Area, A = π r²

= π x 0.0065² = 1.32 x 10⁻⁴ m²

we know,

$L = \dfrac{\mu_0N^2A}{l}$

$L = \dfrac{4\pi \times 10^{-7}\times 90^2\times 1.32\times 10^{-4}}{0.57}$

L = 2.36 μ H

Hence, the inductance of the unstretched cord is equal to 2.36 μ H

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$\Delta E_{a} = |t_{a} - \bar {x}| = |13.54 - 13.44| = 0.01$

$\Delta E_{b} = |t_{b} - \bar {x}| = |13.44 - 13.44| = 0.00$

$\Delta E_{c} = |t_{c} - \bar {x}| = |13.89 - 13.44| = 0.45$

$\Delta E_{d} = |t_{d} - \bar {x}| = |13.41 - 13.44| = 0.03$

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$\Delta E_{f} = |t_{f} - \bar {x}| = |13.54 - 13.44| = 0.10$

Mean Absolute Error, $\Delta E = \frac{\Sigma E_{n}}{6}$

$\Delta E = \frac{0.01 + 0.00 + 0.45 + 0.03 +0.027 + 1.10}{6}$

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