idiopatia0f
2021-12-14
Answered

A long, straight solenoid with a cross-sectional area of 8.00 cm2 is wound with 90 turns of wire per centimeter, and the windings carry a current of 0.350 A. A second winding of 12 turns encircles the solenoid at its center. The current in the solenoid is turned off such that the magnetic field of the solenoid becomes zero in 0.0400 s. What is the average induced emf in the second winding?

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MoxboasteBots5h

Answered 2021-12-15
Author has **35** answers

Expression for magnetic field

$B={\mu}_{0}{n}_{i}$

Where,

$\mu}_{0}=4\pi \times {10}^{-7}\text{}n{A}^{-2$

Given values:

$n=90$ per cm to per $m=9000$

$i=0.350\text{}A$

Placing all known values in equation

$B=3\pi \times {10}^{-7}\times 90\times {10}^{2}\times 0.350\text{}A$

$=3.9\times {10}^{-3}\text{}T$

Expression for average induced emf-

$\u03f5=NA\frac{\mathrm{\u25b3}B}{\mathrm{\u25b3}t}$

Where,

$N=12$

$A=8\text{}c{m}^{2}=8\times {10}^{-4}\text{}{m}^{2}$

$\mathrm{\u25b3}B=3.9\times {10}^{-4}\text{}{m}^{2}$

$\mathrm{\u25b3}t=0.04\text{}s$

Placing known values give us-

$\u03f5=12\times (8\times {10}^{-4}\text{}{m}^{2})\frac{3.9\times {10}^{-3}\text{}T}{0.04\text{}s}$

$=9.3\times {10}^{-4}\text{}V$

Where,

Given values:

Placing all known values in equation

Expression for average induced emf-

Where,

Placing known values give us-

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