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3 years ago

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A small loop of area A = 5.5 mm2 is inside a long solenoid that has n = 919 turns/cm and carries a sinusoidally varying current

i of amplitude 3.08 A and angular frequency 226 rad/s. The central axis of the loop and solenoid coincide. What is the amplitude of the emf induced in the loop?

1 answer:

Tema [17]3 years ago

8 0

Answer: $410.90\times 10^{-6} V$

Explanation:

Given

$A=5.5 mm^2$

$n=919 turns/cm\approx 85400 turns/m$

$\omega =226 rad/s$

According to the Faraday law of induction, induced emf is given by

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

Magnetic field $\phi _B=B\cdot A$

$B=\mu _0ni=\mu _0ni_osin\left ( \omega t\right )$

$B=4\pi \times 10^{-7}\times 85400\times 3.08\times \sin (226t)$

$B=0.3305\sin (226t)$

$\phi _{B}=0.3305\times \sin (226t)\times 5.5\times 10^{-6}$

$\phi _{B}=1.818\times 10^{-6}\sin (226t)$

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

$E=-1.818\times 10^{-6}\times 226\cos (226t)$

$E=-410.90\times 10^{-6}\cos (226t)$

Amplitude of EMF induced $=410.90\times 10^{-6} V$

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Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

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b)

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