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Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.245 A/s ,

Airida [17]

Answer:

Complete question:

c.If the current in the second coil increases at a rate of 0.365 A/s , what is the magnitude of the induced emf in the first coil?

a. $M= 6.53\times10^{-3} H$

b.flux through each turn = Ф = $4.08\times10^{-4} Wb$

c.magnitude of the induced emf in the first coil = e= $2.38\times10^{-3} V$

Explanation:

a. rate of current changing = $\frac{di}{dt}=[tex]M=\frac{1.60\times10^{-3} V}{0.240\frac{A}{s} }}$ [/tex]

Induced emf in the coil =e= $1.60\times10^{-3} V$

For mutual inductance in which change in flux in one coil induces emf in the second coil given by the farmula based on farady law

$e=-M\frac{di}{dt}$

$M=\frac{e}{\frac{di}{dt} }$

$M=\frac{1.60\times10^{-3} }{-0.245}$

$M= 6.53\times10^{-3} H$

b.

Flux through each turn=?

Current in the first coil =1.25 A

Number of turns = 20

using MI = NФ

flux through each turn = Ф = $\frac{6.53\times10^{-3}\times1.25}{20}$

flux through each turn = Ф = $4.08\times10^{-4} Wb$

c.

second coil increase at a rate = 0.365 A/s

magnitude of the induced emf in the first coil =?

using $e=-M\frac{di_{2} }{dt}$

$e= 6.53\times10^{-3} \times 0.365$

magnitude of the induced emf in the first coil = e= $2.38\times10^{-3} V$

4 0

3 years ago

The speed of light is about 3.00 × 10 meters per second. What is the frequency of green light that has a wavelength about 500 na

RideAnS [48]

Answer:

The frequency of the green light is $6x10^{14}Hz$

Explanation:

The visible region is part of the electromagnetic spectrum, any radiation of that electromagnetic spectrum has a speed of $3.00x10^{8}m/s$ in the vacuum.

Green light is part of the visible region. Therefore, the frequency can be determined by the following equation:

$c = \lambda \cdot \nu$ (1)

Where c is the speed of light, $\lambda$ is the wavelength and $\nu$ is the frequency.

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave (that is a magnetic field perpendicular to an electric field).

Then, $\nu$ can be isolated from equation 1

$\nu = \frac{c}{\lambda}$ (2)