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gladu [14]
3 years ago
8

When I move a conductor through a magnetic field, the change in magnetic field creates a net change in magnetic flux. The EMF th

at is induced by this changing flux creates tiny motional currents that are called _______. Eddy Currents Generating Currents Faraday Currents Lenz Currents Quantizing Currents
Physics
1 answer:
iris [78.8K]3 years ago
6 0

Answer:

Eddy currents

Explanation:

When a magnetic flux linked across the coil changes, an induced emf is developed in the coil and hence the induced current is also developed in the coil.

If the induced current in the large conductor, the current is called eddy current. It produces the heating effect.

There are several applications of the eddy current.

(1) Electromagnetic brakes

(2) Induction furnace, etc.

To minimize the effect of eddy current we use choke coils, in which the core is made of laminated thin sheets.

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The AC voltage source is connected to an inductor and a resistor in series. If the frequency of the source is increased the curr
DaniilM [7]

Answer:

If the frequency of the source is increased the current in the circuit will decrease.

Explanation:

The current through the circuit is given as;

I = \frac{V}{Z}

Where;

V is the voltage in the AC circuit

Z is the impedance

Z = \sqrt{R^2 + X_L^2}

Where;

R is the resistance

X_L is the inductive reactance

X_L = ωL = 2πfL

where;

L is the inductance

f is the frequency of the source

Finally, the current in the circuit is given as;

I = \frac{V}{\sqrt{R^2 + (2\pi fL)^2} }

From the equation above, an increase in frequency (f) will cause a decrease in current (I).

Therefore, If the frequency of the source is increased the current in the circuit will decrease.

5 0
3 years ago
For crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate.
Kamila [148]

Answer:

A) E = 4.96 x 10³ eV

B) E = 4.19 x 10⁴ eV

C) E = 3.73 x 10⁹ eV

Explanation:

A)

For photon energy is given as:

E = hv

E = \frac{hc}{\lambda}

where,

E = energy of photon = ?

h = 6.625 x 10⁻³⁴ J.s

λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m

Therefore,

E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}

E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})

<u>E = 4.96 x 10³ eV</u>

<u></u>

B)

The energy of a particle at rest is given as:

E = m_{0}c^2

where,

E = Energy of electron = ?

m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

E = (9.1 x 10^{-31} kg)(3  x  10^8 m/s)^2\\

E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\

<u>E = 4.19 x 10⁴ eV</u>

<u></u>

C)

The energy of a particle at rest is given as:

E = m_{0}c^2

where,

E = Energy of alpha particle = ?

m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

E = (6.64 x 10^{-27} kg)(3  x  10^8 m/s)^2\\

E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\

<u>E = 3.73 x 10⁹ eV</u>

8 0
2 years ago
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