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bezimeni [28]
3 years ago
5

Which change will cause an increase in the electric current produced through electromagnetic induction?using more wire loops in

the solenoidusing the opposite pole of the magnetusing a weaker magnetmoving the solenoid slower?
Physics
2 answers:
miss Akunina [59]3 years ago
8 0

Answer:

using more wire loops in the solenoid

Explanation:

Here we know that rate by Faraday's law of electromagnetism

EMF = -\frac{d\phi}{dt}

Now for solenoid we know that

EMF = - L\frac{di}{dt}

now this EMF of induces need to be increased so here we need to increase

1). By increasing the self inductance(L) of the solenoid coil

2). By increasing the rate of change in flux in the coil by moving it fast

Now here inductance is increased by increasing number of turns in the coil

So here correct answer should be related to the more number of loops of the coil which will increase the electric current in the solenoid

allsm [11]3 years ago
4 0
<span>using more wire loops in the solenoid</span>
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The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu
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A. Lithium

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E=\phi + K

where

E=\frac{hc}{\lambda} is the energy of the incident light, with h being the Planck constant, c being the speed of light, and \lambda being the wavelength

\phi is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

\lambda=190 nm=1.9\cdot 10^{-7}m, so the energy of the incident light is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J

Converting in electronvolts,

E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

\phi = E-K=6.5 eV-4.0 eV=2.5 eV

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

\lambda=510 nm=5.1\cdot 10^{-7} m

So its energy is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J

Converting in electronvolts,

E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

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- Copper work function: \phi = 4.5 eV

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

E=\phi+K=4.5 eV+2.7 eV=7.2 eV

Then the copper is replaced with sodium, which has work function of

\phi = 2.3 eV

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

K=E-\phi = 7.2 eV-2.3 eV=4.9 eV

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