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3 years ago

Explain the following defects of a simple electric cell:

Physics

1 answer:

eimsori [14]3 years ago

4 0

Answer:

Explanation:

The two major defects of simple electric cells causes current supplied to be for short time. These defects are: polarization and local action.

a. Polarization: This is a defect caused by an accumulation of hydrogen bubbles at the positive electrode of the cell. It can be prevented by the use of vent, using a hydrogen absorbing material or the use of a depolarizer.

b. Local Action: This is the gradual wearing away of the electrode due to impurities in the zinc plate. It can be controlled by the amalgamation of the zinc plate before it is used.

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3 years ago

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An electromagnetic wave is traveling through vacuum in the positive x direction. Its electric field vector is given by E⃗ =E0sin

fgiga [73]

Given that,

The electric field is given by,

$\vec{E}=E_{0}\sin(kx-\omega t)\hat{j}$

Suppose, B is the amplitude of magnetic field vector.

We need to find the complete expression for the magnetic field vector of the wave

Using formula of magnetic field

Direction of $(\vec{E}\times\vec{B})$ vector is the direction of propagation of the wave .

Direction of magnetic field = $\hat{j}$

$B=B_{0}\sin(kx-\omega t)\hat{k}$

We need to calculate the poynting vector

Using formula of poynting

$\vec{S}=\dfrac{E\times B}{\mu_{0}}$

Put the value into the formula

$\vec{S}=\dfrac{E_{0}\sin(kx-\omega t)\hat{j}\timesB_{0}\sin(kx-\omega t)\hat{k}}{\mu_{0}}$

$\vec{S}=\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}$

Hence, The poynting vector is $\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}$

7 0

3 years ago

A plane electromagnetic wave, with wavelength 4.1 m, travels in vacuum in the positive direction of an x axis. The electric fiel

marusya05 [52]

(a) $7.32\cdot 10^7 Hz$

The frequency of an electromagnetic waves is given by:

$f=\frac{c}{\lambda}$

where

$c=3.0\cdot 10^8 m/s$ is the speed of light

$\lambda=4.1 m$ is the wavelength of the wave in the problem

Substituting into the equation, we find

$f=\frac{3.0\cdot 10^8 m/s}{4.1 m}=7.32\cdot 10^7 Hz$

(b) $4.60\cdot 10^8 rad/s$

The angular frequency of a wave is given by

$\omega = 2\pi f$

where

f is the frequency

For this wave,

$f=7.32\cdot 10^7 Hz$

So the angular frequency is

$\omega=2\pi(7.32\cdot 10^7 Hz)=4.60\cdot 10^8 rad/s$

(c) $1.53 m^{-1}$

The angular wave number of a wave is given by

$k=\frac{2\pi}{\lambda}$

where

$\lambda$ is the wavelength of the wave

For this wave, we have

$\lambda=4.1 m$

so the angular wave number is

$k=\frac{2\pi}{4.1 m}=1.53 m^{-1}$

(d) $1.03\cdot 10^{-6}T$

For an electromagnetic wave,

$E=cB$

where

E is the magnitude of the electric field component

c is the speed of light

B is the magnitude of the magnetic field component

For this wave,

E = 310 V/m

So we can re-arrange the equation to find B:

$B=\frac{E}{c}=\frac{310 V/m}{3\cdot 10^8 m/s}=1.03\cdot 10^{-6}T$

(e) z-axis

In an electromagnetic wave, the electric field and the magnetic field oscillate perpendicular to each other, and they both oscillate perpendicular to the direction of propagation of the wave. Therefore, we have:

- direction of propagation of the wave --> positive x axis

- direction of oscillation of electric field --> y axis

- direction of oscillation of magnetic field --> perpendicular to both, so it must be z-axis

(f) 127.5 W/m^2

The time-averaged rate of energy flow of an electromagnetic wave is given by:

$I=\frac{E^2}{2\mu_0 c}$