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xxTIMURxx [149]

3 years ago

6

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.5 meters

in the first 2.84 μs after it is released. What is magnitude of the electric field? What is the direction of the magnetic field? Are we justified if ignoring gravity on the electron in this situation?

1 answer:

dem82 [27]3 years ago

8 0

Answer:

The electric and magnetic field are 6.34 N/C and $2.11\times10^{-8}\ T$ .

Explanation:

Given that,

Distance = 4.5 m

Time = 2.84 μs

We need to calculate the acceleration

Using equation of motion

The distance covered by the electron is

$s=ut+\dfrac{1}{2}at^2$

When, the electron at rest

$s = \dfrac{1}{2}at^2$

Where, s = distance

a = acceleration

t = time

Put the value into the formula

$4.5=\dfrac{1}{2}\times a\times(2.84\times10^{-6})^2$

$a=\dfrac{2\times4.5}{(2.84\times10^{-6})^2}$

$a=1.116\times10^{12}\ m/s^2$

We need to calculate the electric field

Using formula of the electric field

$E=\dfrac{F}{q}$

$E=\dfrac{ma}{q}$

Put the value into the formula

$E=\dfrac{9.1\times10^{-31}\times1.116\times10^{12}}{1.6\times10^{-19}}$

$E=6.34\ N/C$

We need to calculate the magnetic field

Using formula of magnetic field

$B = \dfrac{E}{c}$

Put the value into the formula

$B=\dfrac{6.34}{3\times10^{8}}$

$B=2.11\times10^{-8}\ T$

According to the right hand rule,

The direction of magnetic field is outward because the direction of force is upward.

Hence, The electric and magnetic field are 6.34 N/C and $2.11\times10^{-8}\ T$ .

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Answer:

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Explanation:

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2 years ago

In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10

sergij07 [2.7K]

Answer:

A) $\Phi=83.84\times 10^{-9}$

B) $\Phi=0 Wb$

C) $emf=5.4090\times 10^{-4}V$

Explanation:

Given that:

no. of turns i the coil, $n=200$

area of the coil, $a=13.1 \times 10^{-4}\,m^2$

time interval of rotation, $t=3.1\times 10^{-2}\,s$

intensity of magnetic field, $B=6.4\times 10^{-5}\,T$

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

$\Phi=B.a\,cos \theta$ ..................................(1)

$\theta$ is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}$

$\Phi=83.84\times 10^{-9} Wb$

(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴ $\theta=90^{\circ}$

From eq. (1)

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}$

$\Phi=0 Wb$

(C)

According to the Faraday's Law we have:

$emf=n\frac{B.a}{t}$

$emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}$

$emf=5.4090\times 10^{-4}V$

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3 years ago

You are riding in an elevator that is accelerating upward. Suppose you stand on a scale. The reading on the scale is __________.

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Given the following choices;

A) less than your true weight, mg.

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D) could be more or less than your true weight, mg, depending on the value of the speed.

The answer is; C

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8 0

2 years ago

Read 2 more answers

Determine the speed, wavelength, and frequency of light from a helium-neon laser as it travels through polystyrene. The waveleng

klemol [59]

Answer:

Speed:

$2.01x10^{8}m/s$

Wavelength:

$4.24x10^{-7}m$

Frequency:

$4.74x10^{14}Hz$

Explanation:

The speed of the laser as it travels through polystyrene can be determine by means of the equation of the refraction index:

$n = \frac{c}{v}$ (1)

Where c is the speed of light and v is the speed of the laser in the medium.

Therefore, v will be isolated from equation 1

$v = \frac{c}{n}$

$v = \frac{3x10^{8}m/s}{1.490}$

$v = 2.01x10^{8}m/s$

Hence, the speed of the laser has a value of $2.01x10^{8}m/s$

Frenquency:

Since, wavelength is the only one who depends on the media. Therefore the frequency in both medium will be the same.

To determine the frequency it can be used the following equation

$c = \nu \cdot \lambda$ (2)

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

Then, $\nu$ wil be isolated from equation 2.

$\nu = \frac{c}{\lambda}$ (3)

Before using equation 3 it is necessary to express $\lamba$ in units of meters.

$\lambda = 632.8nm . \frac{1m}{1x10^{9}nm}$ ⇒ $6.328x10^{-7}m$

$\nu = \frac{3x10^{8}m/s}{6.328x10^{-7}m}$

$\nu = 4.74x10^{14}s^{-1}$

$\nu = 4.74x10^{14}Hz$

Hence, the frequency of the laser has a value of $4.74x10^{14}Hz$

Wavelength:

To determine the wavelength it can be used:

$v = \nu \cdot \lambda$

$\lambda = \frac{v}{\nu}$

Where v is the speed of the laser through the polystyrene.

$\lambda = \frac{2.01x10^{8}m/s}{4.74x10^{14}s^{-1}}$

$\lambda = 4.24x10^{-7}m$

Hence, the wavelength of the laser has a value of $4.24x10^{-7}m$

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