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The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi

Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

$\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$

Where,

B= Magnetic Field

l = length

$\mu_0$ = Vacuum permeability

$\epsilon_0$ = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

$B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}$

Recall that the speed of light is equivalent to

$c^2 = \frac{1}{\mu_0 \epsilon_0}$

Then replacing,

$B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}$

$B = \frac{r}{2C^2} \frac{dE}{dt}$

Our values are given as

$dE = 2150N/C$

$dt = 5s$

$C = 3*10^8m/s$

$D = 0.440m \rightarrow r = 0.220m$

Replacing we have,

$B = \frac{r}{2C^2} \frac{dE}{dt}$

$B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}$

$B =5.25*10^{-16}T$

Therefore the magnetic field around this circular area is $B =5.25*10^{-16}T$

3 0

2 years ago

A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and

iogann1982 [59]

Answer:

$p_2 = 1.76 atm$

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

p_2 = ?

t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

$\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}$

Putting the values

$\frac{1.15*2.37}{280} =\frac{p_2 *1.68}{304}$

$9.733*10^{-3} = \frac{p_2 *1.68}{304}$

$9.733*10^{-3}*304 = p_2*1.68$

$\frac{9.733*10^{-3}*304}{1.68} =p_2$

$p_2= 1.76 atm$

7 0

3 years ago

Mike walks 200 km in 6 hours.he then walks another 100km in 4 hours .what is his average speed?

kolbaska11 [484]

Average speed is defined as the ratio of total distance covered in total given time

$speed = \frac{distance}{time}$

here we know that total distance that man moved is

$d_1 = 200 km$

$d_2 = 100 km$

so total distance is

$d = d_1 + d_2$

$d = 200 km + 100 km$

$d = 300 km$

now here total time of the motion is

$t_1 = 6 hours$

$t_2 = 4 hours$

total time will be given as

$t = t_1 + t_2$

$t = 6 + 4 = 10 hours$

now by above formula

$v_{avg} = \frac{300}{10}$

$v_{avg} = 30 km/h$

so his average speed is 30 km/h

8 0

3 years ago

How do you calculate change in position? A. initial position times two B. final position plus initial position C. final position

e-lub [12.9K]

The answer is C. Final position minus initial position.

5 0

3 years ago

Why is velocity and not just speed important to these pilots

Alborosie

Velocity is a vector. Therefore, it depends on the direction. Pilots need to know the direction of wind, not just the speed. If the pilot is going South, and there's 5 mph wind going South, they'll be happy, but if the wind is going 5 mph North, they'll be going against the wind.

7 0

3 years ago