The magnetic field perpendicular to a single wire loop of diameter 10.0 cm decreases from 0.50 T to zero. The wire is made of copper and has a diameter of 2.0 mm and length 1.0 cm. How much charge moves through the wire while the field is changing?

I know how to do the calculations, but can someone please explain what is the 10cm diameter and 2mm diameter? Why is there one wire and two diameters? I understand this problem mathematically but not conceptually.

Answers

Answer 1

Hi there!

We can begin by using Lenz's Law:
[tex]\epsilon = -N\frac{d\Phi _B}{dt}[/tex]

N = Number of Loops

Ф = Magnetic Flux (Wb)
t = time (s)

Also, we can rewrite this as:
[tex]\epsilon = -NA\frac{dB}{dt}[/tex]

A = Area (m²)

Since the area is constant, we can take it out of the derivative.

This is a single wire loop, so N = 1.

Now, we can develop an expression for the induced emf.

We can begin by solving for the area:

[tex]A = \pi r^2 \\\\d = r/2 r = 0.05cm \\\\A = \pi (0.05^2) = 0.007854 m^2[/tex]

We can also express dB/dt as:
[tex]\frac{dB}{dt} = \frac{\Delta B}{t} = \frac{0-0.5}{t} = \frac{-0.5}{t}[/tex]

Now, we can create an equation.

[tex]\epsilon = -(1)(0.007854)\frac{-0.5}{t} = \frac{0.003927}{t}[/tex]

To solve the system, we must now develop an expression for current given an emf and resistance.

Begin by calculating the resistance of the copper wire:
[tex]R = \frac{\rho L}{A}[/tex]

ρ = Resistivity of copper (1.72 * 10⁻⁸ Ωm)
L = Length of wire (0.01 m)

A = cross section area (m²)

Solve:
[tex]R = \frac{(1.72*10^{-8})(0.01)}{\pi (0.001^2)} = 5.475 * 10^{-5} \Omega m[/tex]

Now, we can use the following relation (Ohm's Law):

[tex]\epsilon = iR\\\\\epsilon = \frac{Q}{t}R[/tex]

*Since current is equivalent to Q/t.

Plug in the value of R and set the two equations equal to each other.

[tex]\frac{Q}{t}(5.475 * 10^{-5}) = \frac{0.003927}{t}[/tex]

Cancel out 't'.

[tex]Q (5.475 * 10^{-5}) = 0.003927 \\\\Q = \frac{0.003927}{5.475*10^{-5}} = \boxed{71.73 C}[/tex]


Related Questions

An object can not have a charge of?

Answers

Answer:

If an object is electrically neutral it has no net charge becuase it has the same number of protons as it does electrons, which are opposite charges that offset each other. No, that just means that the sum of all its positive and negative amounts of charge equals zero.

Explanation:

A television set is plugged into a 120 V outlet. The television circuit carries a current equal to 0.75 A. What is the overall resistance of the television set?​

Answers

Answer:

R = 160 Ω

Explanation:

A television set is plugged into a 120 V outlet. The television circuit carries a current equal to 0.75 A. What is the overall resistance of the television set?​

V = IR

120 volt = 0.75A * R

R = 160 Ω

Water of mass 3 kg at a temperature of 80 ℃ is added to 5 Kg of water at 5 ℃. Calculate the final temperature of the mixture

Answers

The final temperature of the mixture is 33.123 °C.

What is temperature?

Temperature can be defined as the hotness or coldness of a thing or place.

To calculate the final temperature of the mixture, we use the formula below.

Formula:

Heat gained by the cold water = heat lost by the hot watercm(t₃-t₁) = cm'(t₂-t₃)m((t₃-t₁) = m'(t₂-t₃)......... Equation 1

Where:

m = mass of the cold waterm' = mass of the hot watert₁ = Temperature of the cold watert₂ = Temperature of the hot watert₃ = Temperature of the mixture.

make t₃ the subject of the equation

t₃ = (mt₁+m't₂)/(m+m')............. Equation 2

From the question,

Given:

m = 5 kgm' = 3kgt₁ = 5 °Ct₂ = 80 °C

Substitute these values into equation 2

t₃ = [(5×5)+(3×80)]/(3+5)t₃ = (25+240)/8t₃ = 33.123 °C

Hence, The final temperature of the mixture is 33.123 °C.

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How can you describe the relationship between height and pressure?

Answers

Answer:

p = rho× g × h

Explanation:

p: pressure

rho : density

g : gravity acceleration

h : height

1. (30 pts) Let x(t) = cos(πt/2) be a continuous-time signal,
a. Sketch the signal for -4 b. Find the fundamental period of the signal (if it is periodic).
c. Determine if the signal is odd / even or neither.
d. Compute the energy of the signal for all time.
e. Compute the power of the signal for all time.

Answers

Given that the function of the wave is f(x) = cos(π•t/2), we have;

a. The graph of the function is attached

b. 4 units of time

c. Even

d. 4.935 J/kg

e. 1.234 W/kg

How can the factors of the wave be found?

a. Please find attached the graph of the signal created with GeoGebra

b. The period of the signal, T = 2•π/(π/2) = 4

c. The signal is even, given that it is symmetrical about the y-axis

d. The energy of the signal is given by the formula;

[tex] \frac{1}{2} \cdot \mu^{2} \cdot \omega ^{2} \cdot \: {a}^{2} \times \lambda[/tex]

Which gives;

E = 0.5 × 1.571² × 1² × 4 = 4.935 J/kg

e. The power of the wave is given by the formula;

E = 0.5 × 1.571² × 1² × 4 × 0.25 = 1.234 W/kg

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478 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to compress the gas by a factor of 12.0, starting from its initial volume?

I was thinking of using PV = W formula. Like

478 = P(V/2)
956 = PV

W = P(V/12)
12W/V = P
956 = (12W/V)(V)
956 = 12W
W = 79.66667 J

is this correct? could someone please help?

Answers

Answer:

Explanation:

I don't think so. Think about it. To compress the volume by a factor of 2 it takes 956 Joules.

Now you come along and you want to get the pressure for 1/12 of the volume. It's going to take a huge pressure to do that.

I would suggest that you have to use a modified form of the formula.

PV = 956

You need to compress the volume by 1/6

P(V/6) = 956

6 * PV/6 = 6 * 956

PV = 5736 J

Why did I only take 1/6? Because. 956 represents the pressure needed for 1/2 the volume. You need to multiply 1/2 * 1/6 to get 1/12

Can someone please help me with this assignment, this is due today

Answers

Answer:

did you get it done if not lmk I will help you out tomorrow when I get up

PLEASE HELP!!!!!!! MT TIME FOR MY TEST IS ALMOST OVER!!!

An ideal spring, with a pointer attached to its end, hangs next to a scale. With a 100-N weight attached, the pointer indicates "40" on the scale as shown. Using a 200-N weight instead results in "60" on the scale. Using an unknown weight X instead results in "30" on the scale. The weight of X is:

Answers

Answer:

50 N

Explanation:

Let the natural length of the spring = L

so

100 = k(40 - L)       (1)

200 = k(60 - L)       (2)

(2)/(1):   2 = (60 - L)/(40 - L)

60 - L = 2(40 - L)

60 - L = 80 - 2L

2L - L = 80 - 60

L = 20

Sub it into (1):

100 = k(40 - 20) = 20k

k = 100/20 = 5 N/in

Now

X = k(30 - L) = 5(30 - 20) = 50 N

Draw free body diagrams for the following objects: (12pts)
1A) A coaster sitting under a cup of coffee.
1B) A car slowing down as it approaches a stop sign.
1C) Your test stuck to your fridge by a magnet.
1D) A baseball just before it leaves the bat.

Answers

(a) The force diagram of a coaster sitting under a cup of coffee includes the weight of the coater plus the weight of coffee acting downwards.

(b) The force diagram of a car slowing down as it approaches a stop sign includes force of the car and frictional force opposing the motion.

(c) The force diagram includes the force of the test and action of the fridge which are eqaul and opposite.

(d) The force diagram of baseball before it leaves the bat incudes only the weight of the baseball acting downwards.

Force diagram of coaster sitting under a cup of coffee

The force diagram of a coaster sitting under a cup of coffee includes the weight of the coater plus the weight of coffee acting downwards.

                                        ↑ Fn

                                        Ф                  Fn = W

                                        ↓ W

Where;

W is weight of the coaster plus weight of coffeFn is the normal reactionForce diagram of a car slowing down as it approaches a stop sign

The force diagram inlcudes the applied force and frictional force opposing the motion.

                                      Ff ←  Ф  → F

where;

Ff is the kinetic frictional forceF is force of the car

Force diagram of test stuck to your fridge

The force diagram includes the force of the test and action of the fridge which are eqaul and opposite.

                   

                               Fb ←  Ф → Fa

where;

Fa is the force of the testFb is the force of the fridge

Force diagram of baseball before it leaves the bat

The force diagram includes only the weight of the baseball acting downwards.

                                 Ф

                                  ↓

                                 W = mg

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17) Which object would likely have the greatest velocity
a. a bouncy ball
b. a bowling call
a go-kart
d. a school bus

Answers

Answer: b or d

Explanation: b or d

Answer:

a

Explanation:

F= ma

interestingly

when you increase the mass the acceleration decreases while when the mass decreases the acceleration increases

(man, PHYSICS IS JUST THE BESY)

A: a

object with the smallest mass has largest acceleration

What is the angular momentum at a radius of 2 m with an object of 5 kg at a
velocity of 20 m/s?

Answers

The angular momentum is 200 kg m^2 s^-1

what is angular momentum?

Angular momentum is the product of  linear momentum and the perpendicular distance. Linear momentum is the product of mass and velocity, where radius is the perpendicular distance

Angular momentum = mass * velocity * radius

Angular momentum = 5 * 2 * 20

Angular momentum = 200 kg m^2 s^-1

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how many types air
I need a formula of calculating the s i unti of force​

Answers

The si unit of force is newton.

so, F is eqal to m*g

Answer:

F=ma

Explanation:

force is actually a derived quantity

newton is a unit derived from kg.m/s^2

This is from Newton's second law of motion

F=ma

A bar of soap when weighed in air has a weight of 5,2N . When completely immersed in water , however it has a weight of 3,7N . What is the volume of the bar of soap

Answers

Answer:

A bar of soap when weighed in air has a weight of 5,2N . When completely immersed in water , however it has a weight of 3,7N . What is the volume of the bar of soap

what is one type of compact star with a mass similar to the sun but a diameter similar to earth?

Answers

Explanation:

neutron star, any of a class of extremely dense, compact stars thought to be composed primarily of neutrons. Neutron stars are typically about 20 km (12 miles) in diameter. Their masses range between 1.18 and 1.97 times that of the Sun, but most are 1.35 times that of the Sun.

3
3
If a jogger runs a 10 kilometer race in 60 minutes, what is
her average speed?
A
10 km/hr
B
5 km/hr
С 6 km/hr
D
1.66 km/hr

Answers

If a jogger runs a 10 kilometer race in 60 minutes, her average speed is 10km/hr. Details about average speed can be found below.

How to calculate average speed?

Average speed can be calculated by dividing the distance moved by a body by the time taken. That is;

Average speed = Distance/Time

According to this question, a jogger runs a 10 kilometer race in 60 minutes. The average speed is calculated as follows:

Average speed = 10km/1hr

Average speed = 10km/hr.

Therefore, if a jogger runs a 10 kilometer race in 60 minutes, her average speed is 10km/hr.

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_______ is an SI unit for mass.
A. mass
B. kilogram
C. newton
D. acceleration of gravity
E. weight

Answers

Answer:

B. kilogram

I hope this helps you

:)

Answer:

B. kilogram

Explanation:

When stated in the unit J s, which is equal to kg m2 s1, the kilogram (kg) is defined by considering the fixed numerical value of the Planck constant h to be 6.62607015 1034 when expressed in the unit J s, which is equivalent to kg m2 s1. The United States Prototype Kilogram 20, a platinum-iridium cylinder held at NIST, is the country's principal mass standard. The kilogram was initially known as the Kilogram of the Archives, and it was defined as the mass of one cubic decimeter of water at its greatest density temperature. It was superseded by the International Prototype Kilogram following the International Metric Convention in 1875, which became the unit of mass without reference to the mass of a cubic decimeter of water or the Archives Kilogram. National Prototype Meters and Kilograms were allocated to each country that signed the International Metric Convention. Learn more about the kilogram's history and current definition. The kilogram (kg) is the only SI basic unit whose name and symbol incorporate a prefix for historical reasons. The SI prefix for 1000 or 103 is "kilo." Prefix names and symbols are attached to the unit name "gram," and prefix symbols are attached to the unit symbol "g," to create names and symbols for decimal multiples and submultiples of the unit of mass. Find out more about this historical oddity.

Units of Mass

10 milligrams (mg) = 1 centigrams (cg)

10 centigrams = 1 decigrams (dg) = 100 milligrams

10 decigrams = 1 gram (g)

10 decigrams = 1000 milligrams

10 grams = 1 dekagrams (dag)

10 dekagrams = 1 hectogram (hg)

10 dekagrams = 100 grams

10 hectograms = 1 kilogram (kg)

10 hectograms = 1000 grams

1000 kilograms = 1 megagram (Mg) or 1 metric ton (t)

A body's mass is a measurement of its inertial property, or the amount of stuff it contains. The force imposed on a body by gravity or the force required to maintain it is measured by its weight. On Earth, gravity accelerates a body downward at around 9.8 m/s2. In the context of weights and measurements, weight is frequently used as a synonym for mass. The verb "to weigh," for example, meaning "to ascertain the mass of" or "to have a mass of." Weight should be phased out in favor of mass, and the term mass should be used when mass is indicated. The kilogram is the SI unit of mass (kg). The weight of a body in a given reference frame is defined in science and technology as the force that causes the body to accelerate at the same rate as the local acceleration of free fall in that reference frame. As a result, the newton is the SI unit for the amount weight defined in this way (force) (N).

Twelve identical point charges q are equally spaced around the circumference of a circle of radius R. The circle is centered at the origin. One of the twelve charges, which happens to be on the positive x axis, is now moved to the center of the circle.

A) Determine the magnitude of the total electric force exerted on this charge.
Express your answer in terms of Coulomb's constant k and the variables q and R .
F total = ?

B)Determine the direction of the total electric force exerted on this charge.
Express your answer as an integer.
θ = ? degrees

Answers

Answer:

For B it is 0

Explanation:

I think

. Radiation travels at the speed of light
T or F?

Answers

Answer:

electromagnetic radiation moves at the speed of light

Consider a parallel-plate capacitor with plates of area A and with separation d.
Part A
Find F(V), the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor.
Express your answer in terms of given quantities and ϵ0.
View Available Hint(s)for Part A


F(V)

Answers

The magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor is determined as [tex]\frac{V^2 A \varepsilon _o }{2d^2}[/tex].

Magnitude of the force

The magnitude of the force each plate experiences due to the other plate is determined as follows;

F = U/d

where;

U is potential energy stored in the capacitor

[tex]F = \frac{1}{2} \frac{Q^2}{C} \times \frac{1}{d} \\\\[/tex]

Q = CV

[tex]F = \frac{1}{2} \frac{C^2V^2}{C d} = \frac{CV^2}{2d}[/tex]

where;

C is the capacitance

The capacitance is given as;

[tex]C = \frac{\varepsilon _o A }{d}[/tex]

[tex]F = \frac{\varepsilon _o A }{d} \times \frac{V^2}{2d} \\\\F = \frac{V^2 A \varepsilon _o }{2d^2}[/tex]

Thus, the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor is determined as [tex]\frac{V^2 A \varepsilon _o }{2d^2}[/tex].

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state the precautions that is taken when charging a metal objectexplain why a rubber balloon rubbed will often stick to the wall when it has been rubbed ​

Answers

Answer:

The balloon will attach to the wall because the balloon's negative charges will drive the electrons in the wall to shift to the other side of their atoms, leaving the wall's surface positively charged.

When a penny is dropped, it takes 16 seconds. What is its height

Answers

My guess is 8
16 / 2 = 8

2. All of the following are examples of physical properties except:
A. tearing B. density C. melting point D. boiling point

Answers

All of the following are examples of physical properties except tearing.

What is Physical property?

This is used to describe the state of a physical system and is usually measurable.

Examples include:

DensityMelting point Boiling point

Tearing isn't an example of a physical property which was why option A was chosen as the most appropriate choice.

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Q. 1 MWH is equal to ------- joules
a.3.6*10^10
b.3.6*10^6
c.3.6*10^9
d.3.6

Answers

Correct option is B
1 kWh is equal to 3.6 × 10
6

1 Watt-sec is a Joule a watt expended for a second.
Because there’s 3600 seconds in an hour,
then 3600 Ws = 1 Wh = 3600 Joules = 3.6 kJ
so 1000 Wh = 1 kWh = 1000 × 3.6 kJ = 3.6 MJ = 3.6 × 10
6

Please answer and show formula

Answers

Voltage = 5 volts

[tex]\sf \dfrac{number \ of \ turns \ in \ primary \ coil}{voltage\ in \ primary \ coil} = \dfrac{number \ of \ turns \ in \ secondary\ coil}{voltage\ in \ secondary\ coil}[/tex]

[tex]\hookrightarrow \sf \dfrac{400}{100} = \dfrac{20}{V_2}[/tex]

[tex]\hookrightarrow \sf 400V_2}{} = 20*100[/tex]

[tex]\hookrightarrow \sf V_2 = 5 V[/tex]

A 330-ohm resistor is connected to a 5-volt battery. The current through the resistor is

Answers

Question :-

A 330 Ohm Resistor is connected to a 5 Volt Battery . What is the Current through the Resistor ?

Answer :-

Current of the Battery is 66 Ampere.

Explanation :-

As per the provided information in the given question, The Resistance is given as 330 Ohm . The Voltage is given as 5 Volt . And, we have been asked to calculate the Current .

For calculating the Current , we will use the Formula :-

[tex] \bigstar \: \: \: \boxed{ \sf{ \: Current \: = \: \dfrac{Voltage}{Resistance} \: }} [/tex]

Therefore , by Substituting the given values in the above Formula :-

[tex] \dag \: \: \: \sf {Current \: = \: \dfrac {Voltage}{Resistance} } [/tex]

[tex] \longmapsto \: \: \: \sf {Current \: = \: \dfrac {5}{330} } [/tex]

[tex] \longmapsto \: \: \: \sf {Current \: = \: \dfrac {1}{66} } [/tex]

[tex] \longmapsto \: \: \: \textbf {\textsf {Current \: = \: 66 }} [/tex]

Hence :-

Current = 66 Ampere .

[tex] \underline {\rule {180pt} {4pt}} [/tex]

Additional Information :-

[tex] \Longrightarrow \: \: \: \sf {Voltage \: = \: Current \: \times \: Resistance} [/tex]

[tex] \Longrightarrow \: \: \: \sf {Current \: = \: \dfrac {Voltage}{Resistance} } [/tex]

[tex] \Longrightarrow \: \: \: \sf {Resistance \: = \: \dfrac {Voltage}{Current} } [/tex]

Who came up with the 3 Laws of Motion?

Answers

Answer:

Isaac Newton came up with the 3 Laws of Motion

Answer:

Isaac Newton came up with three laws of motion

Two identical charges are located 1 m apart and feel a 1 N repulsive electric force. What is the charge of each particle.

Answers

The charge on each particles which are 1 m apart and feeling a repulsive force of 1 N is 1.05×10¯⁵ C

Assumption

Let the charge on each particles be q

How to determine the charge Final force (F) = 1 NDistance apart (r) = 1 mElectrical constant (K) = 9×10⁹ Nm²/C²Charge on 1st particle (q₁) = q =? Charge on 2nd particle (q₂) = q =?

The charge on each particle can be obtained by using the Coulomb's law equation as shown below:

F = Kq₁q₂ / r²

F = Kq² / r²

1 = (9×10⁹ × q²) / 1²

1 = 9×10⁹ × q²

Divide both side by 9×10⁹

q² = 1 / 9×10⁹

Take the square root of both side

q =  √(1 / 9×10⁹)

q = 1.05×10¯⁵ C

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If a transverse wave osculates 7 times every second and the speed of the wave is 27 m/s what is the wavelength of the wave

Answers

Explanation:

formula is ˠ=vf

f=1/T

1/7

f=0.14

wavelength=27Ⅹ0.14

=3.78m

OR

7Ⅹ27

=189m

Find the temperature

Answers

Answer:

-------1

Explanation:

beacuse that is what i know

A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume calorimeter. Consequently,
the temperature of the water rose by 4.20 ºC. If the heat capacity of the bomb plus water was 10.4 kJ/ºC,
calculate the molar heat of combustion of methanol.

Answers

Mass of methanol = 1.922g; Change in temperature = 5.14° C; Heat capacity of the bomb calorimeter + water = 8.69kJ/°C. Number of moles.
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