1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Alja [10]
3 years ago
5

Suppose the loop is moving toward the solenoid (to the right). Will current flow through the loop down the front, up the front,

or not at all? Explain in terms of magnetic forces exerted on the charges in the wire of the loop. Describe the direction of the magnetic moment of the loop. Given your answer, do you expect the loop to be attracted toward or repelled from the solenoid?
Engineering
2 answers:
Tems11 [23]3 years ago
6 0

Answer:

See explanation

Explanation:

The magnetic force is

F = qvB sin θ

We see that sin θ = 1, since the angle between the velocity and the direction of the field is 90º. Entering the other given quantities yields

F

=

(

20

×

10

−

9

C

)

(

10

m/s

)

(

5

×

10

−

5

T

)

=

1

×

10

−

11

(

C

⋅

m/s

)

(

N

C

⋅

m/s

)

=

1

×

10

−

11

N

Eddi Din [679]3 years ago
5 0

Answer:

If the loop is moving toward the solenoid (to the right). The current flow through the loop up the front(From the right hand rule).

The magnetic force acting on the charges in the wire of the loop, will induce a current in the opposite direction to change producing it(This is Lenz law). This magnetic force on the charges will set up a magnetic moment in the opposing direction to the loop direction.

It is expected the the loop will be repelled from the solenoid.

Explanation:

Magnetic moment is set on the charges in a loop, under an influence of magnetic force. The directions of these quantities can be observed using the Right-Hand Rule and Lenz's law.

Directions of currents and associated magnetic fields can all be found using only the Right Hand Rule. When the fingers of the right hand are pointed in the direction of the magnetic field, the thumb points in the direction of the current. When the thumb is pointed in the direction of the magnetic field, the fingers point in the direction of the current.

By applying the Lenz's law, it can be seen that the direction of the current will be such as to oppose the change.

You might be interested in
Hi, everyone I'm a high school student in Texas. My engineering teacher is asking us to find an active engineer to complete a li
jenyasd209 [6]

Answer:ill look it up

Explanation:

6 0
2 years ago
You are a designer of a new processor. You have to choose between two possible implementations (called M1 and M2) of the same ar
Kaylis [27]

Answer:

A ) CPI : M1 = 2.4 , M2 = 2.65

B ) MIPS : M1 = 1083, M2 = 1056

C ) The machine that has a better performance based on MIPS is M1 and this is  by 27 million number of instructions per sec

Explanation:

A) The CPI for each machine

CPI = ( Total number of execution cycles ) / ( instruction counter executed )

For Machine 1 ( M1 )

we have to make some assumptions : number of instructions = 10

number of times A was executed = 4 , Number of times B was executed = 2.5 , number of times C was executed = 2.5, Number of times D was executed = 1. and this was based on the frequency given above

hence CPI for M1 =[ ( 1 * 4 ) + ( 3 * 2.5 ) + ( 3 * 2.5 ) + ( 5 * 1 ) ] / 10

       CPI  for M1 = 2.4

For Machine 2 ( M2 )

we have to make some assumptions : number of instructions = 10

number of times A was executed = 4 , Number of times B was executed = 2.  number of times C was executed = 1.5, Number of times D was executed = 2.5 times. and this was based on the frequency given above

Hence CPI  for M1 = [ ( 2 * 4 ) + ( 2 * 2 ) + ( 3 * 1.5 ) + ( 4 * 2.5 ) ] / 10

            CPI for M2 = 2.65

B ) Calculate the native MIPS  ratings for M1 and M2

MIPS = ( instruction counts ) / ( Execution time * 10^6 )

For M1

Assumptions : number of instructions executed = 10

                        each clock cycle = 0.3846 * 10^-9.      frequency = 2.6 Ghz

first we calculate the total execution time which is equal to :

= [ ( 1 * 4 ) + ( 3 * 2.5 ) + ( 3 * 2.5 ) + ( 5 * 1 ) ] * 0.3846 * 10 ^-9

= 9.2304 * 10 ^-9 secs

therefore the MIPS for M1

= 10 / ( 9.2304 * 10^-9 ) * 10^6  = 1083

                                         

For M2

Assumptions : number of instructions executed = 10

                        each clock cycle = 0.3846 * 10^-9.      frequency = 2.8 Ghz

first we calculate the total execution time which is equal to :

= [ (2*4) + (2*2) + (3 * 1.5 ) + ( 4 * 2.5 ) ] * 0.3846 * 10^-9 = 9.4631 * 10^-9 secs

therefor the MIPS for M2

= 10 / ( 9.4631*10^-9) * 10^6 = 1056

C ) The machine that has a better performance based on MIPS is M1 and this is by 27 million number of instructions per sec

8 0
3 years ago
How can input from multiple individuals improve design solutions for problems that occur because of a natural disaster, such as
Alla [95]

Answer:

Map and avoid high-risk zones.

Build hazard-resistant structures and houses.

Protect and develop hazard buffers (forests, reefs, etc.)

Develop culture of prevention and resilience.

Improve early warning and response systems.

Build institutions, and development policies and plans.

Explanation:

5 0
3 years ago
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0 mm (0.20 in
Sonja [21]

The load is 17156 N.

<u>Explanation:</u>

First compute the flexural strength from:  

σ = FL / πR^{3}

   = 3000 \times (40 \times 10^-3) / π (5 \times 10^-3)^3

σ = 305 \times 10^6 N / m^2.

We can now determine the load using:

F = 2σd^3 / 3L

  = 2(305 \times 10^6) (15 \times 10^-3)^3 / 3(40 \times 10^-3)

F = 17156 N.  

7 0
3 years ago
Carnot heat engine A operates between 20ºC and 520ºC. Carnot heat engine B operates between 20ºC and 820ºC. Which Carnot heat en
nikklg [1K]

Answer:

engine B is more efficient.

Explanation:

We know that Carnot cycle is an ideal cycle for all working heat engine.In Carnot cycle there are four processes in which two are constant temperature processes and others two are isentropic process.

We also kn ow that the efficiency of Carnot cycle given as follows  

\eta =1-\dfrac{T_1}{T_2}

Here temperature should be in Kelvin.

For engine A

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{520+273}

\eta =0.63

For engine B

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{820+273}

\eta =0.73

So from above we can say that engine B is more efficient.

4 0
3 years ago
Other questions:
  • Is CO, an air pollutant? How does it differ from other emissions resulting from the combustion of fossil fuels?
    7·1 answer
  • Many people use microwave ovens to cook their food. Which option is best described as an engineering endeavor related to microwa
    9·2 answers
  • 1. What is an op-amp? List the characteristics of an ideal op-amp
    11·1 answer
  • Suppose an underground storage tank has been leaking for many years, contaminating a groundwater and causing a contaminant conce
    8·1 answer
  • suppose we number the bytes in a w-bit word from 0 (less significant) to w/8-1 (most significant). write code for the followign
    11·1 answer
  • The title block generally contains ________.
    12·1 answer
  • Calculate the number of vacancies per cubic meter at 1000∘C for a metal that has an energy for vacancy formation of 1.22 eV/atom
    14·1 answer
  • What have you learned from the previous lesson? Let's try to check your prior knowledge
    9·1 answer
  • Materials to be used to build a watch tower​
    9·1 answer
  • The phase angle in a circuit is 45 degrees what's the power factor of this circuit?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!