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
masya89 [10]
3 years ago
11

A wire loop is suspended from a string that is attached to point P in the drawing. When released, the loop swings downward, from

left to right, through a uniform magnetic field, with the plane of the loop remaining perpendicular to the plane of the paper at all times. Determine the direction of the current induced in the loop as it swings past the locations labeled (a) I and (b) II. Specify the direction of the current in terms of the points x, y, and z on the loop (e.g., x→y→z or z→y→x). The points x, y, and z lie behind the plane of the paper. What is the direction of the induced current at the locations (c) I and (d) II when the loop swings back, from right to left?

Physics
1 answer:
Naddik [55]3 years ago
5 0

Complete Question

The complete question iws shown on the first uploaded image  

Answer:

a

    y \to z \to x

b

  x \to z \to y

Explanation:

Now looking at the diagram let take that the magnetic field is moving in the x-axis

 Now the magnetic force is mathematically represented as

             F =  I L x B

Note (The x is showing cross product )

Note the force(y-axis) is perpendicular to the field direction (x-axis)

Now when the loop is swinging forward

 The motion of the loop is  from   y to z to to x to y

Now since the force is perpendicular to the motion(velocity) of the loop

Hence the force would be from z to y and back to z  

and from lenze law the induce current opposes the force so the direction will be from y to z to x

Now when the loop is swinging backward

   The motion of the induced current will now be   x to z to y

 

You might be interested in
In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue
agasfer [191]

Answer:

a)  p₀ = 1.2 kg m / s,  b) p_f = 1.2 kg m / s,  c)   θ = 12.36, d)  v_{2f} = 1.278 m/s

Explanation:

a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved

a) the initial impulse is

        p₀ = m v₁₀ + 0

        p₀ = 0.6 2

        p₀ = 1.2 kg m / s

b) as the system is isolated, the moment is conserved so

       p_f = 1.2 kg m / s

we define a reference system where the x-axis coincides with the initial movement of the cue ball

we write the final moment for each axis

X axis

        p₀ₓ = 1.2 kg m / s

        p_{fx} = m v1f cos 20 + m v2f cos θ

        p₀ = p_f

       1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ

        1.2482 = v_{2f} cos θ

Y axis  

       p_{oy} = 0

       p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ

       0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ

       0.2736 = v_{2f} sin θ

we write our system of equations

        0.2736 = v_{2f} sin θ

        1.2482 = v_{2f} cos θ

divide to solve

        0.219 = tan θ

         θ = tan⁻¹ 0.21919

         θ = 12.36

let's look for speed

           0.2736 = v_{2f} sin θ

            v_{2f} = 0.2736 / sin 12.36

           v_{2f} = 1.278 m / s

7 0
2 years ago
Two particles collide, one of which was initially moving and the other initially at rest. Is it possible for both particles to b
solmaris [256]

Answer:

Not possible

Explanation:

Unless there's some extra external force to keep both particles at rest after the collision, the momentum must be conserved before and after the collision.

So before the collision, 1 particle is at rest, 1 not ->  total momentum is non-zero

After the collision, both particles are at rest -> total momentum is zero which is different from before.

Therefore this is not possible.

4 0
3 years ago
A wheel rotates at 24 revolutions every 3 minutes. This is equivalent to
mrs_skeptik [129]
8 revolutions every minute
8 0
2 years ago
If an ice skater with a mass of 50 kg was accelerating at 10 m/s and is on her way to hit a
IrinaK [193]

500N

Explanation:

Given parameters:

Mass of skater = 50kg

Acceleration = 10m/s

Unknown:

Force  = ?

Solution;

Force is a push or pull on a body. it is given as the product of mass and acceleration:

               Force = mass x acceleration

    Input the parameters:

                  Force = 50 x 10 = 500N

She hits the wall with a force of 500N

Learn more:

Force brainly.com/question/3820012

#learnwithBrainly

5 0
3 years ago
A disk with a hole has inner radius rin and outer radius rout. the disk is uniformly charged with total charge q. find an expres
Effectus [21]
Here it is.  I used the surface charge density σ as q/Area, and then wrote it out in terms of q at the end

3 0
3 years ago
Other questions:
  • A crane exerts a net force of 900 N upward on a 750-kilogram car as the crane starts to lift the car from the deck of a cargo sh
    13·1 answer
  • Why is the sky blue all the time
    14·1 answer
  • Nancy rides her horse 26km in 142 minutes. What is her average speed in kilometers per hour?
    5·1 answer
  • What is the best way to describe the rate of motion of an object that changes speed several times over a period of time is to ca
    7·1 answer
  • What determines how the different moon phases appear from earth
    14·2 answers
  • How do i figure this out ? Anyone help me please
    14·1 answer
  • How many miles per day can you walk at a MODERATE Intensity level and your heart rate is 170?
    9·1 answer
  • The speed an
    15·1 answer
  • In a game of tug of war, team one pulls to the right with a force of 500 newtons and team two pulls to the left with a force of
    6·1 answer
  • A charge Q is transferred from an initially uncharged plastic ball to an identical ball 10.0 cm away. The force of attraction is
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!