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
Tcecarenko [31]
3 years ago
5

A long straight wire carries current to the right of the page. A rectangular loop is positioned directly under the wire in the p

lane of the page and carries current in the clockwise direction.
The net force exerted by this loop on the straight wire with the current is directed:

a) towards the loop.
b) away from the loop.
c) to the right of the page.
d) to the left of the page.
e) is zero.
Physics
1 answer:
slavikrds [6]3 years ago
8 0

Answer:

Correct option A.

The net force exerted by this loop on the straight wire with the current is directed TOWARDS THE LOOP

Explanation:

The magnetic field exerts a force on a current-carrying wire in a direction given by the right-hand rule 1 (the same direction as that on the individual moving charges). This force can easily be large enough to move the wire since typical currents consist of very large numbers of moving charges.

Given that,

The wire's current is directed towards the right of the page.

The rectangular loops carry current in a clockwise direction.

Since the 'dot' field is increasing hence the induced magnetic field is 'cross', i.e. into the page and by the right-hand rule, the induced current is clockwise.

Then the magnetic field is into the page.

Since was known that

F= iL×B

Note the current is through the wire. Then, the length is in direction of the current.

Note: this equation gives the magnetic force that acts on a length L of a straight wire carrying a current (i) and immersed in a uniform magnetic field (B), that is perpendicular to the wire.

So, the magnetic field is always perpendicular to the current.

So using right hand rule,

F = i(L×B)

The length is to the right i.e. +x direction and the Magnetic field is perpendicular to the plane, i.e. in the +z direction

F = i (L•i × B•k)

F = iLB (i×k)

F = iLB•(-j)

F = -iLB•j

Then, the force is in the negative y-direction i.e. towards the loop.

You might be interested in
Object A is positively charged. Object A and Object B
ozzi

Answer: object B is negatively charged, object C is positively charged and object D is also positively charged

Explanation: since unlike charges attract and like charges repel, for object A which is positively charged and B to attract B must be negatively charged and then for B which is negatively charged and C to attract C must be positively charged and for C and D to repel they have to be of thesame charge which means D is positive as well.

4 0
2 years ago
An archer shoots an arrow 75 m distant target; the bull's-eye of the target is at the same height as the release height of the a
Liono4ka [1.6K]

Answer:

16.25^{\circ}

Explanation:

R = Horizontal range of projectile = 75 m

v = Velocity of projectile = 37 m/s

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Horizontal range is given by

R=\dfrac{v^2\sin2\theta}{g}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{Rg}{v^2}}{2}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{75\times 9.81}{37^2}}{2}\\\Rightarrow \theta=16.25^{\circ}

The angle at which the arrow is to be released is 16.25^{\circ}.

4 0
3 years ago
To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(
DerKrebs [107]

Answer:

0.0549 m

Explanation:

Given that

equation y(x,t)=Acos(kx−ωt)

speed  v = 8.5 m/s

amplitude A = 5.5*10^−2 m

wavelength λ   = 0.5 m

transverse displacement = ?

v = angular frequency / wave number

and

wave number = 2π/ λ

wave number =  2 * 3.142 / 0.5

wave number = 12.568

angular frequency = v k

angular frequency = 8.5 * 12.568

angular frequency = 106.828 rad/sec ~= 107 rad/sec

so

equation y(x,t)=Acos(kx−ωt)

y(x,t)= 5.5*10^−2 cos(12.568 x−107t)

when x =0 and and t = 0

maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))

maximum y(x,t)= 5.5*10^−2  m

and when x =  x = 1.52 m and t = 0.150 s

y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )

y(x,t)= 5.5*10^−2 × (0.9986)

y(x,t) = 0.0549 m

so the transverse displacement is  0.0549 m

5 0
3 years ago
Superman stops a "speeding locomotive" of mass 8.0x103 kg in 4.0 seconds. If the train was originally moving at 40.m/s,
madreJ [45]

Hello!

<em>a) What is the change in momentum</em>?

The change of the momentum is the velocity, because the velocity is reduced to zero.

We can calculate the aceleration, applicating the formula:

\boxed{a=\frac{V-Vi}{t} }

Like you see, the final velocity will be zero, so:

a = \dfrac{0m/s-40m/s}{4s}

a = \dfrac{-40m/s}{4s}

a = -10\ m/s^{2}

Like you see, the aceleration applicate by superman is of <u>-10 m/s^2.</u>

If the aceleration is negattive, means the velocity will decrease.

<em>b.) What is the magnitud of the force wich superman exerts on the train?</em>

For calculate the force applicate for superman, lets applicate the second law of Newton:

\boxed{F=ma}

Lets replace and resolve it:

F = 8x10^3 kg * (-10 m/s^2)

F = 8000 kg * (-10 m/s^2)

F = -80 000 N

The force applicated is of <u>-80000 Newtons.</u>

When the force is negative means the force applicated in the another direction of the object what is traveling.

8 0
2 years ago
There is a uniform magnetic field of magnitude B, pervading all space, perpendicular to the plane of rod and rails. The rod is r
Charra [1.4K]

The right hand rule to find the direction of the magnetic field for a falling bar is:

  • The charge is positive the magnetic field is outgoing, horizontally and towards us.
  • The charge of the bar is negative, the magnetic field is incoming, that is horizontal away from us.

The magnetic force is given by the vector product of the velocity and the magnetic field.

        F = q v x B

Where the bolds indicate vectors, F is the force, q the charge on the particle, v the velocity and B the magnetic field.

In the vector product, the vectors are perpendicular, which is why the right-hand rule has been established, see attached:

  • The thumb points in the direction of speed.
  • Fingers extended in the direction of the magnetic field.
  • The palm is in the direction of the force if the charge is positive and in the opposite direction if the charge is negative.

They indicate that the bar is dropped, therefore its speed is vertical and downwards, it moves to the left therefore this is the direction of the force, we use the right hand rule, the magnetic field must be horizontal, we have two possibilities:

  • If the charge is positive the magnetic field is outgoing, horizontally and towards us.
  • If the charge of the bar is negative, the magnetic field is incoming, that is, horizontal away from us

In conclusion using the right hand rule we can find the direction of the magnetic field for a falling bar is:

  • The charge of the bar is negative, the magnetic field is incoming, that is horizontal away from us.
  • The charge is positive the magnetic field is outgoing, horizontally and towards us.

Learn more about the right hand rule here:  brainly.com/question/12847190

6 0
2 years ago
Other questions:
  • How is a crime scene secured ?
    10·2 answers
  • More massive objects have a large gravitational force or weight. Why do heavier objects accelerate to the ground at the same rat
    10·2 answers
  • Which is a product of photosynthesis?
    8·1 answer
  • In an experiment if doubling the manipulated variable results in doubling of the responding variable the relationship between th
    14·1 answer
  • What happens at the end of silence of the lambs?
    11·1 answer
  • Suppose a miracle car has a 100% efficient engine and burns fuel having an energy content of 40 MJ/L. If the air resistance and
    14·1 answer
  • Gamma rays
    14·1 answer
  • Which of the following objects is in static equilibrium?
    10·1 answer
  • Forces that are equal in size but opposite in direction are called what ?
    10·2 answers
  • How long will it take to travel 2000 Km at 500 Km<br> a.100,000 hr<br> b.2500 Km<br> c.4 hr
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!