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ale4655 [162]
3 years ago
11

There is a uniform magnetic field of magnitude B, pervading all space, perpendicular to the plane of rod and rails. The rod is r

eleased from rest, and it is observed that it accelerates to the left. In what direction does the magnetic field point?
Physics
1 answer:
Charra [1.4K]3 years ago
6 0

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

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Hope this helps.
5 0
3 years ago
A stationary shell is exploded in to three fragments A, B, C of masses in the ratio 1:2:3. A travels
spin [16.1K]

Answer:

20 m/s

Explanation:

If the mass of fragment A is m, then the mass of fragment B is 2m, and the mass of fragment C is 3m.

The velocity of A is 60 m/s at angle 0°.

The velocity of B is 30 m/s at angle 120°.

The velocity of C is v at angle θ.

In the x direction:

Momentum before = momentum after

(m + 2m + 3m) (0) = m (60 cos 0°) + 2m (30 cos 120°) + 3m (v cos θ)

0 = 60m − 30m + 3m v cos θ

0 = 30m + 3m v cos θ

-30m = 3m v cos θ

-10 = v cos θ

In the y direction:

Momentum before = momentum after

(m + 2m + 3m) (0) = m (60 sin 0°) + 2m (30 sin 120°) + 3m (v sin θ)

0 = 0 + 30√3 m + 3m v sin θ

-30√3m = 3m v sin θ

-10√3 = v sin θ

Square the two equations and add together:

(-10)² + (-10√3)² = (v cos θ)² + (v sin θ)²

100 + 300 = v² cos² θ + v² sin² θ

400 = v² (cos² θ + sin² θ)

400 = v²

v = 20

The speed of fragment C is 20 m/s.

7 0
3 years ago
A 9V battery is directly connected to each of 3 LED bulbs. Select the statement that accurately describes this circuit.
dem82 [27]

Answer:

where are the statements?

Explanation:

8 0
3 years ago
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VMariaS [17]

Answer:

a.proton, proton

hope this helped :) have a goodday

6 0
4 years ago
A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi
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We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
dI =r^2*dm
where dm =M(b*dr)/(ab)
Now we find the moment of inertia by integrating from -a/2 to a/2
The moment of inertia is
I= \int\limits^{-a/2}_{a/2} {r^2*dm} = M \int\limits^{-a/2}_{a/2} r^2(b*dr)/(ab)=(M/a)(r^3/3) (from (-a/2) toI=(M/3a)(a^3/8 +a^3/8)=(Ma^2)/12 (a/2))



4 0
4 years ago
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