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SIZIF [17.4K]
3 years ago
13

An electric motor is connected to a battery. The current flows through brushes to a commutator ring, which is attached to a elec

tromagnet turning through a permanent in a magnetic field what is the purpose of the commutator ring
Physics
1 answer:
Rzqust [24]3 years ago
8 0

Answer:

Commutator is a ring which reverse the direction of current in AC circuit so that the coil connected to it will continuous to move in the same direction.

Explanation:

In motors there exist a coil which is rotated due to torque of magnetic field when current flow through it. Since AC current is used to run the motor so we know that AC current changes its direction after half cycle.

So here commutator plays an important role to reverse the direction of current after every half cycle so that the current goes in same direction always into the coil.

This will produce a constant direction torque on the coil so that it will rotate in same sense always.

So commutator role is to provide same direction current to the coil by reversing its direction after every half cycle

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Consider a coaxial cable (like the kind that is used to carry a signal to your TV). In this cable, a current I runs in one direc
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Answer:

0 < r < r_exterior     B_total = \frac{\mu_o I}{2\pi  r}

r > r_exterior            B_total = 0

Explanation:

The magnetic field created by the wire can be found using Ampere's law

        ∫ B. ds = μ₀ I

bold indicates vectors and the current is inside the selected path

           

outside the inner cable

          B₁ (2π r) = μ₀ I

          B₁ = \frac{\mu_o I}{2\pi  r}

the direction of this field is found by placing the thumb in the direction of the current and the other fingers closed the direction of the magnetic field which is circular in this case.

For the outer shell

for the case   r> r_exterior

         

           B₂ = \frac{\mu_o I}{2\pi  r}

This current is in the opposite direction to the current in wire 1, so the magnetic field has a rotation in the opposite direction

for the case r <r_exterior

in this case all the current is outside the point of interest, consequently not as there is no internal current, the field produced is zero

           B₂ = 0

Now we can find the field created by each part

0 < r < r_exterior

          B_total = B₁

          B_total = \frac{\mu_o I}{2\pi  r}  

r > r_exterior

          B_total = B₁ -B₂

          B_total = 0

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Explanation:

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