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Scorpion4ik [409]
2 years ago
14

Two circular coils are concentric and lie in the same plane. The inner coil contains 170 turns of wire, has a radius of 0.0095 m

, and carries a current of 8.9 A. The outer coil contains 150 turns and has a radius of 0.015 m. What must be the magnitude of the current in the outer coil, such that the net magnetic field at the common center of the two coils is zero?
Physics
1 answer:
PIT_PIT [208]2 years ago
4 0

Answer:

Current in outer circle will be 15.826 A

Explanation:

We have given number of turns in inner coil N_I=170

Radius of inner circle r_i=0.0095m

Current in the inner circle I_i=8.9A

Number of turns in outer circle N_o=150

Radius of outer circle r_o=0.015m

We have to find the current in outer circle so that net magnetic field will zero

For net magnetic field current must be in opposite direction as in inner circle

We know that magnetic field is given due to circular coil is given  by

B=\frac{N\mu _0I}{2r}

For net magnetic field zero

\frac{N_I\mu _0I_I}{2r_I}=\frac{N_O\mu _0I_0}{2r_O}

So \frac{170\times \mu _0\times 8.9}{2\times 0.0095}=\frac{150\times \mu _0I_O}{2\times 0.015}

I_O=15.92A

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Two students are watching a person riding a skateboard up and down a ramp. Each student shares what they think about the energy
Setler79 [48]

Answer:

The correct option is;

Raymond: I think the skateboarder has the same total energy at all points on the ramp

Explanation:

The total energy, also known as the total mechanical energy, is the sum of the kinetic and potential energies of the skateboarder

Given that the potential energy is the energy gained due to elevation, the maximum potential energy is obtained at the top of the ramp, while the maximum kinetic energy, which is the energy due to motion, is at the bottom of the ramp where the skateboarder moves fastest.

However, by the energy conservation principle, the kinetic energy of he skateboarder comes from the conversion of the potential energy, such that the total energy is the same at any particular point on the ramp.

6 0
2 years ago
What is the kinetic energy of a 120-cm thin uniform rod with a mass of 450 g that is rotating about its center at 3.60 rad/s?
goldfiish [28.3K]

Answer:

1.05 J.

Explanation:

Kinetic Energy: This is the energy possessed by a body due to its motion. The S.I unit of kinetic energy is Joules (J). The formula of kinetic energy is given as

Ek = 1/2mv²................. Equation 1

Where Ek = kinetic energy, m = mass of the uniform rod, v = liner velocity of the rod.

But,

v = αr .......................... Equation 2

Where α = angular velocity of the rod, r = radius of the circle.

Given: α = 3.6 red/s, r = 120/2 = 60 cm = 0.6 m.

Substitute into equation 2

v = 3.6(0.6)

v = 2.16 m/s.

Also given: m = 450 g = 0.45 kg.

Substitute into equation 1

Ek = 1/2(0.45)(2.16²)

Ek = 1.05 J.

4 0
3 years ago
How did the continental Shelf form? Please help, thank you!! :)
shutvik [7]
The movements of the tectonic plates
6 0
3 years ago
(b) A piece of wood of volume 0.6 m² floats in water. Find the volume
enot [183]

Answer:

Explanation:

Let the volume below water be v . Then

buoyant force = v d g where d is density of water , g is acceleration due to gravity

= v x 1000 x g

weight of wood piece = volume x density of wood x g

= .6 x 600 x g

for equilibrium while floating

buoyant force = weight

= v x 1000 x g  =  .6 x 600 x g

v = .36 m²

volume above water or volume exposed = .6 - .36

= .24 m²

When immersed completely ,

buoyant force = .6 x 1000 x 9.8

= 5880 N

weight of wood

=  .6 x 600 x g

= 3528 N

buoyant force is more than the weight . In order to equalise them for floating with full volume in water

weight required = 5880 - 3528

= 2352 N.

6 0
3 years ago
suppose that you look into a photometer's eyepiece and the fluorescent disks appear to be equal in intensity. If the distance be
d1i1m1o1n [39]
Use the Inverse square law, Intensity (I) of a light is inversely proportional to the square of the distance(d).

I=1/(d*d)

Let Intensity for lamp 1 is L1 distance be D1 so on, L2 D2 for Intensity for lamp 2 and its distance.

L1/L2=(D2*D2)/(D1*D1)

L1/15=(200*200)/(400*400)
L1=15*0.25
L1=3.75 <span>candela</span>

7 0
3 years ago
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