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marissa [1.9K]
3 years ago
8

Assuming a current is flowing, what increases the strength of the magnetic field of a coiled electrical wire? Select all that ap

ply. Straightening the wire changing the direction of the current flow adding more coils to the wire placing a metal bar within the coils stopping the current.
Physics
1 answer:
goblinko [34]3 years ago
8 0

The strength of the magnetic field is increased by adding more coils to the wire and placing a metal bar within the coils.

The given problem is based on the concept and fundamentals of the magnetic strength in a current-carrying coil of wire. The current-carrying wire in a coiled form is known as Solenoid.

And the mathematical expression for the magnetic field in a solenoid is,[B = \dfrac{\mu_{0} \times N \times i}{L}

Here,

B is the strength of the magnetic field.

\mu_{0} is the magnetic permeability of free space.

N is the number of coils.

The strength of the magnetic field will increase by increasing the number of coils and placing the metallic slab (it will increase magnetic permeability) within the coils.

Thus, we can conclude that the strength of the magnetic field is increased by adding more coils to the wire and placing a metal bar within the coils.

learn more about the magnetic field here:

brainly.com/question/19542022

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A balloon of mass M is floating motionless in the air. A person of mass less than M is on a rope ladder hanging from the balloon
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Let the mass of the person be m. Total momentum is conserved (because the exterior forces on the system are balanced), especially the component in the vertical direction.

Given that,

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Ronch [10]

Answer:

Value of electric field along the axis and equitorial axis  E=31.25\ N/c and E = 15.625\ N/c respectively.

Explanation:

Given :

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Magnitude of charges , q=1\ nC = 10^{-9}\ C.

Dipole moment , p=qL=10^{-9}\times 3\times 10^{-3}=3\times 10^{-12} \ C\ m.

Case A) (x,y) = (12.0 cm, 0 cm) :

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E=\dfrac{2kp}{r^3}

Putting all values and r=12\times 10^{-2}\ m.

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Case B) (x,y) = (0 cm, 12.0 cm) :

Electric field of dipole on equitorial axis ,

E = \dfrac{kp}{r^3}

Putting all values and r=12\times 10^{-2}\ m.

We get , E = 15.625\ N/c.

Hence , this is the required solution.

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Answer:

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