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pav-90 [236]
4 years ago
15

A circular coil consists of N = 410 closely winded turns of wire and has a radius R = 0.75 m. A counterclockwise current I = 2.4

A is in the coil. The coil is set in a magnetic field of magnitude B = 1.1 T. The magnetic field points to right
Physics
1 answer:
Svetllana [295]4 years ago
7 0

The complete question is;

A circular coil consists of N = 410 closely winded turns of wire and has a radius R = 0.75 m. A counterclockwise current I = 2.4 A is in the coil. The coil is set in a magnetic field of magnitude B = 1.1 T.

a. Express the magnetic dipole moment μ in terms of the number of the turns N, the current I, and radius

R.

b. Which direction does μ go?

Answer:

A) μ = 1738.87 A.m²

B) The direction of the magnetic moment will be in upward direction.

Explanation:

We are given;

The number of circular coils;

N = 410

The radius of the coil;R = 0.75m

The current in the coils; I = 2.4 A

The strength of magnetic field;

B =1.1T

The formula for magnetic dipole moment is given as;

μ = NIA

Where;

N is number of turns

I is current

A is area

Now, area; A = πr²

So, A = π(0.75)²

Thus,plugging in relevant values, the magnetic dipole moment is;

μ = 410 * 2.4 * π(0.75)²

μ = 1738.87 A.m²

B) According to Fleming's right hand rule, the direction of the magnetic moment comes out to be in upward direction.

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Two identical asteroids travel side by side while touching one another. If the asteroids are composed of homogeneous pure iron a
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Answer:

diameter = 21.81 ft

Explanation:

The gravitational force equation is:

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Where:

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  • M => Mass of asteroid 1
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We also know that the asteroids are identical so their masses are identical:

  • M=m

Since R is the distance between centers of the two asteroids and their diameters are identical (see attachment), we can conclude that:

  • R=d=2r

We don´t know the mass of the asteroids but we know they are composed of pure iron, so we can relate their masses to their density:

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This is going to be helpful because the volume of a sphere is:

  • \frac{4}{3}\pi r^{3}

And know we can write our original force of gravity equation in terms of the radius of the asteroids:

  • F=\frac{GMm}{R^{2} } =\frac{Gmm}{(2r)^{2} } =\frac{Gm^{2} }{4r^{2} }
  • F=\frac{G ( \frac{4}{3}\pi r^{3}ρ)^{2} }{4r^{2} }
  • F= \frac{G(16)\pi ^{2} r^{6} ρ^{2}}{(9)(4)r^{2} } =\frac{G(16)\pi ^{2} r^{4}ρ^{2}  }{36}

Now let´s plug in the values we know:

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  2. G = 6.67(10)^{-11}     gravitational constant
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  • 1= \frac{6.67(10)^{-11} \pi ^{2} r^{4} (491.5)^{2}}{36}

Solve for r and multiply by 2 because 2r = diameter

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Result is d = 21.81 Feet

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Answer:

Explanation:

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Answer:

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