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The magnetic moment m⃗ of a current loop is defined as the vector whose magnitude equals the area of the loop times the magnitude of the current flowing in it (m=IA), and whose direction is perpendicular to the plane in which the current flows. Find the magnitude, F, of the force on the loop from Part A in terms of the magnitude of its magnetic moment. Express F in terms of m, I2, a, d, and μ0.

Respuesta :

Answer:

Explanatiocn:

F= I1*a*uo*I2*((1/2pi(d-0.5a)-(1/2pi(d+0.5a)) since m=I/A, then I= m/A F = (m/A)*a*uo*I2*((1/2pi(d-0.5a)-(1/2pi(d+0.5a)

facundo3141592

We want to find the force on a current loop due to the magnetic moment.

The force is given by:

F = (m/A)*d×(I*μ0)/(√(A*2*pi))

We know that:

m = I*A

where:

I = current

A = area

m = magnetic moment.

We know that the force is given by:

F = I*d×B

Where d represents the "length" of the wire, and B is the magnetic field.

We know that:

B = (I*μ0)/(2*pi*r) where r depends on the distance to the wire, because we are just on the surface, we will have:

r = √(A/2pi)

Then:

B =  (I*μ0)/(2*pi*√(A/2pi)) = (I*μ0)/(√(A*2*pi))

Ok, now we know all the things we need, we can replace them in the force equation to get:

F = I*d×(I*μ0)/(√(A*2*pi))

And we know that:

m = I*A

then:

I = m/A

replacing that we get:

F = (m/A)*d×(I*μ0)/(√(A*2*pi))

If you want to learn more, you can read:

https://brainly.com/question/3160109

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