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In this question, you will calculate the force between two identical magnetic dipoles. Assume that the magnetic dipole moment of both of the dipoles are 1-112 (a) We had shown that the torque that one dipole feels due to a external magnetic field is given by T = × B. Show that the corresponding potential energy is given by U =-i. (b) Even though the previous result is derived for a rotations in a uniform magnetic field, it is true in general, i.e. also for a nonuni form magnetic field. Assuming that the magnetic moment is kept fixed at -: What is the force acting on the magnetic dipole? Hint: Review how the force acting on a particle is related to its potential energy)

Explanation / Answer

Part a)

The potential energy is determined by the work done by the Bampo field to align the dipole, the expression of the work is

dW = d

we substitute the expression of torque = x B = b Sin

dW = B sin d

W = if B sin d

The work is the variation of the energy system

Uf – Ui = B if sin d = B (-)( Cos f – Cos i)

Uf – Ui = B ( Cos i – Cos f)

We must choose the condition for initial energy

Ui= 0 cuando i = 90

with this and making Uf =U

U = - B Cos

remember that the definition of scalar product a.b = a b Cos

U = - . B

Part b)

Let's use the suggestion given,

F = - dU/dz

U = - ( ^k . B ^k)

^ where k is a unitary vector.

Producer climb the only part remaining is the one in the direction of the axis z

^k. ^k = 1 ^k. Î = ^k .^j =0

F = - d ( B ) /dk

F = - B d/dz - d B/dz

and indicate that B = uniform dB/dz = 0

F = - B d/dz

Part c)

For this part we will use the Biot - Savart

dB = o/4 I (ds x ^r )/ r2

Approximate the dipole as a loop, a it is the radius of the coil or wire separation of the dipole

B = o1 I/( 2 (z2 + a 2) 3/2

z = d ^k

B = o I/(2 (d2 + a 2) 3/2

if d >>a

B = o I/2d3

Part d)

The magnetic field created by the dipole 1 in this point is

B1 = o I/(2 (d2 + a 2) 3/2

The force on the dipole 2 is given by the expression calculated in part b

F = - B1 d/dz

= I a b ^k

a b It is the area of the dipole

F = - o I/(2 (d2 + a 2) 3/2 I d (ab ^k)/dz

F = - o I/(2 (d2 + a 2) 3/2 I ab

They are many questions and I run out of time so I leave it up to this part

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