The figure below shows a current loop ABCDEFA carrying a current i = 5.00 A. The

Question

The figure below shows a current loop ABCDEFA carrying a current i = 5.00 A. The sides of the loop are parallel to the coordinate axes shown, with AB = 20.0 cm, BC = 65.0 cm, and FA = 10.0 cm. In unit-vector notation, what is the magnetic dipole moment of this loop? (Hint: Imagine equal and opposite currents i in the line segment AD; then treat the two rectangular loops ABCDA and ADEFA.) Note: be sure to enter the components of your magnetic dipole moment in the order i hat-component first (if any), then j-component (if any), and then k-component (if any).

Explanation / Answer

Given that

The loop carrying a currrent ( i ) = 5.00 A.

The sides of the loop are parallel to the coordinate axes shown, with AB = 20.0 cm, BC = 65.0 cm, and FA = 10.0 cm.

Now let us consider

a =BC =65cm =0.65m

b =AB =20cm =0.20m

c =FA =10cm =0.10m

Now the magnetic dipole moment is given by

u =u1+u2

=-iabk+iacj

=ia(cj-bk)

=(5.00A)(0.65m)[(0.10m)j -(0.20)k]

=[0.325j-0.65k]A.m2

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