A circular loop of wire having a radius of 9.28 cm carries a current of 0.103 A.

Question

A circular loop of wire having a radius of 9.28 cm carries a current of 0.103 A. A vector of unit length and parallel to the dipole moment of the loop is given by 0.60i Overscript EndScripts – 0.80j Overscript EndScripts. If the loop is located in a uniform magnetic field given by Upper B Overscript right-arrow EndScripts = (0.958 T)i Overscript EndScripts + (0.882 T)k Overscript EndScripts, find (a) the x-component, (b) the y-component, and (c) the z-component of the torque on the loop and (d) the magnetic potential energy of the loop.

Explanation / Answer

The magnetic moment of a planar loop carrying constant current is given by = IAn, where n is a normal vector to the loop with direction assigned by the right-hand rule. In other words, we can write = IA', where ' is a unit vector in the direction of , which is given. In general, the torque on a current-carrying loop due to a constant magnetic field B is given by:

                                       = × B
                                         = IA' × B.

To calculate the cross product, you can use any method you like. The determinant method is common, though I like to expand them out using the distributive property and the rules i x j = k, j x k = i, k x i = j, a x b = - b x a, and a × a = 0 unless both vectors have all nonzero components.

= IA' × B
= IA[(0.60i - 0.80j) × (0.958i + 0.882k)]
= IA[(0.5748(i×i) + 0.5292(i × k) - 0.7664(j × i) - 0.7056(j × k)]
= IA(-0.5292j + 0.7664k - 0.7056i)

Using the fact that I = 0.103A and A = (0.0928 m)² = 0.0271 m², we get a torque of:

(-1.97 x 10^(-3)) i + (-1.48 x 10^(-3)) j + (2.14 x 10^(-3) k) N m.

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