The magnetic moment of a current loop is defined m vector = i A vector. The magn

Question

The magnetic moment of a current loop is defined m vector = i A vector. The magnetic moment is a vector in the direction of the A vector for area, determined by the right-hand current rule, as shown A current of I Ampere enclosing any plane area of 1 m^2 causes a magnetic moment of 1 A-m^2 The magnetic moment m vector is analogous to the dipole moment, p vector, of an electrostatic dipole (=qi vector ) in Problem 379(a). Prove that the torque on a current loop in an external B vector field can be written as Calculate the work required to rotate a current loop in an external B field, and thus prove that the potential energy of the loop, when the angle between m vector and B vector is theta, is given by P.E. = mB(1 - cos theta) if we take the P.E. to be zero at the angle of minimum potential energy. One more commonly funds the result written as P.E. (=U) = -m vector. B vector Shows that this form will be obtained if the zero energy position of the loop is taken with A vector (or m vector) perpendicular to B vector,(One reason that physicists like to use m vector in describing current loops is that the torque and energy equations assume simple forms:

Explanation / Answer

a)    Force on a current carrying loop in magnetic field = (ILXB)

     => torque on a current carrying loop in magnetic field = IA X B

As, magnetic moment , m = IA

=> torque on a current carrying loop in magnetic field = m X B

b)   A magnetic dipole moment in a magnetic field will possess potential energy which depends upon its orientation with respect to the magnetic field. Since magnetic sources are inherently dipole sources which can be visualized as a current loop with current I and area A, the energy is usually expressed in terms of the magnetic dipole moment:

=> work done = - m.B

where, magnetic moment , m = IA

=> work required to rotate loop = -mBcos(theta)

=> potential energy of loop = -mBcos(theta) - (- mBcos(0))

                                              =    -mBcos(theta) + mB

                                                = mB(1 - cos(theta))

Here, if theta = 90 degrees or A is perpendicular to B .

=>   potential energy of loop = mB(1 - cos(90))

                                              = mB

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