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State that a currentcarrying loop behaves as a magnetic dipole. Hence writean expression for its magnetic dipole moment . |
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Answer» Solution :The magnetic FIELD from the centre of a circular LOOP of RADIUS R along the axis is given by `vecB = (mu_(0)I)/ 2 R^(2)/((R^(2) + z^(2))^(3/2))hatk` At large distance `z gt gt R`, therefore `R^(2) + z^(2) approx z^(2)`, we have `vecB = (mu_(0)I)/2 R^(2)/z^(3) hatk` .....(1) Let A be the area of the circular loop `A = pi R^(2)`. So rewriting the equation (1)in TERMS of area of the loop , we have `vecB = (mu_(0)I)/ (2 pi) A/z^(3) hatk` ` vecB = (mu_(0))/(4 pi) (2 I A)/z^(3) hatk ` ...(2) COMPARING equation (2) with equation (1) dimensionally, we get `p_(m) = IA` ltbr. where `p_(m) ` is called magnetic dipole moment . In vector notation, `vecp_(m) = I vecA`......(3) This implies that a current carrying circular loop behaves as a magnetic dipole of magnetic moment `vecp_(m)`. So, the magnetic dipole moment of any current loop is equal to the product of the current and area of the loop. |
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