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write equation the magnetic field due to current carrying loop as magnetic dipole on its axial point. |
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Answer» Solution :1. The magnetic field on the AXIS of a circular loop of a radius R, carrying a steady current I, at x distance from centre is given by `B=(mu_(0)IR^(2))/(2(x^(2)+R^(2))^(3/2))""...(1)` 2. And its DIRECTION is along the axis and the right hand thumb rule. Here, x is the distance along the axis from the centre of the loop. 3. For `xgtgtR`, we MAY neglect `R^(2)` term in the denominator. Thus, `B=(mu_(0)IR^(2))/(2x^(2))""...(2)` 4. Multiply right SIDE of EQUATION (2) with `pi` in denominator and numerator, `thereforeB=(mu_(0)I(piR^(2)))/(2pix^(3))` Area of coil A = `piR^(2)` `thereforeB=(mu_(0)IA)/(2pix^(3))` Here, IA = magnetic dipole moment = m `B=(mu_(0)m)/(2pix^(3))` `thereforeB=(2mu_(0)m)/(4pix^(3))` and In vector form, `thereforevecB=mu_(0)/(4pi)((2vecm)/x^(3))` which is formula for magnetic field on point of axis of coil. |
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