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Show that a current carrying solenoid is equivalent to a bar magnet. |
Answer» Solution : The magnitude of magnetic field at a point .P. due to a circular element of THICKNESS .dx. is given by `dB=(mu_0/(4pi)) (2pindxla^2)/(((r-X)^2 + a^2)^(3/2))` The magnitudeof totalmagnetic fieldis given by `B=(mu_0/(4pi)) 2pinIa^2 int_(-1)^1 (dx)/([(r-x)^2 +a^2]^(3/2))` For r > > aand r > > l , `[(r-x)^2 + a^2]^(3//2) ~~ r^3` `thereforeB =(mu_0/(4pi)) (2pinIa^2)/r^3 int_(-l)^l dx "" int_(-l)^ldx = (x)^(-l)^l` i.e.,`B=(mu_0/(4pi)) (2pinIa^2)/r^3 2l` =l + l =2l We note that the productn(2l) I `(pi a^2)` = MAGNETICMOMENT .m. `therefore B=(mu_0/(4pi)) (2m)/r^3` --(1) This expression is similar to the magnetic field at a point on the axis of a short MAGNET. Therefore a bar magnet and a SOLENOID produce similar magnetic fields. |
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