Show the derivation to prove that solenoid acts as a bar magnet

Suppose we have a current carrying solenoid of length 2l.and radius of solenoid=aNumber of turns per unit length=nCurrent passing through solenoid=ILet us consider a point P outside the solenoid at distance r from the centre of solenoid 'O'.i.e. OP=rConsider a small element of thickness dx of solenoid at distance x from O, then number of turns in element=ndxWe know that the magnetic field due to n turns coil at axis of solenoid is given bydB=μ0Ia2(ndx)2[(r-x)2+a2]3/2If r>>a and r>>x,then [(r-x)2+a2]3/2=r3Therefore,dB=μ0Ia2(ndx)2r3The total magnetic field due to entire solenoid is given byB=μ0Ia2(n)2r3x=-lx=l dx=μ0Ia2(n)2r3×2l------(i)The magnetic moment of solenoid, M=NIA=(n×2l)×I×(πr2)Or (n×2l)×I=M(πr2)Putting it in (i), we getB=μ02M4πr3 or μ04π×2Mr3
Thus, it is clear from the above expression that magnetic moment of a bar magnet is equal to the magnetic moment of an equivalent solenoid that produce the same magnetic field or we can say that a solenoid acts as a bar magnet.

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