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Question

Derive an expression for the intensity of magnetic field on broad side on position due to a bar magnet.

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Solution

Let NS be a bar magnet of pole strength m and effective length 2l. Consider a point P on the neutral axis at a distance d from the centre of the magnet.
Let PN=PS=x.
Now, the intensity of field at P due to N-pole is,
B1=μ04πmx2(alongNP)
Similarly, the intensity of field at P due to S pole is given by
B2=μ04πmx2(alongPS)
Since, B1=B2=μ04πmx2
Hence, B1=B2
Hence, by the law of parallelogram of force the resultant will be given by
B=B21+B22+2B1B2cos2θ (Where 2Q is the angle between B1 and B2)
But, B1=B2, hence
B=B21+B21+2B21cos2θ
=2B21+2B21cos2θ
=2B21(1+cos2θ)
=2B21[1+2cos2θ1]
=2B212cos2θ=2B1cosθ
or B=2μ04πmx2cosθ
or B=μ04π2mx21x,(cosθ=1x)
or B=μ04π2mlx3
But, 2ml=M (magnetic moment)
B=μ04π2Mx3
But, by the Pythagoras theorem in right angle ΔPQS
x=d2+12
or x3=(d2+12)3/2
Hence, B=μ04πM(d2+12)3/2
If the magnet is small i.e. d1
B=μ04πMd3
This is the required expression.
The direction of magnetic field is from north pole to south pole and parallel to magnetic axis of the magnet.
666152_628512_ans_915c662be5ec4295a5f5d79cba629271.png

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