1.

If the bar magnet in the above problem is turned around by `180^@`, where will the new null points be located?

Answer» When the bar magnet is turned through `180^@`, neutral points would lie on equatorial line, so that
`B_2=(mu_0)/(4pi)(M)/(d_2^3)=H` …(i)
In the previous question, `B_1=(mu_0)/(4pi)(2M)/(d_1^3)=H` …(ii)
From (i) and (ii), `(mu_0)/(4pi)(M)/(d_2^3)=(mu_0)/(4pi)(2M)/(d_1^3) :. d_2^3=d_1^3/2=((14)^3)/(2)`
`d_2=(14)/(2^(1//3)=11*1cm`


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