A short bar magnet of magnetic moment 5.25 × 10–2 J T–1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on (a) its normal bisector and (b) its axis. The magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
(a) According to question, the resultant field is inclined at 45o with earth's magnetic field. tanθ=BBH θ=45otanθ=1=BBH BH=B=4πr3μoM Given, B= 0.42×10−4T M=5.25×10−2J/T Therefore,0.42+10−4=10−7×r35.25×10−2r3=12.5×10−5=125×10−6r=5×10−2m = 5 cm (b) At axis of magnet, tan45°=1=BHB BH=B=4πr3μo2M =>0.42×10−4=10−7×r32×5.25×10−2=>r3=25×10−5m=>r=6.3cm
