A short bar magnet of magnetic moment $5.25 \times 10^{- 2} {JT}^{- 1}$ 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 45⁰ with earth’s field on (a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.45G. Ignore the length of the magnet in comparison to the distances involved.

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2.8 cm, 5 cm

2 cm, 5 cm

8 cm, 1.2 cm

5 cm, 6.3 cm

answer is D.

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Detailed Solution

$(a) B_{e} = B_{H}, \Rightarrow \frac{μ_{0}}{4 π} \frac{M}{d^{3}} = B_{H} d^{3} = \frac{10^{- 7} \times 5 .25 \times 10^{- 2}}{0 .45 \times 10^{- 4}} = (116 .6) \times 10^{- 6} \Rightarrow d = 5 cm (b) B_{axial} = B_{H} = \frac{μ_{0}}{4 π} \frac{2 M}{d^{3}} d^{3} = \frac{10^{- 7} \times 2 \times 5 .25 \times 10^{- 2}}{0 .45 \times 10^{- 4}} = 233 \times 10^{- 6} \Rightarrow d = 6.3 cm$