A short bar magnet of magnetic movement `5.25xx10^(-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. 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.

Explanation

(a) `(mu_(0))/(4pi) ((m)/(r^(3))) = 0.42 xx 10^(-4)` which gives r = 50cm .
(b) `(2mu_(0))/(4 pi) ((m)/(r_(1)^(3))) = 0.42xx 10^(-4)` which gives r = 6.3cm 10. Below the cable .
`R_(n) = 0.39 cos 35^(@) - 0.2 = 0.12 G`
`R_(v) = 0.39 sin 35^(@) = 0.22 G`
`R = sqrt(R_(h)^(2) + R_(v)^(2)) = 0.25 G`
`theta = tan^(-1) ((R_(v))/(R_(h))) = 62^(@)`
Above the cable .
`R_(h) = 0.39 cos 35^(@) + 0.2 = 0.52 G`
`R_(v) = 0.39 sin 35^(@) = 0.224 G`
`R = sqrt(R_(v)^(2) + F_(v)^(2)) = 0.57 G `
`therefore = tan ^(-1) ((R_(v))/(R_(0))) = 23^(@)`