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A short bar magnet of magnetic moment 5.25 xx 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 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. |
Answer» Solution : Here , `m =5.25 XX 10^(-2) J T^(-1) and B_E = 0.42 G = 4.2 xx 10^(-5) T` (a) As shown in fig. 5.01 (a) , let at a DISTANCE `r_1` along the normal bisector of the magnet the resultant field is inclined at `45^@` with the earth.s field `B_E` `thereforetan 45^@ =(B_(AQ))/(B_E) impliesB_E ` or ` (mu_0)/(4pi) * (m)/(r_1^3) = B_E` `impliesr_2 = [(mu_0)/(4pi) * (m)/(B_E)]^(1/3) =[(10^(-7)xx 5.25xx10^(-2))/(4.2xx10^(-5))]^(1/3)= 5 xx 10^(-2)m = 5 cm` (b) As shown in fig. 5.01 (b) , let at a distance `r_2` along the axis of the magnet , the resultant field is inclined at an angle `45^@` with the earth.s field `B_E` . Then, `B_("AXIAL") =(mu_0)/(4pi) * (2m)/(r_2^3) = B_E tan 45^@ = B_E` `impliesr_2 =[(mu_0)/(pi) * (2m)/(B_E)]^(1/3) =[(10^(-7)xx 2 xx 5.25xx10^(-2))/(4.2 xx 10^(-5))]^(1/3) = 6.3xx10^(-2) cm` |
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