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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 fleld at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved. |
Answer» Solution :(a) Figure shows the situation described in the STATEMENT.  Suppose `overset(to) (B_R)` is the resultant magnetic field at `P_1` which makes angle 45° with `overset(to)(B)`. Now, `TAN45^(@) = (B_e)/( B)` `therefore B_(e) = B` `therefore B_(e) = 0.42 xx 10^(-4) T` `therefore (mu_(0) )/( 4pi) ((m)/( r_(1)^(3) )) =0.42 xx 10^(-4)` `therefore (4pi xx 10^(-7) )/( 4pi) (0.0525)/( r_(1)^(3) ) =0.42 xx 10^(-4)` `therefore r_(1)^(3) = 125 xx 10^(-6)` `therefore r_(1) = 5 xx 10^(-2) ` m (b) Figure drawn as per the statement is as follows.  In above figure, `overset(to) (B)= ` Magnetic field of Earth at point `P_2` `overset(to) (B_a) =` Magnetic field of bar MAGNET at point `P_2`. `overset(to) (B_R) =` Resultant magnetic field at point `P_2` which makes angle 45° with `overset(to)(B)`. Here, `tan45^(@) = (B_a)/( B)` `therefore B_a= B (because tan45^@ =1)` `therefore B_a = 0.42 xx 10^(-4) T` `therefore (mu_0)/( 4pi) ((2M)/( r_(2)^(3)) ) = 0.42 xx 10^(-4)` `therefore ((4pi xx 10^(-7))/( 4pi) ) ((2 xx 0.0525)/( r_(2)^(3) ) ) = 0.42 xx 10^(-4)` `therefore r_(2)^(3) = 250 xx 10^(-6)` `therefore r_(2) = 6.2996 xx 10^(-2)` m |
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