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A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north - south direction . Null point are found on the axis of the magnet at 14 cm from the centre of the magnet . The earth's magnetic field at the place if 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null point (i.e., 14 cm) from hte centre of the magnetic ? (At null points , field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field.) |
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Answer» Solution :`d = 14 cm = 14 X10^(-2) m, B = 0.36G` Given `(10^(-7)xx2m)/d^3=0.36xx10^(-1)(T)` On the equatorial LINE , `B_("equatorial line ")=1/2B_("axial")` `=0.18 xx10^(-4)(T)=0.18G` TOTAL field `=B_("equatorial line ")+H_E` (of earth ) = 0.18 + 0.36 = 0.54 G 
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