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A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth's magnetic Meld at the place is 0.36 G and the angle of dip 18 zero. What is the total magnetic feld on the normal bisector of the magnet at the same distance as the mull-point (i.e., 14 cm) from the centre of the magnet? At rull points, field due to a magnet 1s equal and opposite to the horizontal component of earth's magnetic field.) |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/null-17167" style="font-weight:bold;" target="_blank" title="Click to know more about NULL">NULL</a> point of a given bar magnet is that point in the horizontal surface at which magnetic field due to that magnet is nullified by horizontal component of Earth.s magnetic field. <br/> For null points `P_1 and P_2` of a <a href="https://interviewquestions.tuteehub.com/tag/short-642706" style="font-weight:bold;" target="_blank" title="Click to know more about SHORT">SHORT</a> bar magnet on its axis. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/KPK_AIO_PHY_XII_P1_C05_E02_041_S01.png" width="80%"/> <br/> We know that magnetic field of a bar magnet on its axis is in the direction of its dipole moment (in above <a href="https://interviewquestions.tuteehub.com/tag/figure-987693" style="font-weight:bold;" target="_blank" title="Click to know more about FIGURE">FIGURE</a>, towards right). <br/> Now, as shown in the diagram, if `P_1 and P_2` are the null points on the axis of given bar magnet then at these points, <br/> `B_(a) = B_(<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>)` <br/> `therefore (mu_(0))/( 4 pi) ((2m) /(r^3) )= 0.36b G ""...(1)` <br/> For the points `P_3 and P_4` of a short bar magnet on its equator located at distance r from its centre. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/KPK_AIO_PHY_XII_P1_C05_E02_041_S02.png" width="80%"/> <br/> We know that magnetic field of a bar magnet is in the direction, opposite to its dipole moment (in above figure, towards left). <br/> At points `P_3 and P_4` (and also at all the points on the circle of radius r in the equatorial plane), magnetic field due to bar magnet is, <br/> `B_(e)= ((mu_0 )/(r^3) )` <br/> `= (1)/(2) ((mu_(0) )/( 4 pi )) ((2m)/( r^3) )` <br/> `= (1)/(2) B_a` <br/> `= (1)/(2) (0.36 )` [From equation (1)] <br/> `therefore B_(e) = 0.18 G` <br/> Now, if <a href="https://interviewquestions.tuteehub.com/tag/resultant-1187362" style="font-weight:bold;" target="_blank" title="Click to know more about RESULTANT">RESULTANT</a> magnetic field at points `P_(3) and P_(4)` (and also at all the points on a circle of radius r in the equatorial plane) is `overset(to) (B_R)` then, <br/> `overset(to) (B_R) = overset(to) (B_(e) ) + overset(to)(B_h) ` <br/> `therefore B_R = B_e + B_h ""(because "Here", overset(to) (B_e) || overset(to) (B_h) )` <br/> `= 0.18 + 0.36` <br/> `therefore B_R = 0.54` G <br/> Note : At null points of a bar magnet, resultant magnetic field is zero only along horizontal direction because vertical component of magnetic field of Earth is not balanced at the null point. This is because bar magnet is kept on the horizontal surface.</body></html> | |