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InterviewSolution
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What is the magnetic field along the axis and equatorial line of a bar magnet ? |
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Answer» Solution :Magnetic field at a point along the equatorial line due to a magnetic DIPOLE (bar magnet) Consider a ber magnet. NS. Let N be the north pole and S be the south pole of the bar magnet, each with pole strength `q_(m)` and separated by a distance of 2l. The magnetic field at a point C (lies along the equatorial line ) at a distance r from the geometrical center O of the bar magnet can be computed by keeping unit north pole `(q_(m) C = 1 A m )` at C. The force e`xx`perienced by the unit north pole at C due to pole strength N-S can be computed using Coulomb.s law of magnetism as follows , the force of repulsion between North Pole of the bar magnet and unit north pole at point C (in free space ) is `vec(F_(N)) = - F_(N) cos theta hat(i) + F_(N) sin theta hat(j) ` Where `F_(N) = (mu_(0))/(4pi ) (q_(m))/(r.^(2))` The force of attraction ( in free space ) between southpole of the bar magnet and unit north pole at point C is `vec(F_(S)) = - F_(S) cos theta hat(i) - F_(S) sin theta hat(j)` where , `vec(F_(S)) = (mu_(0))/(q_(m))(q_(m))/(r.^(2))` From equation (1) and equation (2) , the net force at poin C is `vec(F) = vec(F_(N)) + F_(S)`. This net force is equal to the magnetic field at the point C . `vec(B) - (F_(N) + F_(S)) cos theta hat(i)` since, `F_(N) = F_(S)` `vec(B) = - (2mu_(0))/(4pi) (q_(m))/(r.^(2)) cos theta hat(i) = (2 mu_(0))/(4pi ) ( q_(m))/((r^(2) + l^(2))) cos theta hati` In a right angle TRIANGLE NOC as shown in the Figure l cos `theta= ("adjacent")/("hypotenuse") = (l)/(r.) = (l)/((r^(2) + l^(2))^((1)/(2)))` Substituting equation 4 in equation 3 we get `vec(B) = - (mu_(0))/(4pi) (q_(m) xx (2l))/((r^(2) + l^(2))^((3)/(2)) )` Since, magnitude of magnetic dipole moment is `|vec(P_(m))| = P_(m) = q_(m). 2l` and substituting in equation (5 ). The magnetic field at a point C is `vec(B)_("equatorial") = - (mu_(0))/(4pi) (P_(m))/((r^(2) + l^(2))^((3)/(2)) )` If the distance between two poles in a bar magnet are small (lools like short magnet ) when compared to the distance between geometrical center O of bar magnet and the location of point C i.e., r `gt gt ` l, then , `(r^(2) + l^(2))^((3)/(2)) approx r^(3)` Therefore, using equation (7) in equation (6), we get `vec(B)_("equatorial") = - (mu_(0))/(4 pi ) (P_(m))/(r^(3)) hat(i)` Since, `P_(m) hat(i) = vec(P_(m)) . ` in general, the magnetic field at equatorial point is GIVEN by `vec(B_("equatorial")) = - (mu_(0))/(4 pi ) (P_(m))/(r^(3))` Note that magnitude of `vec(B_("axial")) `is twice that of magnitude of `B_("equatorial")` and the direction of `B_("axial ")and B_("equantorial") ` are opposite . 
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