Saved Bookmarks
| 1. |
Assume the dipole model for earth's magnetic field B which is given by B_V= vertical component of magnetic field = (mu_0)/( 4pi ) ( 2m cos theta )/( r^3) B_H= Horizontal component of magnetic field B_H= (mu_0)/( 4pi ) (m sin theta )/( r^3) theta =90^@ - latitude as measured from magnetic equator. (a) Find loci of points for which (i) |overset(to)(B) | is minimum, (ii) dip angle is zero, (iii) dip angle is pm 45^@. |
|
Answer» Solution :(a) Given that, `B_V = (mu_0)/( 4pi ) (2m cos THETA)/( r^3) ""…(1)` `B_H = (mu_0)/( 4pi ) (m sin theta )/( r^3) ""…(2)` Squaring and ADDING equation (1) and (2), `B_(V)^(2) +B_(H)^(2) = ((mu_0)/( 4pi ))^(2) (m^2)/( r^6) [ 4 cos ^(2) theta + sin^(2) theta]` `therefore B^(2) = ((mu_0)/( 4pi ))^(2) (m^2)/( r^6) [ 4 cos^(2) theta +1 - cos^(2) theta ]` `therefore B^(2) = sqrt((B_(V)^(2) + B_(H)^(2) ) ` `= sqrt(((mu_0)/( 4pi ))^(2)(m^2)/(r^6) [ 3 cos^(2) theta +1] )` `= (mu_0)/( 4pi ) (m)/(r^3) [ 3 cos theta^(2) theta + 1]^(1//2) ""...(3)` From above equation, the value of B is minimum if `cos theta = 0`, hence `theta = (pi)/(2)` Thus, B is minimum at the magnetic equator. (b) For angle of dip, `TAN delta = (B_V)/( B_H) = (2 ((mu_0)/( 4pi )(m cos theta )/( r^3) ) )/( ( (mu_0)/( 4pi) (m sin theta )/( r^3) ) )` `-2 ( cos theta)/(sin theta)` `(B_V)/( B_H) = tan delta = 2 cot theta ""...(4)` For dip angle `delta = 0`, then `cottheta =0` `therefore theta= (pi)/(2)` Hence locus is on magnetic equator. (c) `tan delta = (B_V)/( B_H)` For `delta = pm 45^@` `(B_V)/( B_H) = tan (pm 45^@)` `(B_V)/( B_H) = 1` `therefore 2 cot theta =1 ""` [From equation (4)] `therefore cot theta = (1)/(2) ` ` therefore tan theta = 2` `therefore theta= tan^(-1) (2)` is the locus. |
|