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A bar magnet is pivoted at its centre and placed in such a manner that it can freely rotate in horizontal plane. Time period of oscillation for this magnet in earth's magnetic field is measured at two different places. Time period of oscillation is 3s at a place where angle of dip is 30^@and time period is 4s at a place where angle of dip is 60^@. Let B_1 and B_2are the net magnetic fields due to earth at two places in same order. Then B_1//B_2 |
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Answer» `(16)/(9 sqrt3)` `COS delta = (B_H)/(B)` ` rArr B= B_H sec delta ` Hence we can write the following for two places on earth. `B_1/B_2 = (B_(H_1) sec delta_1)/(B_(H_2) sec delta_2)` Time period of oscillation of magnet due to horizontal component can be written as follows: ` T = 2pi sqrt((I)/(MB_H) )` `rArr (T_1^2)/(T_2^2) = (B_(H2))/(B_(H1))` `rArr (B_(H1) )/(B_(H2)) = (T_2^2)/(T_1^2)` ` rArr (B_(H1))/(B_(H2)) = ((4)^2)/((3)^2)` ` rArr (B_(H1))/(B_(H2)) = 16/9`.....(ii) Substituting given angles of dip and value from equation (2) in equation (1) we get the following: `B_1/B_2 = 16/9 xx (sec 30^@)/(sec 60^@) ` `B_1/B_2 = 16/9 xx (2)/(sqrt3 xx 2)` `B_1/B_2 = (16)/(9 sqrt3)` 
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